geometry
circumcircle
geometric transformation
homothety
trigonometry
Euler
reflection

I have already posted this in [url=http://groups.yahoo.com/group/Hyacinthos/message/8510?expand=1]Hyacinthos message #8150[/url], but there was no reaction. Hope that this will be different here. The problem is really beautiful, and I have a rather nice synthetic solution. Here goes the problem:

Let D, E, F be the feet of the three altitudes of a triangle ABC, and let D', E', F' be the midpoints of the segments AD, BE, CF. Let the line E'F' meet the lines AB and CA at the points $B_a$ and $C_a$, let the line F'D' meet the lines BC and AB at the points $C_b$ and $A_b$, and let the line D'E' meet the lines CA and BC at the points $A_c$ and $B_c$.

[b](a) (original problem by A. Angelescu, Gazeta Matematica 35):[/b] If X', Y', Z' are the circumcenters of triangles $AB_aC_a$, $BC_bA_b$, $CA_cB_c$, respectively, then the points D', E', F', X', Y', Z' lie on one circle.

[b](b) (my observations, which lead to a proof of (a)):[/b] Let L be the de Longchamps point of triangle ABC (this is the image of the orthocenter of triangle ABC in the homothety with center at the centroid of triangle ABC and factor -2). Let the lines AL, BL, CL meet the lines BC, CA, AB at the points X, Y, Z, respectively. Then, the points $B_a$, $C_a$ and D lie on the circle with diameter AX. Similarly for the circles with diameters BY and CZ.

  Darij

Denote $a = BC, b = CA, c = AB$ the triangle sides and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ the triangle angles. WLOG, assume that the angle $\gamma > \alpha, \beta$ is the largest one, the relatively trivial case of an isosceles triangle is disregarded. We will also assume that $CH < HF$, i.e., the points $C, H, F', F$ follow on the altitude $CF$ in this order. By considering an isosceles triangle with $CH = HF$,

$CH = \frac{a \sin \alpha}{2} = \frac{a \cos \alpha}{\tan \alpha} = HF$

$\sin^2{\alpha} = 2 \cos^2{\alpha},\ \ \cos^2{\alpha = \frac 1 3,\ \ \cos \gamma = 1 - 2 \cos^2{\alpha} = \frac 1 3,\ \ \gamma \doteq 70.53^o}$

we see that if $CH < HF$, $\cos \gamma < \frac 1 3$ and $\gamma > 70.53^o$ is the largest triangle angle. The case when $CH > HF$, while $\gamma$ remains the largest triangle angle, is treated similarly.

[color=red][b]Step 1:[/b][/color] Let $(O_B), (O_E)$ be the circumcircles of the triangles $\triangle BE'F, \triangle EE'F$ and let $A', B'$ be their intersections with the sides $CA, BC$ other than the points $E, B$, respectively. Using properties of the orthic triangle $\triangle DEF$, the triangles $\triangle EFA' \sim \triangle BFB'$ are similar, because the angles $\angle A'EF = \angle B'BF = \beta$ are equal and the angles

$\angle BB'F = \angle BE'F = 180^o - \angle FE'E = \angle EA'F$

are also equal. The similarity coeficient of these 2 triangles is

$\frac{EF}{BF} = \frac{a \cos \alpha}{a \cos \beta} = \frac{\cos \alpha}{\cos \beta}$

The angles $\angle EFA' = \angle EE'A'$ spanning the same arc $FA'$ of the circle $(O_E)$ are equal and the angles $\angle BFB' = \angle BE'B'$ spanning the same arc $BB'$ of the circle $(O_B)$ are also equal. Due to similarity of the triangles $\triangle EFA' \sim \triangle BFB'$, these angles are equal to each other, hence, the angles $\angle EE'A' = \angle BE'B'$ are vertical, which means that the points $A'E'B'$ are collinear. The triangles $\triangle AFA' \sim \triangle DFB'$ are also similar, because the angles $\angle FAA' = \angle FDB' = \alpha$ are equal and the angles $\angle AA'F = 180^o - \angle EA'F = 180^o - \angle BB'F = \angle DB'F$ are also equal. The similarity coefficient of these 2 triangles is also

$\frac{AF}{DF} = \frac{b \cos \alpha}{b \cos \beta} = \frac{\cos \alpha}{\cos \beta}$

Denote $\omega = \angle BE'F$. Using the sine theorem for the triangle $\triangle BFE'$,

$\frac{BF}{FE'} = \frac{\sin \omega}{\sin{(90^o - \alpha)}} = \frac{\sin \omega}{\cos \alpha}$

The segment $FE'$ is the median of the triangle $\triangle BEF$ from the vertex $F$. Hence

$FE'^2 = \frac 1 4(2EF^2 + 2BF^2 - BE^2) =$

$= \frac{a^2}{4}(2\cos^2{\alpha} + 2 \cos^2{\beta} - \sin^2{\gamma}) = \frac{a^2}{4}(\cos^2{\alpha} + \cos^2{\beta} - 2 \cos \alpha \cos \beta \cos \gamma)$

$\sin^2{\omega} = \frac{BF^2}{FE'^2}\ \cos^2{\alpha} = \frac{4 \cos^2{\alpha} \cos^2{\beta}}{\cos^2{\alpha} + \cos^2{\beta} - 2 \cos \alpha \cos \beta \cos \gamma}$

$\cot^2{\omega} = \frac{1 - \sin^2{\omega}}{\sin^2{\omega}} = \frac{\cos^2{\alpha} + \cos^2{\beta} - 4 \cos^2{\alpha} \cos^2{\beta} + 2 \cos^2 {\alpha} \cos^2{\beta} - 2 \sin \alpha \cos \alpha \sin \beta \cos \beta}{4 \cos^2{\alpha} \cos^2{\beta}} =$

$= \frac{\cos^2{\alpha} (1 - \cos^2{\beta}) + \cos^2{\beta} (1 - \cos^2{\alpha}) - 2 \sin \alpha \cos \alpha \sin \beta \cos \beta}{4 \cos^2{\alpha} \cos^2{\beta}} =$

$= \frac 1 4 (\tan^2{\alpha} + \tan^2{\beta} - 2 \tan \alpha \tan \beta) = \frac 1 4 (\tan \alpha - \tan \beta)^2$

Let $C'$ be the intersection of the altitude $CF$ with the circle $(O_B)$ other than the point $F$. Since the angle $\angle BFC' = 90^o$ is right, $BC' = 2O_BB$ is a diameter of the circle $(O_B)$ and

$FC' = \sqrt{BC'^2 - BF^2} = \sqrt{4O_BB^2 - BF^2}$

Radius of the circle $(O_B)$ is equal to $O_BB = \frac{BF}{2 \sin \omega}$ and the power of the vertex $C$ to this circle is

$CB \cdot CB' = CF \cdot CC',\ \ \ CB (CB - BB') = CF (CF + FC')$

$BB' = CB - \frac{CF (CF + \sqrt{4O_BB^2 - BF^2})}{CB} =$

${= a - a (\sin^2 \beta + \sin \beta \cos \beta \cot \omega}) = a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega)$

Let the line $A'B'$ cut the altitude $AD$ at a point $D''$. Using Menelaus' theorem for the triangle $\triangle ACD$ cut by the line $A'B'$,

$\frac{A'A}{A'C} \cdot \frac{B'C}{B'D} \cdot \frac{D''D}{D''A} = 1$

$\frac{D''D}{D''A} = \frac{A'C}{A'A} \cdot \frac{B'D}{B'C} = \frac{\cos \beta}{\cos \alpha} \cdot \frac{A'C}{BD'} \cdot \frac{B'D}{B'C} = \frac{\cos \beta}{\cos \alpha} \cdot \frac{A'E + EC}{BC - BB'} =$

$= \frac{\cos \beta}{\cos \alpha} \cdot \frac{BB' \frac{\cos \alpha}{\cos \beta} + a \cos \gamma}{a - BB'} = \frac{BB' \cos \alpha + a \cos \beta \cos \gamma}{(a - BB') \cos \alpha} =$

$= \frac{a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) \cos \alpha + a \cos \beta \cos \gamma}{[a - a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega)] \cos \alpha} =$

$= \frac{(\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) \cos \alpha + \cos \beta \cos \gamma}{(\sin^2{\beta} + \sin \beta \cos \beta \cot \omega) \cos \alpha} = \frac{\cot \beta - \cot \omega - \cot \beta + \tan \alpha}{\tan \beta + \cot \omega} =$

$= \frac{\tan \alpha - \cot \omega}{\tan \beta + \cot \omega} = \frac{\tan \alpha - \frac 1 2 (\tan \alpha - \tan \beta)}{\tan \beta + \frac 1 2 (\tan \alpha - \tan \beta)} = \frac{\frac 1 2 (\tan \alpha + \tan \beta)}{\frac 1 2 (\tan \alpha + \tan \beta)} = 1$

Thus we proved that $D''A = D''D$, i.e., $D'' \equiv D'$ is the midpoint of the altitude $AD$ lying on the line $A'B'$. Since the midpoint $E'$ of the altitude $BE$ also lies on the line $A'B'$, the points $A' \equiv A_c$ and $B' \equiv B_c$ are identical. As a result, the quadrilateral $CA_cFB_c$ is cyclic, because

$\angle A_cFB_c = \angle A_cFE +  \angle EFD + \angle DFB_c =$

$= (180^o - 2 \gamma) + \angle A_cFE +  \angle AFA_c = 180^o - 2 \gamma + \gamma = 180^o - \gamma$

This means that the circumcircle $(Z')$ of the triangle $\triangle CA_cB_c$ passes through the foot of the altitude $CF$. Consequently, this circle is centered on the triangle midline $A_1B_1 \parallel AB$, where $A_1, B_1$ are the midpoints of the sides $BC, CA$. Similarly, quadrilaterals $AB_aDC_a, BA_bEC_b$ are also cyclic, the circles $(X'), (Y')$ pass through the altitude feet $D, E$ and they are centered on the triangle midlines $B_1C_1 \parallel BC, C_1A_1 \parallel CA$, where $C_1$ is the midpoint of the remaining triangle side $AB$. The power of the orthocenter $H$ to the circlec $(X'), (Y'), (Z')$ is $HA \cdot HD = HB \cdot HE = HC \cdot HF$, i.e., the orthocenter is the radical center of these 3 circles.

To be continued.

Yetti

[color=red][b]Step 2:[/b][/color] Earlier, we found that

$BB_c = a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) = a \left(\cos^2{\beta} - \sin \beta \cos \beta\ \frac{\tan \alpha - \tan \beta}{2}\right)$

By cyclic exchange, we obtain

$CC_a = b \left(\cos^2 \gamma - \sin \gamma \cos \gamma\ \frac{\tan \beta - \tan \gamma}{2}\right)$

Comparing the segments $CC_a, AA_c$, we have

$DB_c = BD - BB_c = c \cos \beta - a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) =$

$= a \left(\frac{\sin \gamma}{\sin \alpha} \cos \beta - \cos^2{\beta} + \sin \beta \cos \beta \cot \omega \right) =$

$= a (\cos^2{\beta} + \cot \alpha \sin \beta \cos \beta  - \cos^2{\beta} + \sin \beta \cos \beta \cot \omega) =$

$= a \sin \beta \cos \beta \left(\cot \alpha + \frac{\tan \alpha - \tan \beta}{2}\right)$

Removing the angle $\alpha$ from the expression for $AA_c$,

$AA_c = \frac{\cos \alpha}{\cos \beta}\ DB_c = b \sin \alpha \cos \alpha \left(\cot \alpha + \frac{\tan \alpha - \tan \beta}{2}\right) = $

$= b \left(\cos^2{\alpha} + \frac{\sin^2{\alpha}}{2} - \sin \alpha \cos \alpha\ \frac{\tan \beta}{2}\right) = $

$= \frac b 2 [1 + (\cos \beta \cos \gamma - \sin \beta \sin \gamma)^2\ +$

${+ (\sin \beta \cos \gamma + \cos \beta \sin \gamma)(\cos \beta \cos \gamma - \sin \beta \sin \gamma) \tan \beta}] =$

$= \frac b 2 [1 + \cos^2{\beta} \cos^2{\gamma} - 2 \sin \beta \sin \gamma \cos \beta \cos \gamma + \sin^2{\beta}\sin^2{\gamma}\ +$

$+ (\sin \beta \cos \beta \cos^2{\gamma} - \sin^2{\beta} \sin \gamma \cos \gamma + \cos^2{\beta} \sin \gamma \cos \gamma - \sin \beta \cos \beta \sin^2{\gamma}) \tan \beta] =$

$= \frac b 2 (1 + \cos^2{\beta} \cos^2{\gamma} - 2 \sin \beta \sin \gamma \cos \beta \cos \gamma + \sin^2{\beta}\sin^2{\gamma}\ +$

$+ \sin^2{\beta} \cos^2{\gamma} - \sin^2{\beta} \tan \beta \sin \gamma \cos \gamma + \sin \beta \cos \beta \sin \gamma \cos \gamma - \sin^2{\beta} \sin^2{\gamma}) =$

$= \frac b 2 (1 + \cos^2{\gamma} - \sin \beta \sin \gamma \cos \beta \cos \gamma - \sin^2{\beta} \tan \beta \sin \gamma \cos \gamma) =$

$= \frac b 2 [2 \cos^2{\gamma} + \sin \gamma \cos \gamma (\tan \gamma - \sin \beta \cos \beta - \sin^2{\beta} \tan \beta)] =$

$= b \left(\cos^2{\gamma} - \sin \gamma \cos \gamma  \frac{\tan \beta - \tan \gamma}{2}\right) = CC_a$

Thus we proved that the segments $AA_c = CC_a$ are not only equal, but either both inside or both outside of the triangle side $CA$ (this follows from the signs of the tangents). Similarly, we could show that the segments $BB_c = CC_b, AA_b = BB_a$ are equal and either both inside or both outside of the triangle sides $BC, AB$. As a result, the powers of the midpoint $C_1$ of the side $AB$ to the circles $(X'), (Y')$ are equal. For example, if the points $A_b, B_a$ are both outside of the side $AB$,

$C_1A \cdot C_1B_a = C_1A (C_1B + BB_a) = C_1B (C_1A + AA_b) = C_1B \cdot C_1A_b$

and consequently, the radical axis of the circles $(X'), (Y')$ passes through the midpoint $C_1$. Similarly, the radical axes of the circle pairs $(Y'), (Z')$ and $(Z'), (X')$ pass through the midpoints $A_1, B_1$ of the sides $BC, CA$.

To be continued.

Yetti

[color=red][b]Step 3:[/b][/color] The angles $\angle B_cA_cF \equiv \angle E'A_cF = \angle E'EF \equiv \angle BEF$ and the angles $\angle A_cB_cF \equiv \angle E'B_cF = \angle E'BF \equiv \angle EBF$ spanning the same arcs $E'F$ of the circles $(O_B), (O_E)$ are equal. Hence, the triangles $\triangle ABH \sim \triangle EFB \sim \triangle AFD \sim \triangle A_cFB_c$ are all similar, because

$\angle FAD \equiv \angle BAH = 90^o - \beta,\ \ \angle B_cA_cF = \angle BEF = 90^o - \angle FEA = 90^o - \beta$

$\angle FBE \equiv \angle ABH = 90^o - \alpha,\ \ \angle A_cB_cF = \angle EBF \equiv \angle EBA = 90^o - \alpha$

The segments $FE', HC_1$ are the corresponding medians of the similar triangles $\triangle EFB \sim \triangle ABH$, hence, the angles $\angle FC_1H \equiv BC_1H = \angle FE'E = 180^o - \angle FE'B = \omega$ are equal. The angles $\angle FE'B = \angle FB_cB = 180^o - \omega$ spanning the the same arc $BF$ of the circle $(O_B)$ are also equal. Let the circle $(Z')$ intersect the line $AB$ at a point $Z$ different from $F$. Since the angle $\angle CFZ \equiv CFB = 90^o$ is right, $CZ$ is a diameter of the circle $(Z')$. The angles $\angle FZC = \angle FB_cC = 180^o - \angle FB_cB = \omega$ are also equal. Thus the right angle triangles $\triangle CFZ \sim \triangle HFC_1$ are centrally similar with the similarity center $F$ and consequently, the lines $HC_1 \parallel CZ$ are parallel. As a result, the circumcenter $Z'$ of the triangle $\triangle CA_cB_c$ is identical with the intersection of the midline $A_1B_1$ with the line $CZ \parallel HC_1$ parallel to the radical axis of the circles $(X'), (Y')$. Since the cevians $A_1H, B_1H, C_1H$ of the medial triangle $\triangle A_1B_1C_1$, the pairwise radical axes of the circles $(X'), (Y'), (Z')$, meet at the orthocenter $H$ of the original triangle $\triangle ABC$, by central similarity of these 2 triangles with the similarity center at the triangle centroid $G$ and similarity coefficient -2, the cevians $AX, BY, CZ$ of the triangle $\triangle ABC$ parallel to $A_1H, B_1H, C_1H$ meet at a point $L$ on the line $GH$ (Euler line) opposite to the orthocenter $H$, such that $\frac{GH}{LH} = 2$ (Longchamps point, never heard of it) or $\frac{OH}{OL} = 1$, i.e., the reflection of the orthocenter $H$ in the circumcenter $O$. Since the angles $\angle ADX = \angle BEY = \angle CFZ = 90^o$ spanned by the appropriate arcs $AX, BY, CZ$ of the circumcircles $(X'), (Y'), (Z')$ are right, the circumcenters $X', Y', Z'$ are the midpoints of the cevians $AX, BY, CZ$.

The right angle triangle $\triangle CFZ$ is half the size of the right angle triangle $\triangle CF'Z'$, hence, the triangles $\triangle CFZ \sim \triangle HFC_1$ are also similar with the similarity coefficient

$\frac{CF'}{HF} = \frac{CF}{2HF} = \frac{a \sin \beta}{2 a \cos \beta \cot \alpha} = \frac{\tan \alpha \tan \beta}{2}$

$Z'F' = \frac{\tan \alpha \tan \beta}{2}\ C_1F = \frac{\tan \alpha \tan \beta}{2} \left(a \cos \beta - \frac c 2\right) = \frac{a \tan \alpha \tan \beta}{4} \left(2 \cos \beta - \frac{\sin \gamma}{\sin \alpha}\right) =$

$= \frac{a \tan \alpha \tan \beta}{4} \left(2 \cos \beta - \cos \beta - \frac{\sin \beta}{\tan \alpha}\right) = \frac{a \sin \beta}{4}(\tan \alpha - \tan \beta)$

$A_1Z' = A_1F' - Z'F' = \frac{a \cos \beta}{2} - \frac{a \sin \beta}{4}(\tan \alpha - \tan \beta) =$

$= \frac{a}{4}[\cos \beta (1 - \tan \alpha \tan \beta) + \cos \beta (1 + \tan^2{\beta}) ] = \frac{a}{4 \cos \beta} [\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1]$

$A_1F' \cdot A_1Z' = \frac{a \cos \beta}{2} \cdot \frac{a}{4 \cos \beta} [\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1] =$

$= \frac{a^2}{8} [\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1] = \frac{a^2}{8} \left(1 - \frac{\cos \beta \cos \gamma}{\cos \alpha}\right)$

Let $Y$ be the intersection of the circle $(Y')$ with the line $CA$ other that the points $E$. For similar reasons as above, the right angle triangles $\triangle BE'Y' \sim \triangle BEY \sim \triangle HEB_1$ are similar, the similarity coefficient of $\triangle BE'Y' \sim \triangle HEB_1$ equal to

$\frac{BE'}{HE} = \frac{BE}{2HF} = \frac{a \sin \gamma}{2 a \cos \gamma \cot \alpha} = \frac{\tan \gamma \tan \alpha}{2}$

$Y'E' = \frac{\tan \gamma \tan \alpha}{2}\ B_1E = \frac{\tan \gamma \tan \alpha}{2} \left(\frac b 2 - a \cos \gamma\right) = \frac{a \tan \gamma \tan \alpha}{4} \left(\frac{\sin \beta}{\sin \alpha} - 2 \cos \gamma\right) =$

$= \frac{a \tan \gamma \tan \alpha}{4} \left(\frac{\sin \gamma}{\tan \alpha} + \cos \gamma - 2\cos \gamma\right) = \frac{a \sin \gamma}{4} \left(\tan \gamma - \tan \alpha\right)$

$A_1Y' = AE' + Y'E' = \frac{a \cos \gamma}{2} + \frac{a \sin \gamma}{4}(\tan \gamma - \tan \alpha) =$

$= \frac{a}{4}[\cos \gamma (1 + \tan^2{\gamma) + \cos \beta (1 - \tan \gamma \tan \alpha) ] = \frac{a}{4 \cos \gamma} [\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) + 1]}$

$A_1E' \cdot A_1Y' = \frac{a \cos \gamma}{2} \cdot \frac{a}{4 \cos \gamma} [\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) + 1] =$

$= \frac{a^2}{8} [\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) + 1] = \frac{a^2}{8} \left(1 - \frac{\cos \beta \cos \gamma}{\cos \alpha}\right)$

Thus we proved that $A_1E' \cdot A_1Y' = A_1F' \cdot A_1Z'$, i.e., that the quadrilateral $E'F'Y'Z'$ is cyclic. In an entirely similar way, we can show that $B_1F' \cdot B_1Z' = B_1D' \cdot B_1X'$ and $C_1D' \cdot C_1X' = C_1E' \cdot C_1Y'$, i.e., that the quadrilaterals $F'D'Z'X', D'E'X'Y'$ are also cyclic. This still leaves the possibility that the circumcircles of these 3 cyclic quadrilaterals are different form each other and from the circumcircle of the triangle $\triangle D'E'F'$. Assume that the circumcircles $(P_A), (P_B)$ of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X'$ intersecting at the points $F', Z'$ are different from each other. Since $C_1D' \cdot C_1X' = C_1E' \cdot C_1Y'$, the point $C_1$ has the same power to the circles $(P_A), (P_B)$, which means that it lies on their radical axis $F'Z' \equiv A_1B_1$, i.e., that the midpoints $A_1, B_1, C_1$ of the triangles sides $BC, CA, AB$ are collinear, which is clearly impossible. As a result, the circumcircles of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X', D'E'X'Y'$ are identical. This concludes the proof of the problem proposition.

Perhaps we should mention the case, when the angle $\gamma = 90^0$ is right. The triangle $\triangle D'E'F'$ then degenerates into a line segment, its circumcircle into the midline $A_1B_1$ of the right angle triangle $\triangle ABC$, the circles $(X'), (Y')$ degenerate into the lines $CA, BC$ and the circle $(Z')$ becomes identical with the triangle 9-point circle. Hence, the circumcenters $X', Y'$ move to infinity and the circumcenter $Z'$, the midpoint of the segment $CC_1 \equiv OH$ lies on the midline $A_1B_1$, the degenerate circumcircle of the degenerate triangle $\triangle D'E'F'$.

To be continued.

Yetti

[color=red][b]Bonus:[/b][/color] Earlier, we proved that the circumcenters $X', Y', Z'$ lie of the (extended) triangle midlines $B_1C_1, C_1A_1, A_1B_1$. The midpoints $D', E', F'$ of the altitudes $AD, BE, CF$ also lie on these midlines. The cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\triangle A_1B_1C_1$ cut its (extended) sides at the points $X', Y', Z'$ in the ratios $\frac{X'B_1}{X'C_1},\ \frac{Y'C_1}{Y'A_1},\ \frac{Z'A_1}{Z'B_1}$.

$Z'B_1 = A_1B_1 - Z'A_1 = \frac c 2 - \frac{a}{4 \cos \beta} \left[\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1\right] = \frac a 4 \left(\frac{2 \sin \gamma}{\sin \alpha} - \cos \beta (1 - \tan \alpha \tan \beta) - \frac{1}{\cos \beta}\right) =$

$= \frac a 4 \left(2 \cos \beta + 2 \sin \beta \cot \alpha - \cos \beta + \tan \alpha \sin \beta - \frac{1}{\cos \beta}\right) =$

$= \frac{a \sin \beta}{4} (2 \cot \alpha + \tan \alpha - \tan \beta) = \frac{a \sin \beta}{4 \sin \alpha \cos \alpha} \left[\cos^2{\alpha}(1 - \tan \alpha \tan \beta) + 1\right]$

$\frac{Z'A_1}{Z'B_1} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta } \cdot \frac{\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1}{\cos^2{\alpha} (1 - \tan \alpha \tan \beta) + 1}$

By cyclic exchange (and paying attention to the signs of the tangents),

$\frac{X'B_1}{X'C_1} \cdot \frac{Y'C_1}{Y'A_1} \cdot \frac{Z'A_1}{Z'B_1} =$

$= \frac{\cos^2{\gamma} (1 - \tan \beta \tan \gamma) + 1}{\cos^2{\beta} (1 - \tan \beta \tan \gamma) + 1} \cdot \frac{\cos^2{\alpha} (1 - \tan \gamma \tan \alpha) + 1}{\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) + 1} \cdot \frac{\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1}{\cos^2{\alpha} (1 - \tan \alpha \tan \beta) + 1} =$ [color=white].[/color]$1$

because all terms cancel out, for example,

$(?)\ \ \cos^2{\gamma} (1 - \tan \beta \tan \gamma) + 1 = \cos^2{\alpha} (1 - \tan \alpha \tan \beta) + 1$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = (\sin \gamma \cos \gamma - \sin \alpha \cos \alpha) \tan \beta = \frac{(\sin \gamma \cos \gamma - \sin \alpha \cos \alpha) (\sin \alpha \cos \gamma + \cos \alpha \sin \gamma)}{\cos \beta}$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = \frac{(\sin^2{\gamma} - \sin^2{\alpha}) \cos \alpha \cos \gamma + (\cos^2{\gamma} - \cos^2{\alpha}) \sin \alpha \sin \gamma}{\cos \beta}$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = \frac{(\cos^2{\gamma} - \cos^2{\alpha}) (\sin \alpha \sin \gamma - \cos \alpha \cos \gamma)}{\cos \beta}$

and the last equation is an obvious identity. By Ceva's theorem, the cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\triangle A_1B_1C_1$ concur at a point $Q$, so-called Yetti's point of the triangle $\triangle ABC$ (hey, everybody's got a triangle point and there's plenty more available :rotfl:).

[color=blue][b]Proposition:[/b][/color] Prove that the common circumcenter $P$ of the triangles $\triangle D'E'F', \triangle X'Y'Z'$ lies on the line $QH$ connecting Yetti's point $Q$ with the triangle orthocenter $H$.

Yetti

[b]Note:[/b] All trigonometric formulas in the entire proof check with the Sketchpad.

http://forumgeom.fau.edu/FG2004volume4/FG200405index.html

    The above article by JPE  may bring some clearity into the subject roughed-up by previous
    replies by our own Yetti.   A complete quad where the 4th side is a line thru midpoints of
    altitudes in a triangle made by the other 3 lines should yield some interesting properties.

    Maj.  Pestich

I still hope that in some geo problem the Fox-Talbot theorem will be used.
   Is it possible this is the one?

    Maj.  Pestich

[quote="pestich"]    The above article by JPE  may bring some clearity into the subject roughed-up by previous
    replies by our own Yetti.   A complete quad where the 4th side is a line thru midpoints of
    altitudes in a triangle made by the other 3 lines should yield some interesting properties.

    Maj.  Pestich[/quote]

Let's make one thing clear, comrade Pestich. You do not own me.

As for the problem, I solved it, not roughed up. If the solution was quick and easy, those guys at Hyacintos would take it apart in no time, trust me. If you do not understand the concept of solution (something you never will), then shut up. It seems that you just cannot let go and are looking for another trashing. Please seek help.

Yetti

[quote="pestich"]My dear Yetti,

  You do not have to rough me up.  I was the first one to agree that it was a wonderful solution indeed.
  Put your customary doubts about your own (I had to use this word again, sorry) solutions aside: this
  time it is a bona fide one.  I'd say it is  even more than a solution to a tough problem (those people at Hya-group
  wouldda never imagined the circumcenters on midlines), it is a novel to read, with a hero, villans, chapters,
  paragraphs and such.  But you are a bit lazy, and if I do not tell you, nobody will. Where are  the table of contents,
  index, and referrences?

   Sincerely,

   Maj.  Pestich[/quote]

Now you are highly embarrassing :blush:  :blush:  :blush: ... Please, don't try to be funny, try to be serious and prove something, because that makes you hilarious. I fondly remember your exquisite proof that [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=387&forum=DCForumID6#11]all quadrilaterals with perpendicular diagonals are concyclic[/url]  :rotfl:. Trust me, I am not the only one on this forum, who knows that Golland = Pestich. Or when you [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=391&forum=DCForumID6#17]projected two parallel lines to infinity[/url]  :rotfl:. Or when you repeatedly made a claim about [url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6#1]poles and polars[/url] equivalent to $a+b = b+a\ \Rightarrow\ a = b$  :rotfl:. But let's make a table, if that's what you miss. It is the table of [b]proofs without words[/b] by no other than comrade Pestich.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16824&start=11]projections which lie on a circle[/url], reply 11
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18084&start=6]Perpendicular projections and acute-angled triangle[/url], replies 6, 8, 10, 12
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19133]Interesting geometry[/url], replies 4, 6
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19178]The multisolution problem![/url], reply 6
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16213]Poland 3[/url], replies 3, 9
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19770]IMO ShortList 1999, geometry problem 2[/url], reply 5
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3028]Pizza everybody!!![/url], reply 11
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14741]posted before? [a circumcircle tangent to the incircle][/url], reply 4
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19500]GEOMETRY[/url], replies 4, 6
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19718]Cyclic[/url], reply 5
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22633]Thailand TST[/url], replies 7, 11, 14
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23066]concurrent [PD, QE, RF concur at the incenter of ABC][/url], reply 4
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19468]Inradii problem[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5799]Concurrence of AA', BN and CM in the Gergonne configuration[/url], reply 17 [moderator edit: i've merged this topic and deleted pestich's post. darij]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28877]Special triangle[/url], replies 6, 9
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29333]HELP!! [related to the nine-point circle of triangle ABC][/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29782]Triangle center[/url], replies 4, 10
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30002]MN||OO'[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30036]Triangle + Circle problem[/url], replies 4, 9, 12
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30195]cyclic[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31420]Concurent[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31077]Locus problem [2][/url], reply 2
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18426]Please prove elementarily[/url], reply 2
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31921]convex quadrilateral and centroids[/url], reply 8
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32251]an easy problem!!!![/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32412]Perpendicular Distances[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33050]actually a physics problem[not hard][/url], reply 2
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33174]Orthopole on incircle[/url], replies 3, 5
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33149]Please solve my problem!!![/url], reply 7
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33278]Circumcenter of ABC is same as incenter of CMN[/url], reply 2
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32582]acute triangle[/url], reply 4
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33205]Please help to prove this[/url], replies 3, 5, 7, 9
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33514]cyclic quadrilateral[/url], replies 8, 13
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33948]Angle bisector[/url], reply 4, 7
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34000]Circumcenter on radical axis[/url], replies 3, 6
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34316]classical triangle geo - points on circle[/url], reply 4
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30611]Simson line property[/url], reply 5
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36132]ugly geometry[/url], reply 6
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37964]Unique[/url], reply 3
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4450]constant angle[/url], reply 3

That's about 22% - not bad for proofs without words, although these are a bit sketchy and I don't even think that you are correct in a single one. No claims are made as to the completeness of this table, it may and will grow at any time.

Looking forward to more stuff from you, but please, include some math content (just to make it funny).

:rotfl:  :rotfl:  :rotfl:
Yetti

jesus christ , jesus christ yetti
i havent seen a stonger one in calculating than you in my life even never heard of
and this skill of you is so valuable but please dont take these in a  bad manner but
do you think that all of these are necessary  :?:  :?:  :?:  :?:
in some cases some uses of geometrical ideas would really make the calculations shorter, but in all the ways no one can disagree your abilities :omighty:
:omighty: :omighty: :omighty: :omighty: :omighty:

No, I do not think that all these calculations are necessary. The fact that everything came out in closed forms (i.e., I did not have to resort anywhere to numerical calculations) indicates that the problem has underlying symmetries, which can be used in the solution (reflections in lines?). After I identified the basic steps of the proof, I was not able to see those symmetries, hence, I used trigonometry. But trigonometry itself is a (powerful) [b]geometric[/b] tool. If the sine and cosine functions are defined as the ratios of the opposite and adjacent legs of a right angle triangle to the hypotenuse, geometric proofs of any trigonometric formulas do exist. Trigonometry works even if nothing else does, i.e., if the problem does not have underlying symmetries and you are willing to use numerical methods to get the result. Moreover, after the solution is found by any means, for example by trigonometry, it might be easier for someone else to see the underlying symmetries. In this particular case, it is a mute proposition, because the problem was not posted in the Unsolved Problems section and Darij said that he already has a synthetic solution.

I will give you an example from my line of work. Suppose I have to transport a particle beam in an accelerator over huge distances in a beamline of a rather small diameter without losing too much by hitting the beamline walls. The beamline has to be evacuated and any increase in the beamline diameter calls for larger magnets, the principal item in the cost of building an accelerator. So, the beamline diameter really has to be rather small. The phase space volume of the beam is the "volume" in the space with one coordinate being the beam displacement from the beamline axis and the other coordinate the particle radial velocity toward or away from the beamline axis, proportional to the particle divergence, the angle between the particle path and the beamline axis. This phase space volume is called the beam emittance and it looks like an ellipse tilted to these coordinate axes. From Liouville's theorem of analytical mechanics, it follows that this space volume remains constant, i.e., it cannot be decreased by any focusing element in the beamline. The ellipse can be only rotated or "squashed" without decreasing its area. Consequently, to get a good beam, which can be successfully transported over those huge distances, the beam emittance has to be small to begin with, as the beam emerges from the ion source. Also, the beam emittance is an important beam parameter to measure. The measurement methods depend on whether the beam can be "consumed" for the measurement (for a while), or whether it has to be done on the fly and usually, they are very ingenious. The beam focusing elements are almost exclusively magnetic quadrupole lenses, because at relatively low particle energies, they overcome the focusing capabilities of other elements, such as electrostatic lenses, etc. The action of a magnetic quadrupole lens on the beam is described by a matrix operator with elements that are either normal or hyperbolic trigonometry functions. You can have tens to hundreds of these lenses, which rotate and squash the emittance ellipse in various ways. One of the non-destructive methods of measuring the beam emittance consists of slowly moving a thin wire across the beam and measuring the (very small) beam current. In principle, this yields the position of the vertical tangent to the emittance ellipse at this spot. Several thin wires at several different places in between the quadrupole lenses can be moved simultaneously, thus obtaining several vertical tangents to several different ellipses, which are all images of some original ellipse transformed by the quadrupole lenses. Using the inverse operators for the quadrupole lenses, these tangents can be transformed back to one spot and in this transformation, they become rotated and no longer vertical. To get the beam emittance, an ellipse has to be inscribed into these tangents and its area calculated. For this, at least 3 different tangents are necessary (because the ellipse center is known), but since all measurements carry some errors, more than 3 tangents are measured and the best possible ellipse is inscribed into these lines by the least squares method. This is an interesting application of geometry and I doubt that it could be done without both the normal and hyperbolic trigonometry.

So, [color=blue][b]three cheers for trigonometry !!![/b][/color]  :thumbup: :thumbup: :thumbup:

Regards, Yetti

Smeenk show (in Crux Mathematicorum, the problem nr. 2379/98) that the lines MD', NE', PF' are conccurent, where M, N, P are the middlepoints of the sides BC, CA, AB respectively. This fact may be used ?

After a few hours, I finally got a purely synthetic solution to Darij 's observation. It seems that I am getting very close to proving the 3 circumcenters lying on the circumcirlce of $D'E'F'$. I will post the solution in a next few days. I am very busy right now.

Finally, everything is clear. And here is the solution (guess this is the longest I have ever posted). I won't write just only the solutions of this problem but also others nice theorems relating to this wonderful sketch.
  So, let me first redenote some points so that the problem accords with my figure. (I am sure you don't mind :) ).
  Let $ABC$ be a triangle whose circumcenter and orthocenter are $O$ and $H$. Let L be the de Longchamp point, i.e the point on $HO$ such that $\frac{OL}{OH} = -1$
  Let $A'B'C'$ be the orthic triangle and $MNP$ be the medial triangle. Call $D, E, F$ the midpoints of $AA', BB', CC'$. Suppose the line $EF$ intersects the line $AB$ and $AC$ at $A_b$ and $A_c$ respectively. Analogously, we define the point $B_a$, $B_c$, $C_a$, $C_b$.
  Let $O_a, O_b, O_c$ be the circumcenters of $AA{}_bA{}_c, BB{}_aB{}_c, CC{}_aC{}_b$ respectively.
  All right ! Let 's do some small observation of the de Longchamp point for a moment. Well, by calling $T$ the antipode of $A$ in the circle $(O)$.,  we have $AHTL$ is a parallelogram since the diagonals bisect each other. On the other hand, we have $BHCT$ is a parallelogram as its opposite sides are parallel. Hence $H, M, T$ are collinear and therefore $HM // AL$. Suppose we call $O_a'$ and $A_1$  the intersections of $AL$ with $NP$ and $BC$, then surely we have $O_a'$ is the midpoint of $AA_1$. Thus if we call $U$ the midpoint of $HM$, since $HM // AA_1$,we obtain that $A', U, O_a'$ are collinear. (Later, we shall see that this point $O_a'$ is the point $O_a$). So let 's official state it as a theorem :
   [b]  Theorem 1[/b] $A, U, O_a'$ are collinear.
  Let 's next consider the point $U$. Since $\measuredangle{MEH} = \measuredangle{MFH} = \measuredangle{MA'H} = 90^0$, we obtain five points $H, E, A', M, E$ lie on the circle $(U)$ with diameter $HM$. Therefore $\measuredangle{A'EF} = \measuredangle{A'HF} = \measuredangle{A'BA_b}$. This implies that $BA'EA_b$ is cylic. By similar argument, we obtain that $CA'FA_c$ is cyclic and we have a theorem
      [b]Theorem 2[/b] $BA'EA_b$ is inscribed in $(b)$ and $CA'FA_c$ is inscribed in $(c)$
  Let 's derive some more properties from this cool theorem. From theorem 2, we have $\measuredangle{A'A{}_bA{}_c} = \measuredangle{A'A_bE} = \measuredangle{A'BE} = \measuredangle{A'AA_c}$. Hence $AA_bA'A_c$ is cyclic, i.e the circumcircle $(O_a)$ of $AA{}_bA{}_c$ passes through $A'$. This means that $O_a$ lies on $NP$. Now let 's prove that this point lies on $AL$ by showing that  $AO_a // HM$. Let $A_1$ be the second intersection different from $A'$ of the circle $(O_a)$ and $BC$. Then $AA_1$ is the diameter of $(O_a)$. Hence $\measuredangle{A{}_1A{}_cC} = 90^0$. Now according to theorem 1 and 2, we have $\measuredangle{A'AA_1} = \measuredangle{A'A{}_cA{}_1} = \measuredangle{A'A{}_c{}_C} - 90^0 = \measuredangle{A'FC} - 90^0 = \measuredangle{A'FM} = \measuredangle{A'HM}$. This yields that $HM // AA_1$, and therefore $AO_a$ passes through the de Longchamp point $L$. And Darij 's observation is already proved. Let 's state it
     [b] Theorem 3[/b] The circumcircle $(O_a)$ of $AA{}_bA{}_c$ passes through $A'$ and $A, O_a, L$ are collinear.
     In my next posts, we will give some more observations on the relations of this circle with the circle $(U)$, $(b)$ and $(c)$.

OK. Let 's continue with our work.
 We have $\measuredangle{A_bA'F} = \measuredangle{A_bA'E} + \measuredangle{EA'F} = \measuredangle{A_bBE} + \measuredangle{EHF} = \measuredangle{ABB'} + \measuredangle{B'AB} = 90^0$. Hence $A_bA'FC'$ is cylic. Similarly, we have $A'EB'A_c$ is cylic.
    [b]Theorem 4[/b] $A_bA'FC'$ and $A'EB'A_c$ are cylic.
  By applying the Menelaus 's theorem to triangle $ABB'$ and three collinear points $A_b, E, A_c$, we have $\frac{AA_b}{AA_c} = \frac{A_bB}{B'A_c}$. On the other hand, we have $\frac{A_bB}{A'A_b} = \frac{sin{ \measuredangle{BA'A_b}}}{sinB} = \frac{sin{ \measuredangle{BEA_b}}}{sinB} = \frac{sin{ \measuredangle{B'A'A_c}}}{sin{ \measuredangle{A'B'A_c}}} = \frac{B'A_c}{A'A_c}$
   [hide="Reason"] From theorem 4, we have $ \measuredangle{B'A'A_c} =  \measuredangle{B'EA_c} =  \measuredangle{HEA_c} =  \measuredangle{BEA_b}$ [/hide]
  Hence $\frac{A'A_b}{A'A_c} = \frac{A_bB}{B'A_c}$. Thus  $\frac{AA_b}{AA_c} = \frac{A'A_b}{A'A_c}$. This implies that $AA_bA'A_c$ is a harmonic quadrilateral. We will see its application later to prove the original problem.
    [b]Theorem 5 [/b] $AA_bA'A_c$ is a harmonic quadrilateral.
  We will finish our proof in the next post.

Let 's consider the relationship between the two circles $(b)$ and $(c)$. We have $\measuredangle{A'O_aA_c} = 2.\measuredangle{A'AA_c} = 2.\measuredangle{A'AC} = \measuredangle{A'NA_c}$. Thus $A'O_aNA_c$ is cylic. Hence $\measuredangle{O_aA'A_c} = \measuredangle{O_aNA} = \measuredangle{PNA} = \measuredangle{BCA} = \measuredangle{A'CA_c}$. This yields that $A'O_a$ touches the circle ${(c}$ at $A'$. Similarly we have $A'O_a$ touches the circle ${(b}$ at $A'$. And theorem 6 follows
   [b]Theorem 6[/b] $(b)$ and $(c)$ touches each other and the line $A'O_a$ at $A'$.
   All right, we are 90% done.
   Let $l$ be the line joining the centers of $(b)$ and $(c)$. Then $l$ passes through $A'$ and is perpendicular to $O_aA'$. Hence $l$ touches the circle $(O_a)$ at $A'$
   Recall theorem 1 that $A', U, O_a$ are collinear. Thus, the circle $(U)$ touches the circle $(O_a)$ at $A'$. Therefore theorem 7 follows
   [b]Theorem 7[/b] The circle $(O_a)$ and $(U)$ touches each other and the line $l$ at $A'$.
   Again, recall from theorem 5 that $AA_bA'A_c$ is harmonic, we obtain that the tangents at $A$ and at $A'$ (i.e the line $l$) are concurrent with the line $A_bA_c$ at $R$. Since $DO_a$ is the perpendicular bisector of the segment $AA'$, we obtain that $DO_a$ passes through $R$. Therefore $RA'^2 = RE.RF = RA_b . RA_c = RD. RO_a$, we obtain that $D, E, F, O_a$ lie on a circle. By similar argument, it follows that
  [b] Theorem 8 [/b] $D, E, F, O_a, O_b, O_c$ lie on a circle.
  Note that from the equality above, we also obtain that $A_bDO_aA_c$ is cyclic. Since $O_aA_b = O_aA_c$, we have
  [b]Theorem 9[/b] $DA'$ is the bisector of $\measuredangle{A_bDA_c}$.
  All right ! Let me finish this post by introducing some more nice properties relating to this sketch in theorem 10
  [b]Theorem 10[/b]
[i](i)[/i] $R$ is the excenter of similitude of the two circles $(b)$ and $(c)$.
[i] (ii)[/i]$(b)$ and $(U)$ is orthogonal, $(c)$ and $(U)$ is orthogonal.
                             [i](iii)[/i] The circumcircle of $ADA_b$ touches $AC$ at $A$, The circumcircle of $ADA_c$ touches $AB$ at $A$
  [hide="Hint"] The (i) and (ii) questions, we use angle chasing. The (iii) question, we will use a nice property of harmonic quadrilateral. Notice that $D$ is the midpoint of $AA'$ [/hide]
  Khoa Lu Nguyen.

Dear Mathlinkers,
A new proof of "the Mineur's circle" has been put on my website

http://perso.orange.fr/jl.ayme

see contenu, vol. 3
Sincerely
Jean-Louis


A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?

[hide="Solution"]
The general solution is messy and I haven't worked it out very well at all, so I will present a non-rigorous process that I used to arrive at a specific answer.  (It is right, though, I checked the AMC website ;) )

Number the squares $0, 1, 2, 3, ... 2^{k} - 1$.  In this case $k = 10$, but we will consider more generally to find an inductive solution.  Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$.

Now, consider the strip of length $1024$.  The problem asks for $s_{941, 10}$.  We can derive some useful recurrences for $s_{n, k}$ as follows:  Consider the first fold.  Each square $s$ is now paired with the square $2^{k} - s - 1$.  Now, imagine that we relabel these pairs with the indices $0, 1, 2, 3... 2^{k-1} - 1$ - then the $s_{n, k}$ value of the pairs correspond with the $s_{n, k-1}$ values - specifically, double, and maybe $+1$ (if the member of the pair that you're looking for is the top one at the final step).

So, after the first fold on the strip of length $1024$, the $941$ square is on top of the $82$ square.  We can then write

$s_{941, 10} = 2s_{82, 9} + 1$

(We add one because $941$ is the odd member of the pair, and it will be on top.  This is more easily visually demonstrated than proven.)  We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two):

$s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127-82, 6} = 8s_{45, 6}$
$s_{45, 6} = 2s_{63-45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31-18, 4} + 1 = 4s_{13, 4} + 1$
$s_{13, 4} = 2s_{15-13, 3} + 1 = 2s_{2, 3}$

We can easily calculate $s_{2, 3} = 4$ from a diagram.  Plugging back in,

$s_{13, 4} = 9$ (this can be verified from a diagram as well)
$s_{45, 6} = 37$
$s_{82, 9} = 296$
$s_{941, 10} = \boxed{593}$

Whew.
[/hide]

$1024^{th}$ post!   :P  I couldn't post again until I solved this problem, it was too good of an opportunity.

Yeah, that's what I got too by reconstructing the sequence from smaller cases and figuring out the position square 941 is in.

PS. In 7th period right now posting this :P.

[quote="t0rajir0u"](We add one because $941$ is the odd member of the pair, and it will be on top.  This is more easily visually demonstrated than proven.)  We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two)[/quote]
Just curious how would you prove this?


function
integration
calculus
derivative
algebra
domain
algebra proposed

does there exist a function $f$ such that $f$ is defined on an interval. so that $f'$=$f^{-1}$
(where $f^{-1}$=inverse of $f$.for example if$f(x)=2x$ then the inverse is $\frac{x}{2}$)

from $f'(x) = f^{-1}(x)$ we derive two equations \[ f(x) +C=\int f'(x)\,dx = \int f^{-1}(x)\,dx \] \[ f(f'(x))=f(f^{-1}(x))=x \]
therefore we get\[ x+C = \int f^{-1}(f('x))\,dx \]
and hence\[ f^{-1}(f('x))= 1 \]
since $f^{-1}$ needs to be injectiv (otherwise $f$ wouldn't be a function) we find (let $f: N\to M$ ) \[ f'(N) = {a} \qquad\text{,where \ensuremath{a} is a constant} \]
therefore it is only possible, that\[ f(x) = ax+b \]
but then $f'(x) = a$ and $f^{-1}(x) = \frac{x-b}{a}$, so \[ \forall x \in N: a = \frac{x-b}{a} \iff x= a^2+b \]
hence there is no such interval if u dont consider $[x_0,x_0]$ as an interval

if you do, then one'd have to find an $f$ so that\[ \exists x \in \mathbb R : f'(x)= f^{-1}(x) \]
(but i dont want to do that right now ;) )

Huh.

I see an extremely messy solution by writing $f(f'(x)) = x$ in terms of Taylor Series $f(x) = a_0 + a_1 x + \frac{a_2}{2!} x^2 + ...$ and $f'(x) = a_1 + a_2 x + \frac{a_3}{2!} x^2 + ...$.   :(

@dashmiz: is my solution right anyway?
@anybody: is my proof valid? (i did alot of little (or big) mistakes on functions in the last days ;) )

i do not think it is completly right.
i would check it man :)

[quote="Kalle"]In general if we have that $(f^{-1})'(f(x)) \ne 0$ it is true that $f'(x) =\frac{1}{(f^{-1})'(f(x))}$. Given that $f^{-1} = f'$ we get the differential equation $f'(x) =\frac{1}{f''(f(x))}$

I don't know how to work with differential equations, but I'm sure someone who has done a lot of calculus could tell us why there are no solutions to this thing.

edit: Oh, we have that $f''(f(x))f'(x) = 1$. This is a seperable equation, so we now get $f'(f(x)) + C = x$. However, we know that $f' = f^{-1}$ so that equation says that $C = 0$, so all the information we have gotten is that $f'(f(x)) = x$, which we already knew.

peeta, I don't understand how you get  $x + C = \int f^{-1}(f'(x))dx$. Please clarify.[/quote]

oh sorry ... i skipped a step or two there:

we already got\[ f(x) + C = \int f^{-1}(x)\,dx \]
plugging in $f'(x)$ for $x$ (note: $f' \in D_f$ because $f'=f^{-1}\in D_f$) we get\[ f(f'(x)) + C = \int f^{-1}(f'(x))\,dx \]
since we already know that $f(f'(x)) = x$ we get\[ x + C = \int f^{-1}(f'(x))\,dx \]

[quote="peeta"]
plugging in $f'(x)$ for $x$ (note: $f' \in D_f$ because $f'=f^{-1}\in D_f$) we get\[ f(f'(x)) + C = \int f^{-1}(f'(x))\,dx  \][/quote]
This is where your error is. You cannot make substitutions like that. Try for example $x^2 = \int_0^x 2xdx$. Now make essentially any substitution like you just did and it won't work. Of course what you might actually mean is that we find the primitive and then plug in $f'(x)$ instead of $x$, but then your differentiation later in the text won't work (you must use chain rule).

and about you  peeta
kalle said right.you can not just substitute evrything.(remembering $f(g(x))'$) :D

isn;t there any complete solution for that?

[quote="dashmiz"]but Agrippina .i think you'v made a mistake.
check it once more.
and about you  peeta
kalle said right.you can not just substitute evrything.(remembering $f(g(x))'$)[/quote]
yeah thats true i understood that argument, i was just too much into "functional equations" (not diff equ), where one can normally do that ;) im thinking about a new (diffrent) solution, but not today ;)
(but Agrippina seems to have a method to find the solutions, although i dont think either, that his function is a solution, it is pretty close to one as it seems ;) )

[quote="Agrippina"]Hey, what about $f(x)=\phi^{-\frac{1}{\phi}}x^{\phi}$?  $f^{-1}(x)=f'(x)=\phi^{-\frac{1}{\phi^2}}x^{-\frac{1}{\phi}}$, where $\phi$ is the golden ratio.[/quote]

You're onto something here.  Let's try something more general.  $f(x) = ax^b \implies f'(x) = abx^{b-1}, f^{-1} (x) = a^{ \frac{-1}{b} } x^{-b}$ so we need to solve

$ab = a^{ \frac{-1}{b} }$
$b-1 = -b$

This gives $2b = 1 \implies b = \frac{1}{2}$ and $a^3 = 2 \implies a = \sqrt[3]{2}$, so our function is $\boxed{ f(x) = \sqrt[3]{2} \sqrt{x}}$.  I think.  :)

t0rajir0u, rather than $f^{-1} (x) = a^{ \frac{-1}{b} } x^{-b}$, you should have $f^{-1} (x) = a^{ \frac{-1}{b} } x^{\frac{1}{b}}$.  (For example, if $f(x)=x^2$ then $f^{-1}(x)=x^{\frac{1}{2}}$, not $x^{-2}$.)  This produces the function I posted before.

To check, if $f(x)=\phi^{-\frac{1}{\phi}}x^{\phi}$, then $f'(x)=\phi^{1-\frac{1}{\phi}}x^{\phi}$, while $f^{-1}(x)=[\phi^{\frac{1}{\phi}}x]^{\frac{1}{\phi}}$.  Since $1-\frac{1}{\phi}=\frac{1}{\phi^2}$ and $\phi-1=\frac{1}{\phi}$, we have $f'(x)=f^{-1}(x)$.

But as for [i]all[/i] solutions, I haven't a clue.  :lol:

Oh.  Sorry.  I was doing that during lunch and someone else was pelting me with questions about metaphors  :D

but as you mentioned i think the answer to all solutions need higher mathematics.
i would ask it if i find some one.

[quote="Agrippina"]
To check, if $f(x)=\phi^{-\frac{1}{\phi}}x^{\phi}$, then $f'(x)=\phi^{1-\frac{1}{\phi}}x^{\phi}$:lol:[/quote]

 Please , see the derivative again. The exponent of $x$ is $\phi - 1$ and not $\phi$ . I think that many years ago I had proved- if the problem was really the same -  that f can not have as domain the set of real numbers, but I have forgotten the proof.

[u] Babis[/u]

but stergiu  what if the domain(as was mentioned)is a domain.
i think it is then true. :?:

Let me try to fix mine.

We will assume the function is of the form $f(x) = ax^b$ since such functions have the property that both their inverses and derivatives are power functions.

$f'(x) = ab x^{b-1}$
$f^{-1}(x) = \left( \frac{x}{a} \right)^{ \frac{1}{b} } = a^{ \frac{-1}{b} } x^{ \frac{1}{b} }$

$b-1 = \frac{1}{b} \implies b^2 - b = 1 \implies b = \phi$
$a \phi = a^{ \frac{-1}{\phi} }$

$\phi^2 = \phi + 1$
$\phi = 1 + \frac{1}{\phi}$
$\frac{-1}{\phi} = 1 - \phi$

$a \phi = a^{1 - \phi}$
$a^{\phi} \phi = 1$
$a = \left( \frac{1}{\phi} \right)^{ \frac{1}{\phi} } = (\phi - 1)^{\phi - 1}$

$\boxed{ f(x) = (\phi-1)^{\phi - 1} x^{ \phi } }$

 :D

Edit:  So, er, yes, this is equivalent to the previous solution posted, but even so!

As shown , if the domain is an interval , then there exists such a function.For  domain IR , I had seen the problem in an old Romanian book(unsolved) and I spent much time to prove the non existence.

          Babis

It was posted before :) : http://www.mathlinks.ro/Forum/viewtopic.php?t=37916 .


Let $\displaystyle y = f(x) = \frac{a^{x-1}+5}{a^{x-3}+3a}$ be a function, where a is a positive real parameter.
Determine the value of a, such that the range of the function doesn’t contain any odd integer numbers?

does $x$ have to be integer ? Or all real number ?

what about $a=1$   :)


Several times when I have been making documents in Texnic Center, the computer will say there are three errors even though there aren't any. Also, if I add errors and click Ctrl-F7, the number of errors doesn't change.
I have been able to make the 3-error message go away by adding errors intentionally and deleting them, but this only works sometimes.  Has anyone else gotten this false error message?  What can be done to prevent or get rid of it?
Can some1 help? (My challenge set is due Thurs  :( )!!!!

Why don't you tell us what these errors are so we can advise?
[quote="baldcypress"]the computer will say there are three errors even though there aren't any. Also, if I add errors and click Ctrl-F7, the number of errors doesn't change.
I have been able to make the 3-error message go away by adding errors intentionally and deleting them, but this only works sometimes.  Has anyone else gotten this false error message?[/quote]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url] since we are not mind-readers.

There are two types of errors that TexnicCenter gives:

1. The important ones are on the last line which says:
LaTeX-Result: [i]k[/i] Error(s), [i]l[/i] Warning(s), [i]m[/i] Bad Boxe(s), [i]n[/i] Page(s)
and you need to get [i]k[/i] and [i]l[/i] to be 0.

2. Higher up in the output window TexnicCenter tells you if it has found citation, bibdata and bibstyle commands in the aux file. If it hasn't then it says there are errors. Obviously if you have no bibliographic data then these errors aren't real errors at all.

This is only from my own experience, but I highly suspect the problem is that you are not closing the PDF (if you're using something else to view it, I don't know) of the previously compiled version before compiling the next version (after you edit some more) of the document.

Sorry I don't know the correct terminology.

Huh?  I can always view the PDF, then re-compile without errors.

What leoxnlin says happens to me too. When you keep your PDF viewer with the compiled LaTeX document open and then you build your LaTeX document, it gives you an error saying that it can't write on the compiled document.

Wait, what compiler are you using?  I'm using TeXShop (for OSX) so it might be different for me.

That explains it. I use TeXnicCenter lol.

Hi sunehra,

it depends also on the pdf viewer you are using. For instance the [url=http://blog.kowalczyk.info/software/sumatrapdf/]Sumatra PDF viewer[/url] would work like expected. If you use the Adobe Reader you could adjust the output profile in a way that the reader will be closed before compilation by a DDE command and will be opened again.

Stefan

Ok, thanks! I didn't know that. So with other PDF viewers, they don't lock the file or something so that they can't be accessed by other programs?

Adobe 8.0 had this lockng mechanism which was a bug. Versions before and after that are fine. Even if you have version 8.0 there is a workaround. In Build, Define Output Profiles, Viewer add [DocOpen("%bm.pdf")] to the 2 instances of [DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] so it reads [DocOpen("%bm.pdf")][DocOpen("%bm.pdf")][FileOpen("%bm.pdf")].

One advantage of the Sumatra pdf viewer is that it supports [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfsync]pdfsync[/url] which allows you to go to specific places in the pdf and back to the relevant source code. I haven't tried it so I don't know how well it works or if TexnicCenter supports it.


Can anyone translate the Polish MO 2009 Round 2?

First day.

1. $ a_1 \geq a_2 \geq \ldots \geq a_n > 0$. Prove the inequality
$ a_1a_2\ldots a_{n\minus{}1} \plus{} (2a_2\minus{}a_1)(2a_3\minus{}a_2) \ldots (2a_n\minus{}a_{n\minus{}1}) \geq 2a_2a_3 \ldots a_n.$

2. There are given integers numbers $ a, b$ such that $ a>b>1$, $ ab\plus{}1$ is divisible by $ a\plus{}b$ and $ ab\minus{}1$ is divisible by $ a\minus{}b$. Prove that $ a < b \sqrt{3}$.

3. Disjoint circles $ o_1, o_2$, with centers $ I_1, I_2$ respectively, are tangent to the line $ k$ at $ A_1, A_2$ respectively and they lie on the same side of this line. Point $ C$ lies on segment $ I_1I_2$ and $ \angle A_1CA_2 \equal{} 90^{\circ}$. Let $ B_1$ be the second intersection of $ A_1C$ with $ o_1$, and let $ B_2$ be the second intersection of $ A_2C$ with $ o_2$. Prove that $ B_1B_2$ is tangent to the circles $ o_1, o_2$.

Second day.

4. Segment $ AB$ is diameter of the circle $ o$, outscribed on the convex quadrilateral $ ABCD$. Let $ E$ be the intersection of diagonals of this quadrilateral and let $ P$ be the intersection of tangents to the circle $ o$ at points $ C, D$. Prove that $ PC\equal{}PE$.

5. Find all integer numbers $ n \geq 4$ which satisfy the following condition: From every $ n$ different $ 3$-element subsets of $ n$-element set it is possible to choose $ 2$ subsets, which have exactly one element in common.

6. For every integer $ n \geq 3$ find all sequences of real numbers $ (x_1, x_2, \ldots, x_n)$ such that
$ \sum_{i\equal{}1}^{n}x_i \equal{} n$ and $ \sum_{i\equal{}1}^{n} (x_{i\minus{}1}\minus{}x_i\plus{}x_{i\plus{}1})^2 \equal{} n$,
where $ x_0\equal{}x_n$ and $ x_{n\plus{}1}\equal{}x_n$.


A draftsperson makes a scale drawing of a 100 meter $ \times$ 30 meter building, where 1 centimeter represents 2.5 meters. How many centimeters are in the smaller dimension of the drawing of the building?

Complying with the scale, $ \frac{30}{2.5}\equal{}\fbox{12}$ centimeters.


inequalities
LaTeX
inequalities unsolved

please help me with the following problems.

1. for any convex quadrilateral ABCD, AC^2+BD^2<=AB^2+BC^2+CD^2+DA^2
(i have a proof of this in "Inequalities" by Radmila Bulajich Manfrino and two more writers but i want another proof.)

2. for positive reals a,b,c such that abc=1, ab+bc+ca>=a+b+c
(i proved this when i was little(!) but i cant find it, so...)

cant someone solve these easy problems? i've waited for so long and i cant wait any more :mad:

ok here's my solution:
(ab+bc+ca)/(a+b+c)>=[(a^b)(b^c)(c^a)]^[1/(a+b+c)] by weighted AM-GM. now if i assume that a<b<c,
rhs>=[(a^a)(b^a)(c^a)]^[1/(a+b+c)]=1.
my question is, is it safe to assume a<b<c? oh and can someone please show me how to post message using latex?

[quote="tim1234133"]Three hours is hardly too long to wait...

To post in LaTeX, write pretty much how you would do but put dollar signs around the maths.

For example: [code]$ab+bc+ca=a+b+c$[/code]
gives $ab+bc+ca=a+b+c$.

To learn some of the commands (like a^b would be $a^{b}$) hover you cursor over LaTeX that someone has written. e.g.

$\frac{ab+bc+ca}{a+b+c}\geq abc^{\frac{1}{a+b+c}}$ for example.

In the second question, you can divide through by $1=abc$ to get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq a+b+c$ and if we were to change the variables to $a_{1}= \frac{1}{a}, b_{1}=\frac{1}{b}, c_{1}=\frac{1}{c}$ we would again have $a_{1}b_{1}c_{1}=1$, and so if the inequality is true, then we should have equality specifically.  :maybe:

It is ok to assume that $a \leq b \leq c$ but not $a<b<c$.[/quote]

write where?? please explain where i should go first.

[quote="tim1234133"][quote]write where?? please explain where i should go first.[/quote]

Are you talking about writing Latex? Just write it in the body of your post, as you type. There is no need to go anywhere, the forum software deals with it.[/quote]

oh thank you very much tim1234133. now i understand. i'm rellllly greatful to you for your help.

[quote="cefer"][quote="nayel"]now i understand the second problem but what about the first one? :huh:  which states "for any convex quadrilateral $ABCD$,
$AC^{2}+BD^{2}\le AB^{2}+BC^{2}+CD^{2}+DA^{2}$"[/quote]

Let $X$ and $Y$ be midpoints of $AC$ and $BD$,respectively.Then you can easily prove that $XY^{2}= AB^{2}+BC^{2}+CD^{2}+DA^{2}-AC^{2}-BD^{2}$.[/quote]

i think that there is $4XY^{2}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

We say that a subgroup of $S_n$ has the property T if it is abelian and for any $i,j\in 1,2,...,n$ there is a permutation $\sigma$ in this subgroup such that $\sigma(i)=j$. Prove that any subgroup with the property T is cyclic if and only if $n$ is squarefree.

Just one implication for now (if $n$ is square-free then blahblahblah :)):

Let $\sigma$ be any permutation of a $T$-group $G\le S_n$ (by "a $T$-group" I mean a group with the property $T$), which can be written as the product $c_1\ldots c_k$ of disjoint cycles. Since for any other $\pi\in G$ we have $\pi\sigma\pi^{-1}=\sigma$, and for every $i,j$ there is a $\pi$ taking an element in $c_i$ to an element in $c_j$, it means that all the $c_i$'s have the same length $\frac nk$. This means that every element of $G$ has as its order a divisor of $n$, namely $\frac nk$.

There is a homomorphism $\varphi$ from $G$ to $S_k$, sending $g\in G$ to the permutation of $c_1,\ldots,c_k$ that $g$ induces on $\sigma$ by conjugation. If $k\ge 2$, then this homomorphism is non-trivial, and $\mbox{Im}\varphi$ is a non-trivial $T$-subgroup of $S_k$. This means that there is an element $\bar\sigma'$ of $\mbox{Im}\varphi$ having as its order a non-trivial divisor of $k$ (from the result at the end of the paragraph above), which means that an element $\sigma'\in G$ in the inverse image of $\bar\sigma'$ has an order which is not coprime with $k$. We thus have $\sigma\in G$ with order $\frac nk$, and $\sigma'\in G$, with an order which is not coprime with $k$.

If $n$ is square-free, we have $\left(\frac nk,k\right)=1$, so the element $\sigma\sigma'\in G$ has an order strictly larger than $\frac nk$ (when $k\ge 2$). This means that the maximal order of an element in $G$ is, in fact, $n$, and this occurs precisely when the element has $k=1$ cycles, i.e. when it's an $n$-cycle.

We have shown that if $n$ is square-free, every $T$-subgroup $G$ of $S_n$ must have an $n$-cycle, and from here it's easy, since the commutator of an $n$-cycle in $S_n$ is precisely the group generated by it.

The other implication isn't hard either, but I wanted to find a nice, short construction :).

Assume $n$ is not square-free and take some divisor $k$ of $n$ s.t. $\left(k,\frac nk\right)\ne 1$. Write the elements $\{1,2,\ldots,n\}$ as a $k\times \frac nk$ table, and let $\pi_1,\pi_2$ be the permutations in $S_n$ having the rows, and, respectively, the columns of the table as their disjoint cycles. I'm sure it's not hard to see that the group generated by $\pi_1,\pi_2$ is a non-cyclic subgroup of $S_n$ with property $T$.

Simply if $ n $ is not square-free we can take some natural $ a,b > 1 $, such
that $ ab=n $ and $ gcd(a,b) > 1 $. Then consider a group $ G =
\mathbb{Z}_{a} \oplus \mathbb{Z}_{b} $, which is not cyclic and
action of $ G $ on itself.

I hope it is correct.

I really think it's the same thing, man :).


algebra
polynomial
Vieta
algebra proposed

The product of two of the four zeros of the quadric equation

$ x^4\minus{}18x^3\plus{}kx^2\plus{}200x\minus{}1984\equal{}0$

is $ \minus{}32$.Find $ k$

Just use Vieta's theorem and we get $ k\equal{}86$:
suppose $ x_1,..,x_4$ ae the roots and we have $ x_1x_2\equal{}\minus{}32$ then;
$ p\plus{}q\equal{}18$
$ 62\plus{}pq\minus{}32\equal{}k$
$ \minus{}62p\plus{}32q\equal{}200$
where $ q\equal{}x_3\plus{}x_4$ and $ p\equal{}x_1\plus{}x_2$.


Hi!!

I'm creating Big Brother!!!

[i](scattered applause)[/i]

[size=9]I'm sorry for bringing down Amazing Race...[/size]

1.
2.
3.
4.
5.
and so on and so forth... (maybe 10-15)

There are challenges... THIS WAS DEFINITELY NOT COPIED

Head of Household (HoH): This position is the most powerful in the house. It takes place at the start of every week, and everyone competes except for the previous week's HoH. The HoH has the responsibility to nominate two Houseguests for eviction, the worst place to be in the house. If you are still nominated (see Power of Veto) by the Eviction vote, the rest of the house will vote, and potentially send you home.

Power of Veto (PoV): This competition takes place after the HoH has made nominations. There are (at the beginning of the game) 6 competitors: The HoH, the two nominees, and three HGs selected by random draw. The winner of this competition has the ability to Veto one of the HoH's nominees. They do not have to, however. If a nominee is Vetoed, then the HoH is required to put up a replacement nominee.
There are also several ground rules, which must be followed at all times by everyone:

1. Communication, both inside the thread and out, is allowed at all times, except the following:
a. HoH Competitions
b. PoV Competitions, unless otherwise specified
c. Eviction Votes
d. Any other time I tell you so
2. I expect that most of the game should take place outside this thread: I also require that any discussion groups be made known to me.
3. You are NOT allowed to discuss the game with anyone not in the game, unless you are told otherwise.
4. Once you are evicted, you will be sent special instructions about your future role in the game. As a rule, though, don't talk to anyone still in the game once evicted.
5. Use your good judgment: don't do anything you think I would disallow.
6. This is a game: have fun, and make it fun for others. No flaming, etc.

If you ever have any questions about what you are and aren't allowed to do, send me a pm, or ask in-thread if you don't care that others know. I will never reveal anything someone pms me that isn't a competition.

AND I'm bringing back the Coup d'Etat with a little twist... :lol: SO JOIN!!

[size=200]IF YOU ARE GOING TO JOIN, MAKE SURE YOU ARE ACTIVE FOR THE NEXT MONTH OR TWO OR THREE OR...[/size]

okay i will join! i dropped outof amzing race cuz i didnt liek the idea very much  :blush:

1. Maybach
2. Nachosaurus
3. EggyLv.999
4. FlyAgaric
5. spaceguy524
6.
7.
and so on and so forth...

HAHA this is so funny most of these peeps just exited the old big brother

/in

I don't watch the show, but this looks interesting.

1. Maybach
2. Nachosaurus
3. EggyLv.999
4. FlyAgaric
5. spaceguy524
6. westiepaw
7.
8.
9.
10.
11.
12??
13??
14??
15??

I'll join.
MWAHAHHAHAHAheheheheheuheuheuheuahemahemaheamHUUURRAUUGH.
Gagging just after laughing is not fun. :(

1. Maybach
2. Nachosaurus
3. EggyLv.999
4. FlyAgaric
5. spaceguy524
6. westiepaw
7.[hide=" "]If you are reading this, you are granted an extra clue to the Coup d'Etat. The Coup d'Etat will give you a special power in this game. Keep looking for clues in messages such as this one. Good luck! [/hide]meewhee009
8. 1=2
9.
10.
11.
12??
13??
14??
15??

Each of you must PM me a paragraph (6+ sentences)[hide=" "]Good! You're starting to notice things! [/hide]about how you will go about winning the game and your personality.
[b]
WRITE IN COMPLETE SENTENCES.

PM this ASAP [/b]

For people who have joined early, there's a helpful thing!

[b]Coup d’Etat [/b]

Yep, the Coup d’Etat is back!!

The Coup d’état is a power only available to one person in the entire game. That is, this is the only time it will be offered, and only one[hide=" "]or not...[/hide]of you can win it. If won, you will acquire a great power in the house, but you won’t know what it is until you win. Also, no one else will know who won, or its power, until it is used. I will tell you that it could completely turn your game around...The way to win it is to submit to me the correct answer to a puzzle. Occasionally, I will post a new clue. You get only one guess the entire game, so answer wisely. You may guess it at ANY time, including during other competitions, and during voting periods. I will continue posting clues even after it is won. If you answer, I will tell you either you won or you didn’t: not if someone else already won, or the validity of your answer. PM me your answers. Good luck!

And our winner is[hide="..."]

I don't know yet. FlyAgaric has not PMed me yet. Whatever...

There was a 2-2 tie.

So, now, the winner is[hide="..."]

[b]westiepaw[/b] and [b]meewhee009[/b]

Thank you all for another successful season and a successful first game I have modded.[/hide][/hide]

A [b]TIE?![/b] This is completely unacceptable! I declare a mis-vote! Another vote or another test shall be scheduled!

Nope. Wont do. Prod FlyAgaric. We cant leave big brother with a tie

[hide="WE HAVE A WINNER"]
[hide="westiepaw"]
[hide="you"]

not won. Congratulations [b]meewhee009[/b]!!!!!!!!!!!

Nice season.[/hide][/hide][/hide]

:(
Oh.
My.
Gosh.
I would like to thank my family,
all of AoPS,
all of the contestants on Big Brother,
Andrewy,
my imaginary dog,
and ice cream. If it wasn't for ice cream I wouldn't have been here right now.

[b]1=2[/b] was in fact in Season 2, but was evicted first.

I thank everyone again for a great season.

[quote="1=2"]Wait, I was???
I don't remember that.
I must have been really unpopular back then.[/quote]

Wow...

You forgot...

 :rotfl:

Also... [size=200]VISIT MY BLOG!!!
[/size]
Siggy...

[quote="piegraph"]is this game over?
[size=200]in 14 pages?![/size][/quote]

yeah...

piegraph, that post made me think that  you just wanted to raise your post count...

and this isn't one of your RPG games.


This forum is great and I want to know if there are other math forums similar to this one.

Also, I'm looking for any Biology, Physics, and Chemistry related forums.

If you know any, can you give me some suggestions to which one to go to?

Yeah I've been wondering if there were any sites like this one also.  I have found this one [url]http://www.scienceforums.net/forums/[/url]  It has science and math mostly science though.  It looks really good.  The only problem is that a lot of the stuff is a bit on the advanced side for me (at least what Ive looked at).  I'll keep looking at it because I'm sure I can learn a lot from it just like I have here.

yea I just joined that forum...I'm lost on it. very lost.


HMMT Harvard college MIT algebra polynomial AMC

Source: Harvard-MIT Math Tournament - Algebra 2003 #10

Suppose that P(x) is a polynomial such that P(1) = 1 and

P(2x)/P(x+1) = 8 - 56/(x+7)

for all real x for which both sides are defined (i.e. P(x+1) ≠ 0 and(x+7) ≠ 0).  Find P(-1)

I would be willing to wager that in the initial problem, the given condition was supposed to be P(0) = 1, not P(1) = 1.  Try the problem that way.

But then, the problem doesn't work, cuz it becomes 1/P(1) = 0, so it's not a polynomial.

Whoops, good call.  Hmm.

Okay, well, lets play around with it anyways.
P(2x)/P(x+1) = 8 - 56/(x+7) = 8x/(x+7)
8x*P(x+1) = (x+7)*P(2x).  Now, let P(x) = a*x^n + . . . + c.  We can see by checking the coefficient of x^(n+1) in that equation that P is a cubic polynomial.  So P(x) = ax^3 + bx^2 + cx + d.

8x*P(x) = 8ax^4 + 8bx^3 + 8cx^2 + 8dx
(x+7)*P(2x) = (x+7)*(8ax^3 + 4bx^2 + 2cx + d) =
 = 8ax^4 + (4b+56a)x^3 + (2c + 28b)x^2 + (d + 14c)x + 7d.

These two expressions are equal for all x (except possibly -7 and the roots of P(x+1), but those won't matter).  Therefore they have equal coefficients.  Thus:
7d = 0
8d = 14c + d
8c = 2c + 28b
8b = 56a + 4b
8a = 8a

This seems problematic to me.  Someone else catch an error on my part?  Hmm.

According to the document I have, the problem states P(1) = 1.

But I still don't think any such polynomial exists.  Can you verify my work, Stephen?

I think you changed a P(x+1) into a P(x) JBL.
But doing what you did would give you the solution as ax^3 + 8ax^2 + 12ax, or ax(x+2)(x+6). Then you'd substitute in the given value and work out a, and then find P(1).

I'd like to know the actual question though.. sometimes there is an easier way to attack a problem.

I found the question, and it was a bit different indeed.

Suppose P(x) is a polynomial with P(1)=1 and

P(2x)/P(x+1)=8-56/(x+7)

for all real x for which both sides are defined. Find P(-1).


conics
ellipse
trigonometry
ratio
geometry
geometric transformation
homothety

Let $Z$ be a circle with centre $O$. Let $A$ and $B$ be two points on this circle $Z$ such that $AB$ is a diameter of $Z$. Select a point $C$ on the segment $AB$. We construct another circle, say $Y$, with diameter $AC$. Let $D$ be a point on the circumference of $Z$ such that $DCB$ is an isosceles triangle with $CD=BD$. Draw a circle $X$ (inside $Z$) which is tangent to $Z$, to $Y$ and to the segment $CD$. Let $E$ be the centre of this circle $X$. Show that $EC$ is perpendicular to $AB$.

We can look at it backwards: let's show that a circle tangent to the two circles passing through $A$ and with its center $E$ satisfying $EC\perp AB$ must also be tangent to $DC$.

Let $F$ be the projection of $D$ on $AB$. If $r$ is the length of the radius of the circle $x$, then we need to show $\frac{EC^2}{r^2}=\frac{AD^2}{DF^2}$.

$E$ lies on the ellipse having $O$ and $O'$ (the center of $y$) as foci and having $AF$ as its large axis. Call $a,c$ its half of the large axis and distance from the center to a focus, respectively (usual notations). The computation is now fairly simple. If it's unclear, I'll do it here.

I have quite long solution.

Consider inversion with center at p. $C$. circle $Z$ becomes new circle $Z$, circle $Y$ becomes line $Y$ and $Y\perp AB$. Since $X$ tangents $Z$, $Y$ and $CD$ before inversion we conclude new circle $X$ tangents lines $EA$ and $CD$ and circle $Z$. It implies that $F$ lies on bisector of $AED$.
Let $h$ be perpendicular to $AB$ at point $C$. We need to show that $F$ lies on $h$.
We will prove [b]equivalent[/b] statement: Let $F$ be point of intersection of bisector $AED$ and $h$, then distances from $F$ to $AE$, $ED$ and $Z$ are equal, i.e. $FG=FK$ and $O'F=R(Z)+FG$ [b](0)[/b], where $R(Z)$ is a radius of new $Z$. (It means that circle this center $F$ and radius $FG$ tangents lines $AE$, $EC$ and circle $Z$)

First of all, we state that $CB=DB$ [b](1)[/b].
Indeed, before inversion we had $DC=DB$, so after inversion new length $BC$ is $\frac{R^2}{BC}$ and new length $DB$ is $\frac{R^2\cdot DB}{CD\cdot CB}=\frac{R^2}{BC}$, where $R$ is the radius of inversion.

Thus we have $\angle CBD=\angle DCB$ $\Rightarrow$ $\angle ACE=\angle ACH=\angle HAC$ and triangle $HAC$ is isosceles. Since $AEC$ is right triangle we obtain that $HC=EC/2$ [b](2)[/b].

Since $EF$ is bisector of $AEC$ and $AE||CF$ we conclude $\angle CEF=\angle AEF=\angle EFC$ $\Rightarrow$ triangle $EFC$ is isosceles and $CF=EC$ [b](3)[/b].

Since $EF$ is bisector and $AE||CF$ we obtain $GF=GK=AC$ [b](4)[/b].

Suppose WLOG radius $Z$ is 1.

Let $\angle CBD=x$. Then from right triangle $ABD$ we see that $BD=2\cos x$. From [b](1)[/b] we conclude $BC=2\cos x$. It follows that $DC=4\cos x\sin\frac{x}{2}$. Since triangles $HAC$ and $CBD$ are similar we have $HC=BC\cdot \frac{AC}{DC}=2\cos x\cdot\frac{2-2\cos x}{4\cos x\sin\frac{x}{2}}$. So from [b](2)[/b] and [b](3)[/b] we obtain $FC=4\sin\frac{x}{2}$ (we use $1-\cos x=2\sin^2\frac{x}{2}$).

Last step. Using Piphagorean theorem for $O'CF$ we obtain $O'F^2=O'C^2+FC^2$. Accordingly to [b](0)[/b] (and using [b](4)[/b]) we need to show that
\[O'F^2=(1-2\cos x)^2+16\sin^2\frac{x}{2}\mbox{ and }(1+FK)^2=(1+(2-2\cos x))^2\mbox{ are equal.}\]
It is very simple trigonometry exercise.

grobber, sorry if I am super slow, but could you explain more of your solution? (i.e. Why does that ratio do the job?) And I don't really see the computations easy...

Thanks in advance.  :P

Don't worry. I expected this, and it's not because you would be "super slow" or anything, but because my answer realy was cryptic :).

If we let $E'$ be the projection of $E$ on $DC$, we need to show that $E'$ lies on $x$, or, in other words, that $EE'=r$. This means that we must show that a right triangle having $EC$ as a hypothenuse and $r$ as one of its legs is similar with the right triangle $ADF$, with $r$ corresponding to $DF$ (thus the ratios; I put squares because it's easier to compute $r^2,EC^2$, etc.).

The computations aren't easy either, I exaggerated when I said that :). $FD,AD$ are, however, easy to compute in terms of $a,c$: $DF^2=FA\cdot FB=FA\cdot FC=2a\cdot 2c$ and $AD^2=AF^2+DF^2=(2a)^2+2a\cdot 2c$, so $\frac{AD^2}{DF^2}=\frac{a+c}c\ (*)$.

$EC^2$ is simply computed from the equation of the ellipse: $\frac{EC^2}{a^2-c^2}+\frac{O''C^2}{a^2}=1$, where $O''$ is the center of the ellipse, i.e. the midpoint of $OO'$. We use the fact that $O''C=|2c-a|$.

Consider now the inversion of pole $A$ which sends $C$ to $B$. The circles $y,z$ are turned into lines through $B,C$ respectively and perpendicular to $AB$, and $x$ is turned into a circle tangent to these two parallel lines, which can also be obtained from $x$ through a homothety of center $A$. If $r'$ is the radius of the image of $x$, then $r'=\frac{BC}2=2a-2c$ and $\frac{r'}r=\frac{AF}{AC}=\frac{2a}{2c}$. We thus get $r=c(1-\frac ca)\ (**)$.

We can now use $(*),(**)$ to show the equality of those two ratios mentioned in my first post.

Now I think my solution is not so long... :)

Thanks grobber, your solution is indeed valid. (Sorry for the delay)  :)

Dear Mathlinkers,
finally, you can see a completely synthetical proof of this nice san Gaku on my site :
http://perso.orange.fr/jl.ayme    vol. 4  La fameuse san Gaku de lma prefecture de Gumma (1803)
Sincerely
Jean-Louis


function
quadratics
algebra
domain
calculus
calculus computations

Hello,
I am working on a project for functional iterations. I study the function $ y\equal{}\frac{x\plus{}10}{x\plus{}1}$. I know from the cob-web plot that the value $ x\equal{}sqrt(10)$ is a fixed point. My question is, how do i prove mathematically that this function has a UNIQUE fixed point and how do i know if this point is attracting or repeling?
Maybe all this sound very elementary, but i haven't done something simar until now. Thanks for any help.

The equation $ x \equal{} \frac {x \plus{} 10}{x \plus{} 1}$ can be rearranged to a quadratic equation, and that quadratic equation has [i]two[/i] roots: $ \pm\sqrt {10}.$

A fixed point of $ x \equal{} f(x)$ is (at least locally) attracting if $ |f'(x)| < 1$ at that point.

In this case $ f'(x) \equal{} \frac { \minus{} 9}{(x \plus{} 1)^2}.$ Evaluating that at both $ \sqrt {10}$ and $ \minus{} \sqrt {10}$ shows that the fixed point at $ \sqrt {10}$ is attracting and the fixed point at $ \minus{} \sqrt {10}$ is repelling.

The only way this would have a unique fixed point would be if you started by restricting the domain.

Now: suppose we restrict ourselves to $ [0,\infty).$ We can show that $ f$ maps this domain into itself; we can show by the intermediate value theorem that there is at least one fixed point (or explicitly solve for that point), and the very fact that $ f$ is a decreasing function on $ [0,\infty)$ means there cannot be more than one fixed point.

There is a sequence of points $ \{\minus{}1,\minus{}\tfrac{11}2,\minus{}\tfrac{31}9,\dots\}$ such that if you start the iteration at any of those points, some member of the sequence will become undefined. Otherwise, the sequence should converge to $ \sqrt{10}.$

I assume then that also the condition that you state there is also the other way ? (for the point being attractive)

If $ |f'(x)|>1,$ the point is repelling. If $ |f'(x)|\equal{}1,$ we need a more delicate analysis.

I don't understant why $ \left[ 0,\infty \right)$ is mapped to itself..I think that when $ x \geq 0$ then $ f(x) \in \left( 0,10 \right)$. Where is my mistake?
Thanks

Into, not onto. We only require that $ f([0,\infty))\subseteq[0,\infty)$, not that it equal $ [0,\infty).$

Thanks for all your answers. There is something final i would like to ask. The definition of a repelling point is that it is a fixed point p, for which the sequence of iterations diverge when starting from any value sufficiently close, but not equal to p.
When i evaluate this seqence in Matlab for different starting values, even if i start from a value near $ \minus{}\sqrt{10}$, i still see that the series converge to $ \sqrt{10}$. Shouldn't the members of the sequence diverge, by the definition,since $ \minus{}\sqrt{10}$ is a repelling point?
Thanks

Saying that $ \minus{}\sqrt{10}$ is repelling simply means that sequences won't converge to $ \minus{}\sqrt{10}.$ Any sequence that starts out close to that point will, sooner or later, cease to be close to $ \minus{}\sqrt{10}.$ That doesn't tell you that such a sequence will diverge - it just tells you that the repelling point will eventually become irrelevant to the convergence or divergence of that sequence. The sequence will just go somewhere else, and perhaps be captured by the influence of a different fixed point.


There's a marathon in all of the other forums, so I thought I'd bring it here:

[size=150]1. Answer a question
2. Post a question about music or art
 3.:play_ball:  :lol: [/size]

I'll start:

Who's your favorite composer?

Yay, a music and art marathon!! :lol:

My favorite composer's Franz Liszt.

What instrument do you play and how long have you been playing it?

[quote="236factorial"]Piano, 4 years

What is your favorite piece (what a typical question)?[/quote]

Beethoven's Moonlight. First movement and Third movement are completely different, and they both RULE  :)  :lol:

what's your favorite music period?

Romantic era.

What's your least favorite musical time period?

Baroque period

Do you like quicker songs or slower songs?

[quote="236factorial"]Baroque period[/quote]
Haha, me too. :D

I like quicker songs - they're more fun.

What's the hardest piece you've ever played?

moonlight sonata-not only was it hard to get, it was hard for my ears to listen to it. :roll:

favorite artist?

Da Vinci!!!

Sorry for double posting, but this topic's gotta keep going.

$Hendrix$

compare and constrast soloing versus written instrumentals

I'll assume that by "soloing," you mean improvising.

Improvisation can be really expressive if done well (I'm listening to Miles Davis right now :D), but so are most well-written pieces.  I would say that Improvisation demonstrates a greater knowledge and intuition involving harmonic structure and relative sound, as well as a spontaneous creativity that verbatim rehearsal can't emulate, but I think the ideas of written pieces are sometimes clearer and more mature than those embodied by improv.

Your favorite impressionist?  (in music or art, or both, if you please)[/b]

Debussy (I love Golliwog's cakewalk).

Do you think that piano is harder or easier than other instruments?

I guess it is easier.

My question : where exactly on mathlinks is it appropriate to discuss movies?

Movies you just saw is a topic at games and fun

but I'd say : movies are a very popular form of art?

Movies are a form of art. By previous statement of the board owners, discussions along the lines of "What is your favorite band?" belong on fun and games, and discussions of art as art forms (and film is an art form) belong here.

Can everyone sing?

Thank you.
However experience leads to me thinking that here most music people reside here : piano players, classical music fans

I must say I know nearly nothing about that.

Most people discuss films in fun and games, like harry potter, star wars etc...

So when I would want to ask a specific question about movies, I would have to do it here, where most movie experts won't notice my question?

Or am i seeing things incorrectly?


Let $x,y,z \in R$ so that $x+y+z=3$. Prove that $ x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2} \geq 3xyz$.

made by Ghenghiu R.

cheers! :D :D :cool:

We want to prove $x^2y^2+y^2z^2+z^2x^2 \geq xyz(x+y+z)$. This follows directly from Muirhead's theorem, as $\sum_\text{sym} x^2y^2 \geq \sum_\text{sym} |x^2yz| \geq \sum_\text{sym} x^2yz$.

Somehow I wonder why this is posted into the "Unsolved Problems" section while the proof is more or less banal. If we "de-homogenize" the inequality, we get $x^2 y^2 + y^2 z^2 + z^2 x^2 \geq xyz \left(x+y+z\right)$ for three arbitrary real numbers x, y, z (the condition x + y + z = 3 is not necessary anymore). Now, this inequality is trivial if we use Muirhead; if we don't use Muirhead, it is not difficult either:

$x^2 y^2 + y^2 z^2 + z^2 x^2$
$= \frac12 \left(x^2 y^2 + x^2 z^2 + y^2 z^2 + y^2 x^2 + z^2 x^2 + z^2 y^2\right)$
$= \frac12 \left(x^2 \left(y^2 + z^2\right) + y^2 \left(z^2 + x^2\right) + z^2 \left(x^2 + y^2\right)\right)$
$\geq \frac12 \left(x^2 \cdot 2yz + y^2 \cdot 2zx + z^2 \cdot 2xy\right)$
$= x^2 yz + y^2 zx + z^2 xy = xyz \left(x+y+z\right)$.

Darij

It's $a^2+b^2+c^2 \geq ab+bc+ca$ again!!! where $a=xy, b=yz, c=zx$.

 :D  :D  :D

Well,in fact we can take the condition $ a^2+b^2+c^2=3 $ to make it more stronger,however it is still too trivial. :)


number theory
greatest common divisor
least common multiple
combinatorics proposed
combinatorics
games
Combinatorial games

Two players play a game as follows: there $n > 1$ rounds and $d \geq 1$ is fixed. In the first round A picks a positive integer $m_1$, then B picks a positive integer $n_1 \not = m_1$. In round $k$ (for $k = 2, \ldots , n$), A picks an integer $m_k$ such that $m_{k-1} < m_k \leq m_{k-1} + d$. Then B picks an integer $n_k$ such that $n_{k-1} < n_k \leq n_{k-1} + d$. A gets $\gcd(m_k,n_{k-1})$ points and B gets $\gcd(m_k,n_k)$ points. After $n$ rounds, A wins if he has at least as many points as B, otherwise he loses.

For each $(n, d)$ which player has a winning strategy?

B always has a winning strategy.
Let $L$ be the least common multiple of $m_1+1 , m_1+2 , ... , m_1 + (n-1)d$, which are preciesly all numbers such that A can possibly choose during the game. Firstly, let B choose $n_1 = L+m_1$.
Each turn, if A chooses $m_{k-1}+c$, for some $1\leq c\leq d$, then let B choose $n_{k-1}+c$.
Then, we always have $n_k = m_k + L$. But since $m_k|L$, we have $\gcd (m_k , n_k )=m_k$. Therefore, B earns $m_2 + ... + m_n$ points at the end.
If A wants to win this game, we must have $\gcd (m_2 , n_1)+\ldots+\gcd (m_n, n_{n-1})\geq m_2 + ...+m_n$,
so we must have $\gcd (m_k , n_{k-1}) = m_k$, or equivalently, $m_k | n_{k-1}$. In particular, we must have $m_2 | n_1$, which is impossible, since $m_2 | L$, but $m_2 \not| m_1$


function
trigonometry
calculus
derivative
complex analysis

(1) Find a power series of $ z \plus{} 2i$ whose sum is the function $ \frac {1}{1 \minus{} z}$, and calculate its radius of convergence.

(2) (a) Find the power series about 0 for the function $ f(z) \equal{} e^{z^3}$, and give its radius of convergence.

(b) using the series obtained in part (a), compute $ f^{(20)}(0)$ and $ f^{(21)}(0)$.

(3) Find the Maclaurin series for each of the following functions, together with the respective radius of convergence:

(a) $ \cos^3z$

(b) $ (z^2\plus{}1) \sin z$

(c) $ e^z \cos z$

(d) $ \sin(2z)$

(e) $ \cos(z^3)$

(f) $ z \sinh(z^2)$

(g) $ \frac{1}{(2\minus{}z)^3}$

1. Let $ w \equal{} z \plus{} 2i$ so the function may be expressed as $ \frac{1}{1 \plus{} 2i \minus{} w} \equal{} \frac{1 \minus{} 2i}{5} \frac{1}{1 \minus{} \frac{1 \minus{} 2i}{5}w} \equal{} \frac{1 \minus{} 2i}{5} \Big(1 \plus{} \frac{1 \minus{} 2i}{5}w \plus{} \Big(\frac{1 \minus{} 2i}{5}\Big)^2w^2 \plus{} \cdots \Big)$ $ \equal{}$ $ \sum_{n \equal{} 0}^{\infty} \Big(\frac{1 \minus{} 2i}{5}\Big)^{n\plus{}1} w^n$. The radius of convergence is the distance from the center, $ \minus{}2i$, to the nearest singularity or branch point, so $ \sqrt{5}$.

2. Substitute $ z^3$ into the power series for $ \exp z$; it will converge everywhere. The $ n$th derivatives of $ f$ evaluated at $ 0$ will vanish for $ n$ not a multiple of $ 3$ and for $ n \equal{} 3k$ will be something like $ \frac{(3k)!}{k!}$.


i recall there being 63 users on for the max

Yes, that was after 21 November. Unfortunately we lost much of the data from 21 November after in the attack, so some posts and data was lost. See the global annoucements.


calculus
real analysis
real analysis theorems

I just got Spivak's book a couple days ago, I've flipped through it and was looking through the Index. Does Spivak cover Related Rates or Optimization? I did not find it at all, so I am curious and wondering if it's under some other name. Thanks!

[quote="kunny"]See  P.204[/quote]Aw thank you, now that mystery has been solved.


I saw a solution posted in another thread, but I wanted to ask why this solution did not work for me:

[hide]You can set the sides equal to $a^{2}-b^{2}$, $2ab$, and $a^{2}+b^{2}$.

Therefore, $(a-b)(a+b)(ab) = 3[2a(a+b)]$

This cancels to $(a-b)(b)=6$, giving you [size=125][b]4[/b] [/size]answers. [/hide]

The Pythagorean triple generating formulae do not generate all triples.  They only generate all [i]primitive[/i] triples.  For example, $(9, 12, 15)$ cannot be generated.

The [b]full[/b] Pythagorean triple generating formulae are $c(a^{2}-b^{2}), 2abc, c(a^{2}+b^{2})$ where $c$ cannot be represented as a sum of two squares.


For any positive integer $ n$, let $ N_n$ be the set of integers from 1 to $ n$, i.e. $ N_n \equal{} \{1, 2, 3, \cdots, n\}$. Now assume that $ n \ge 10$. Determine the maximum value of $ n$ such that the following inequality
\[ \text{min}_{\substack{
a, b \in A \\
a \neq b
}} |a \minus{} b| \leq 10
\] holds for each $ A \subset N_n$ with $ |A| \ge 10$.

[hide="Solution"]
Assume $ n >\equal{} 100$ Clearly we have a contradiction since there will always be the following set which will not hold

\[ \{ 1, 12, 23, 34, 45, 56, 67, 78, 89, 100 \} \]

When $ n \equal{} 99$

Consider the $ 9$ pigeon holes

$ [1,11], [12, 22], [23, 33], \ldots, [89, 99]$

If $ |A| >\equal{} 10$ clearly one hole must contain more than one number, and their difference will be less than 10

Therefore $ \boxed{99}$ suffices[/hide]

But mustn't the condition hold for all subsets of $ N_n$? In that case the answer would be 11 right  :P  Or am I misunderstanding the problem

This means that the number of elements in $ A$ is greater than or equal to 10 right?

Yes, so only those sets with at least 10 elements have to satisfy the given inequality.

Shucks I misunderstood... I get it now thanks  :)


topology
advanced fields
advanced fields unsolved

Hello,
Here is a question, which is probably not so hard, but I'm not sure how answer:

Let $ H_1, H_2$ be Hilbert spaces and $ A_1 \subset H_1, A_2 \subset H_2$ their subsets.
Suppose there is a transformation $ T: A_1 \to A_2$ such that for all $ x_1, x_2 \in A_1$,
$ < Tx_1, Tx_2 > \equal{} < x_1, x_2 >$. (The corresponding inner/scalar product in H_2 and H_1.)
Then there exists a unitary extension of T (to all H_1). (It is proved.)
The question is:
In case $ \bar{span{A_1}} \equal{} H_1$ (that is [b]closure[/b] of span of A_1 equals to H_1),
prove that this extension is unique.

How do we know it's unique?

Thanks!

[hide="Hint"]If there are two candidates $ T_1$ and $ T_2$, then define $ A'\equal{}\{x\in H_1|T_1(x)\equal{}T_2(x)\}$. Now show that the closed set $ A'$ is dense in $ H_1$.[/hide]


Ενα ενδιαφερον ολοκληρωμα: :wink:

[img]http://glyk.kwikphp.com/forum/img/Maths/mimetex1.gif[/img]

Την λύση θα την γράψω στο φόρουμ μου(ένα φόρουμ για μαθητές τρίτης λυκείου) [url]http://glyk.kwikphp.com/forum/viewtopic.php?id=19[/url]

Mailo..na se rotiso kati..ti rolo pezi pu to iporizo tu deyteru olokliromatos ine grameno me aytai tin morfi?afu grafete os..$Inx$ :huh: .....Meta..arkei na 8esis....
$x=e^{u^{2}}$
+telioses.... :wink: (an dn exo kani  kana la8os...)

 :D

Mallon mperdeuei tous mathites etsi kai isws gi'auto na egine i metatropi auti. Ki egw etsi tin antimetwpisa RAKAR kai esteila ti lisi sto parapanw forum tou kainouriou filou mas. Meta eriksa mia matia oti to en logw oloklirwma (meta pou tha to ftiakseis) einai mia askisi tou diagwnismou tou ASEP tou 2002 (akrivws idia). Nomizw arketa einai arketa vati gia kapoion pou gnwrizei oti apo mono tou to oloklirwma $\int e^{x^{2}}dx$ den iparxei (se analitiki morfi), opote anagastika tha prepei na xrisimopoiisei me kapoio tropo kai to 2o oloklirwma mplekontas ta metaksi tous.

Alexandros

Cretanman,ontos,dn ixa di tn lisi sto alo forum....(etsi akrivos tin elisa+ego)
Ontos to olokliroma ayto....(anaferome sto$\int e^{x^{2}}dx$ )ine arketa <diasimo>... +tis perisoteres fores i emfanisi tu+mono (logo tu oti dn exi analitiki morfi) ine ena kinitro gia to yaksimo mias idiomorfis lisis.... :wink:

 :D

[quote="cretanman"]Mallon mperdeuei tous mathites etsi kai isws gi'auto na egine i metatropi auti. Ki egw etsi tin antimetwpisa RAKAR kai esteila ti lisi sto parapanw forum tou kainouriou filou mas.
Alexandros[/quote]

Ναι την μετατροπή αυτή $\sqrt{lnx^{2}+ln\frac{1}{x}}$ την έκανα εγώ για να μπερδεύει τους μαθητές  :roll:

Πολύ καλή η λύση σας, αλλά υπάρχει και άλλος τρόπος, σχεδόν ίδιος, αρκεί κανείς να παρατηρήσει ότι οι συναρτήσεις $e^{x^{2}}$ και $\sqrt{lnx}$ είναι αντίστροφες.

ΥΓ. Αλεξαδρε σε ευχαριστώ που έγραψες την λύση στο φορουμ μου  :P

[quote="mailo"]Αλεξανδρε, μπορεις να εξηγησεις πως βγηκε αυτο:

$\left[xe^{x^{2}}\right]_{0}^{1}=1$[/quote]

Kata lathos. To diorthwsa idi. Fisika einai iso me $e$.

Alexandros

Αλέξανδρε, έκανα ανάλυση στη λύση που έγραψες για να είναι πιο κατανοητή. Επίσης έγραψα και άλλη μια λύση η οποία είναι ίδια με την δικιά σου άλλα με διαφορετικό τρόπο:

[url]http://glyk.kwikphp.com/forum/viewtopic.php?pid=39#p39[/url]


function
number theory proposed
number theory

I think that the solution of this require an explicit formula for the sumation. Please some body knows a formula?
Is know that C(0;0) + C(2;1)x + C(4;2)x^2 + ... = 1/(1-4x)^1/2  (generating functions)
But a formula for finitely many terms is unknow for me.


For every positive integer $ n,p$, prove that

$ (1\plus{}\frac{2006p^2}{1\plus{}2007p})(1\plus{}\frac{2006p^2}{2(2\plus{}2007p)})...(1\plus{}\frac{2006p^2}{n(n\plus{}2007p)}<\frac{(2007p)!}{p!(2006p)!}$.

Is it possible to replace $ \frac{(2007p)!}{p!(2006p)!}$ with some smaller number?


trigonometry
rotation
geometry
geometric transformation
reflection

I've got a couple problems from homework that I'm not sure how to solve. I've tried manipulating them some and doing substitutions, but math isn't one of my big strengths :(

If anyone could help solve these, it'd be greatly appreciated.

#1: Verify the identity (trig proof): cos(pi+x)=-cosx

#2: Write the product as a sum or difference: 4sinxcos3x

Thanks in advance! :roll:

[url=http://www.math123xyz.com]Interactive Multimedia Math Resources[/url]

For the first one use $ \cos(\alpha \plus{} \beta)\equal{}\cos \alpha \cos \beta \minus{}\sin \alpha \sin \beta$

For second one use $ \sin \alpha \cos \beta\equal{}\frac {1}{2} (\sin (\alpha \minus{} \beta)\plus{}\sin(\alpha \plus{}\beta))$

For #1, you can also note that since:

$ \cos \equal{} \minus{} \cos(\pi \minus{} x)$

So,

$ \cos(\pi \plus{} x)$
$ \equal{} \minus{} \cos(\pi \minus{} (\pi \plus{} x))$
$ \equal{} \minus{} \cos( \minus{} x)$
$ \equal{} \minus{} \cos(x)$

Carbon57, the identity you cite is equivalent to the identity desired.

#1 has a very intuitive interpretation in terms of the unit circle:  the point $ (x, y) \equal{} (r \cos \theta, r \sin \theta)$, when rotated $ 180^{\circ}$, becomes the point $ ( \minus{} x, \minus{} y) \equal{} (r \cos (\theta \plus{} \pi), r \sin (\theta \plus{} \pi))$.  Note that a rotation by $ 180^{\circ}$ is just a reflection through the origin.


A body of mass m was slowly hauled up the hill.(Fig. 1.29) by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction k.
I have trouble understanding the solution of this problem since it involves quite a bit of calculus. So, I would like to see a more elementary solution, if possible, for this problem. The reason I don't understand much about the solution from the book is because I don't have any background of calculus in Physics. I meant I don't know if integral sth turns out to be sth else except for rectilinear motion.
Here's a solution as well as the figure( all included in the following file):

I agree the solution is a bit confusing but I'll see if I can explain it to you:
It seems to me that the book is using $ A$ as its notation for work.  Work is defined as the dot product of force and distance: $ A\equal{}\vec{F}\cdot\vec{r}$, which is a scalar.  The book is taking the amount of work done at each infinitesimal point and then adding them together, which is what integration is doing.  So $ A\equal{}\int\vec{F}\cdot d\vec{r}$, which basically means that the total work is the sum of the work done at each point on the path.

You may want to learn some basic differentiation and integration before you continue with physics-you'll need it.

[quote="JRav"]I agree the solution is a bit confusing but I'll see if I can explain it to you:
It seems to me that the book is using $ A$ as its notation for work.  Work is defined as the dot product of force and distance: $ A \equal{} \vec{F}\cdot\vec{r}$, which is a scalar.  The book is taking the amount of work done at each infinitesimal point and then adding them together, which is what integration is doing.  So $ A \equal{} \int\vec{F}\cdot d\vec{r}$, which basically means that the total work is the sum of the work done at each point on the path.

You may want to learn some basic differentiation and integration before you continue with physics-you'll need it.[/quote]
I understand the last part in the book's solution. However, I can't quite get why $ dA\equal{}\vec{F}\cdot d\vec{r}$. Shouldn't it be the integral of the whole thing:$ d vec{F}.vec{r}$??
@JRav: I learned differentiation and integration already though, but I just learn them at the most basic level. However, I have no clue what does it mean in Physics and Irodov's book shows a lot of solutions like this one. Ca you show me some meanings of these integral and differentiation in Physics? Like in rectilinear motion, differentiation of $ s(t)\equal{}v(t)$ and integral of $ v(t)\equal{}s(t)$. I just learned the meaning of integral and differentiation for rectilinear motion and haven't learned any other meanings for other things(like trajectory, circular motion,...) yet :( .

It's still integration and differentiation like you learned.  That is just notation.  $ \int\vec{F}\cdot\,d\vec{r}$ simply means the integral of the product of force, and some infinitesimal distance.  So let's say we wanted the work done in moving an object in some gravitational field, of mass $ m$, gravitational constant $ g$, and distance $ h$.  You should already suspect that the answer is just $ mgh$ but here's how:
$ \vec{F}$ is the force we need to exert, which is just the weight, $ mg$.  The infinitesimal distance is $ dy$ in the vertical direction.  We are integrating this vertical distance from 0 to $ h$ so we have
$ W\equal{}\int\vec{F}\cdot\,d\vec{r}\equal{}\int_{0}^{h}mg\,dy\equal{}mg\bigg|_{0}^{h}\equal{}mgh\minus{}0\equal{}mgh$
Does that make any sense?

[quote="JRav"]It's still integration and differentiation like you learned.  That is just notation.  $ \int\vec{F}\cdot\,d\vec{r}$ simply means the integral of the product of force, and some infinitesimal distance.  So let's say we wanted the work done in moving an object in some gravitational field, of mass $ m$, gravitational constant $ g$, and distance $ h$.  You should already suspect that the answer is just $ mgh$ but here's how:
$ \vec{F}$ is the force we need to exert, which is just the weight, $ mg$.  The infinitesimal distance is $ dy$ in the vertical direction.  We are integrating this vertical distance from 0 to $ h$ so we have
$ W \equal{} \int\vec{F}\cdot\,d\vec{r} \equal{} \int_{0}^{h}mg\,dy \equal{} mg\bigg|_{0}^{h} \equal{} mgh \minus{} 0 \equal{} mgh$
Does that make any sense?[/quote]
Thank you very much JRav! Your explantation helps me a lot. :)  Can you show me more integral's application for other things such as the net accleration in circular motion? Irodov shows that it is equal to $ \sqrt{angular acceleration^2\plus{}linear acceleration^2}$, which I kind of assuming that he got it from vector component in the xy-plane when an object moves around the circle without friction.

Well yes, that's the magnitude of total acceleration.  But I wouldn't use an integral to determine "net acceleration".  When you get into E&M, there are a lot of applications of integrals.