geometry
circumcircle
trigonometry
geometric transformation
homothety
rotation

Let $\omega$ be the nine-point circle of a triangle $ABC$. Prove that there exist precisely three points $P$, $Q$ and $R$ on the circumcircle of triangle $ABC$ whose Simson lines with respect to the triangle $ABC$ are tangent to $\omega$. Prove also that the triangle $PQR$ formed by these three points is homothetic to the Morley triangle of the triangle $ABC$ (and the triangle with points of touching as the vertices is also homothetic to the Morley triangle).

WLOG, assume that the relative size of the angles $\alpha = \angle CAB$, $\beta = \angle ABC$, $\gamma = \angle BCA$ is $\beta < \alpha < \gamma$ (that's how I drew the triangle). Denote the Morley equilateral triangle $\triangle DEF$, where D is the intersection of the appropriate trisectors of the angles $\beta$,$\gamma$, E the intersection of the appropriate trisectors of the angles $\gamma$, $\alpha$ and F the intersection of the appropriate trisectors of the angles $\alpha$, $\beta$. The angles formed by 2 trisectors from the same vertex of the $\triangle ABC$ with the sides of the equilateral triangle $\triangle DEF$ are:

$\angle CDE = \frac{\beta + \pi}{3}$, $\angle CED = \frac{\alpha + \pi}{3}$

$\angle AEF = \frac{\gamma + \pi}{3}$, $\angle AFE = \frac{\beta + \pi}{3}$

$\angle BFD = \frac{\alpha + \pi}{3}$, $\angle BDF = \frac{\gamma + \pi}{3}$

An easy proof is to draw an arbitrary equilateral triangle $\triangle DEF$, take the angles $\alpha + \beta + \gamma = \pi$, build the above angles on the sides of this equilateral triangle and show, for example, that the angles $\angle FAB = \frac{\alpha}{3}$, $\angle FBA = \frac{\beta}{3}$. Using sine theorem for the triangles $\triangle AEF$, $\triangle BDE$,

$\frac{EF}{AF} = \frac {\sin \left ( \frac{\alpha}{3} \right )}{\sin \left ( \frac{\gamma + \pi}{3} \right )}$, $\frac{DE}{BF} = \frac {\sin \left ( \frac{\beta}{3} \right )}{\sin \left ( \frac{\gamma + \pi}{3} \right )}$, $\frac{AF}{BF} = \frac {\sin \left ( \frac{\beta}{3} \right )}{\sin \left ( \frac{\alpha}{3} \right )}$

Since in addition, $\angle AFB =  2\pi - \frac{\pi}{3} - \frac{\alpha + \pi}{3} - \frac{\beta + \pi}{3} = \pi - \frac{\alpha + \beta}{3}$, from the sine theorem for the triangle $\triangle ABF$, it follows that $\angle FAB = \frac{\alpha}{3}$, $\angle FBA = \frac{\beta}{3}$.

We are going to "construct" the Simson line perpendicular to the side EF of the Morley triangle $\triangle DEF$. Simple angle chase reveals the this direction forms with the side AB the angle

$\pi - \frac{\beta + \pi}{3} + \frac{\alpha}{3} - \frac{\pi}{2} = \frac{\pi}{6} + \frac{\alpha - \beta}{3}$

Let a line through the vertex A and forming this angle with the side AB intersect the circumcircle $(O)$ of the triangle $\triangle ABC$ at a point K and let a normal from the point K to the side BC intersect the circumcircle at another point L. The point L is the pole of the desired Simson line. The line KL perpendicular to the side BC forms the angle $\frac{\pi}{2} - \beta$ with the side AB and consequently, the angle $\angle AKL$ is equal to

$\angle AKL = \frac{\pi}{2} - \beta - \left ( \frac{\pi}{6} + \frac{\alpha - \beta}{3} \right ) = \frac{\pi - (2\beta + \alpha)}{3} = \frac{\gamma - \beta}{3}$

Continuing the angle chase around the circumcircle, the angle $\angle AKB = \angle ACB = \gamma$ (spans the same circumcircle arc AB) and

$\angle KBC = \frac{\pi}{2} - (\gamma - \angle AKL) =  \frac{\pi}{2} - \left ( \gamma - \frac{\gamma - \beta}{3} \right ) = \frac{\pi}{2} - \frac{2\gamma + \beta}{3}$

$\angle KBA = \beta + \angle KBC = \beta + \frac{\pi}{2} - \frac{2\gamma + \beta}{3} = \frac{\pi}{2} - \frac{2(\gamma - \beta)}{3}$

The angles $\angle KLA = \angle KBA$ are equal, because they span the same circumcircle arc KA, and the angle $\angle AOL = 2\angle AKL$ is the central angle of the circumcircle arc AL. Hence

$\angle KLO = \angle OLA - \angle KLA = \frac{\pi - \angle AOL}{2} - \angle KLA =  \frac{\pi}{2} - \angle AKL - \angle KBA =$

$= \frac{\pi}{2} - \frac{\gamma - \beta}{3} - \left ( \frac{\pi}{2} - \frac{2(\gamma - \beta)}{3} \right ) = \frac{\gamma - \beta}{3} = \angle KLA$

Consequently, the line AK (having the same direction as the Simson line perpendicular to the Morley triangle side EF) and the line LO (from the pole of this Simson line through the circumcenter O) form the same angles with their transversal KL and they are therefore parallel. As a result, we know that the three Simsons lines perpendicular to the three sides DE, EF, FD of the Morley triangle are parallel to the lines from their poles through the circumcenter O.

According to Honsberger's Episodes (Chapter 11):
[b]1. [/b]The orthopole of a line lies on the Simson line perpendicular to it.
[b]2. [/b]If a line intersects the circumcircle, the two Simson lines corresponding to the two intersections meet at the orthopole of the line.
[b]3. [/b]The orthopole of a line passing through the circumcenter lies on the 9-point circle.

Let the line LO intersect the circumcircle $(O)$ at a point M diametrally opposite to L. The two Simson lines having poles L and M intersect at the orthopole of this line, which lies on the 9-point circle, and they are perpendicular to each other. Since the Simson line with the pole L is parallel to LM, the Simson line with the pole M is perpendicular to it. Altogether, there are three Simson lines perpendicular to the sides DE, EF, FD of the Morley triangle, respectively, and they mutually form angles $\frac {\pi}{3}$. Likewise, there are three Simson lines perpendicular to first three, which meet them on the 9-point circle. All that remains is to show one of the first three Simson lines passes through the center N of the 9-point circle. The Simson line perpendicular to it will then be a tangent to the 9-point circle and similarly for the other two pairs of Simson lines. For this, another theorem from Honsberger's Episodes (Chapter 5) comes handy:

[b]4. [/b]A Simson line passes through the midpoint of the segment from its pole to the orthocenter and this midpoint lies on the 9-point circle.

The 9-point circle is the circumcircle of the medial triangle, hence, its radius is half of the circumradius. In addition, the orthocenter H is the external homothety center of the circumcircle $(O)$ and the 9-point circle $(N)$ (the centroid G is their internal homothety center). Therefore, the midpoint S of the segment HL and the pole L are centrally similar with the same homothety center H and the same coefficient $\frac 1 2$. Since the line LM and the Simson line perpendicular the EF are parallel and since they pass through the corresponding points L, S, they are themselves centrally similar with the same homothety center and coefficient. Since the line LM passes through the circumcenter O, the parallel Simson line must pass through the center N of the 9-point circle $(N)$.

Q.E.D.

Edit: Honsberger's name and some other spelling errors fixed. Thanks.

to yetti. Nice solution. I had some similar ideas concerning Simson lines passing through the center of nine point circle. And my solution also uses the same well-known facts about the homothety with center in the orthocenter that maps the circumcircle to the nine point circle. But anyway, I suppose that our solutions are different. So, I will post my solution soon.

One more solution is in the article of L. Emelyanov and T. Emelyanova "The Morley theorem. 100 years later." In this article the authors give two open questions:

1. It is easy to prove that the Morley triangle and the original triangle are perspective. What is the center of the perspective?

2. What is the center of homothety of the triangle PQR (which vertices have Simson lines touching to the nine point circle) and the Morley triangle?

These two questions are still open. So, if someone have any ideas...

So, here my solution is.

We will use the following well-known facts.

(1) Let $ABC$ and $A'B'C'$ be triangles such that $AB \parallel A'B'$, $BC \parallel B'C'$ and $CA \parallel C'A'$. Then the triangles $ABC$ and $A'B'C'$ are either homothetic or coincide by parallel translation.

(2) Let $ABC$ be an arbitrary triangle, $H$ be its orthocenter, $\omega$ be its nine point circle, $\Omega$ be its circumcircle with center $O$. Then the homothety with center $H$ and coefficient $1/2$ maps $\Omega$ to $\omega$. In particular, the center of $\omega$ is a midpoint of the segment $OH$.

(3) Let $A'B'C'$ be the Morley triangle of a triangle $ABC$. Then $\angle{C'B'A}=60^{\circ}+\angle{C}/3$.

(4) Let $A'B'C'$ be the Morley triangle of a triangle $ABC$. Then the angle between the lines $BC$ and $B'C'$ is equal to $|\angle{C}/3-\angle{B}/3|$.

(5) Let $PQ$ be a chord of $\Omega$ such that $PQ \perp AC$. Then the Simson line of the point $P$ is parallel to the line $BQ$.

(6) If $X$ and $Y$ are diametrically opposite points then the Simson lines of these points are perpendicular to each other and $\omega$ passes through its point of intersection.

(7) If $P \in \Omega$ then the Simson line of $P$ passes through the midpoint of the segment $PH$.

Back to the original problem, we have to show that there exist precisely three points which Simson lines tangent to $\omega$. Let $P$ be one of such points and $l_P$ be its Simson line. Then, according to (6), for the diametrically opposite point $P'$ we have: $l_{P'} \perp l_P$ and $l_P \cap l_{P'} \in \omega$, that is $l_{P'}$ passes through the center of $\omega$. Observe that the converse is also true, that is if the Simson line of some point passes through the center of $\omega$ then the Simson line of the diametrically opposite point tangents to $\omega$. Therefore it is sufficient to prove that there exist precisely three points which Simson lines pass through the center of $\omega$.

Let $O$ be the center of circumcircle $\Omega$ of the triangle $ABC$ and $H$ be the orthocenter.

[b]Lemma.[/b] Let $P \in \Omega$. Then the Simson line of $P$ passes through the center of $\omega$ if and only if $PO \parallel QB$, where $Q$ is a point at $\Omega$ such that $PQ \perp AC$.

[i]Proof.[/i] According to (7), the Simson line of the point $P$ passes through the midpoint of the segment $PH$. If it, in addition, passes through the midpoint of $OH$ (that is, according to (2), through the center of $\omega$), then it contains the medial line of the triangle $OHP$, and hence is parallel to $PO$. Conversely, if it's parallel to $PO$, then it contains the medial line of the triangle $OHP$, and hence passes through the midpoint of the segment $OH$, that is through the center of $\omega$.

To complete the proof it is sufficient to apply the claim (5). $\square$

And now let $P$ be a point running over the circle $\Omega$. We are going to investigate the behaviour of the point $Q$ (such that $PQ \perp AC$). Let $P$ ran from the point $P_1$ to the point $P_2$ the arc with angular measure $\varphi$, that is the line $OP$ rotated about the point $O$ by the angle $\varphi$. Let $Q_1$ and $Q_2$ be the corresponding to $P_1$ and $P_2$ positions of $Q$. Then $P_1Q_1 \parallel P_2Q_2$, and hence the arcs $P_1P_2$ and $Q_1Q_2$ are equal. But it is easy to see that the line $BQ$ rotated about $B$ by the angle $\varphi/2$, in opposite direction (for instance, if the line $OP$ rotates clockwise, then the line $BQ$ rotates counter-clockwise).

So, during the point $P$ was running all the circle $\Omega$, the line $OP$ rotated by $360^{\circ}$. During the same time the line $BQ$ rotated by $180^{\circ}$ (when the points $B$ and $Q$ were the same, we had to consider the tangent to $\Omega$ at $B$). The question is: "How many times was there a position when these lines ($PO$ and $BQ$) were parallel to each other?". The answer is: 3 times. Indeed, observe that the centers of rotations don't matter. Therefore, without loss of generality, we may assume that the centers are the same. Then, instead of rotation both lines in opposite directions, we may consider rotation only one line by the angle $360^{\circ}+180^{\circ}=3 \cdot 180^{\circ}$. And now it is extremely obvious that the position with $PO \parallel BQ$ occurs exactly three times. So, we have proved that there are exactly three points which Simson lines tangent to $\omega$.

Now let's prove the second part of the problem (about the homotheticness with the Morley triangle). For this goal we will find the location of the points which existance we have just proven.

Without loss of generality we may assume $\angle{BCA} \geq \angle{CAB}$. Let the point $B$ be a start position of the point $P$. In this start position the angle between the lines $OP$ and $BQ$ is equal to $\angle{OBH}=\gamma-\alpha$, where $\alpha=\angle{CAB}$, $\beta=\angle{ABC}$ and $\gamma=\angle{BCA}$. During the rotation of the line $OP$ counter-clockwise by the angle $\varphi$, the angle between the lines $OP$ and $BQ$ increases by $3\varphi/2$. Therefore by the first desired position of the point $P$, the line $OP$ have to rotate by the angle $2(\gamma/3-\alpha/3)$. Call this point $P_1$. Then to obtain the second desired position, the line $OP$ have to
rotate more by the angle $2 \cdot 180^{\circ}/3=120^{\circ}$. And then more by $120^{\circ}$ to obtain the third desired position. Denote these points by $P_2$ and $P_3$. Clearly, the triangle $P_1P_2P_3$ is equilateral. The rest is to prove that its sides are parallel to the sides of the Morley triangle. But it is trivial. Indeed, the angle between the lines $P_1P_3$ and $BC$ is equal to $\smile{P_3C}/2-\smile{BP_1}/2=(\smile{BA}/2+\smile{AC}/2-\smile{BP_1}/2-\smile{P_1P_3}/2)-\smile{BP_1}/2=(\gamma+\beta-\gamma/3+\alpha/3-120^{\circ})-(\gamma/3-\alpha/3)=\ldots=\beta/3-\gamma/3$, that is equal to the angle between the lines $B'C'$ and $BC$ (see the claim (4)), where $A'B'C'$ is the Morley triangle. Consequently, $P_1P_3 \parallel B'C'$. Similarly we obtain that $P_1P_2 \parallel A'B'$ and $P_2P_3 \parallel C'A'$. So, we are done.

We have considered the triangle $P_1P_2P_3$ which vertices have Simson lines passing through the center of the nine point circle. And now let us consider the triangle $P_1'P_2'P_3'$ which vertices have Simson lines touching $\omega$ (this triangle was called $PRQ$ in the problem setting). We have proved earlier that the central symmetry with center $O$ maps $P_1'P_2'P_3'$ to $P_1P_2P_3$. It means that we also have $P_1'P_3' \parallel B'C'$, $P_1'P_2' \parallel A'B'$ and $P_2'P_3' \parallel C'A'$. Applying now the property (1), we can easily obtain desired fact (that is the triangle $P_1'P_2'P_3'$ is homothetic to the triangle $A'B'C'$).

The rest is to prove that the triangle with vertices in the points of touching is also homothetic to the triangle $A'B'C'$. Consider a homothety with center $H$ and coefficient $1/2$. According to (2), it maps $\Omega$ to $\omega$. Therefore the point $P_1'$ coincide with some point $X_1 \in \omega$. On the other hand, by the claim (7), $X_1$ lies at the Simson line of $P_1'$. But the Simson line of $P_1'$ tangents to $\omega$, and hence $X_1$ is a point of touching.

Similarly, we can prove that the points of touching $X_2$ and $X_3$ are the images of the points $P_2'$ and $P_3'$ through our homothety. So, the triangle with vertices in the points of touching is homothetic to the triangle $P_1'P_2'P_3'$, which is homothetic to the Morley triangle as proven earlier.

So, all the parts of the problem are proven now.

[quote="yetti"]WLOG, assume that the relative size of the angles $\alpha = \angle CAB$, $\beta = \angle ABC$, $\gamma = \angle BCA$ is $\beta < \alpha < \gamma$ (that's how I draw the triangle). Denote the Morley equilateral triangle $\triangle DEF$, where D is the intersection of the appropriate trisectors of the angles $\beta$,$\gamma$, E the intersection of the appropriate trisectors of the angles $\gamma$, $\alpha$ and F the intersection of the appropriate trisectors of the angles $\alpha$, $\beta$. The angles formed by 2 trisectors from the same vertex of the $\triangle ABC$ with the sides of the equilateral triangle $\triangle ABC$ are:

$\angle CDE = \frac{\beta + \pi}{3}$, $\angle CED = \frac{\alpha + \pi}{3}$

$\angle AEF = \frac{\gamma + \pi}{3}$, $\angle AFE = \frac{\beta + \pi}{3}$

$\angle BFD = \frac{\alpha + \pi}{3}$, $\angle BDF = \frac{\gamma + \pi}{3}$

An easy proof is to draw an arbitrary equilateral triangle $\triangle DEF$, take the angles $\alpha + \beta + \gamma = \pi$, build the above angles on the sides of this equilateral triangle and show, for example, that the angles $\angle FAB = \frac{\alpha}{3}$, $\angle FBA = \frac{\beta}{3}$. Using sine theorem for the triangles $\triangle AEF$, $\triangle BDE$, $\triangle ABF$,

$\frac{EF}{AF} = \frac {\sin \left ( \frac{\alpha}{3} \right )}{\sin \left ( \frac{\gamma + \pi}{3} \right )}$, $\frac{DE}{BF} = \frac {\sin \left ( \frac{\beta}{3} \right )}{\sin \left ( \frac{\gamma + \pi}{3} \right )}$, $\frac{AF}{BF} = \frac {\sin \left ( \frac{\beta}{3} \right )}{\sin \left ( \frac{\alpha}{3} \right )}$

Since in addition, $\angle AFB =  2\pi - \frac{\pi}{3} - \frac{\alpha + \pi}{3} - \frac{\beta + \pi}{3} = \pi - \frac{\alpha + \beta}{3}$, it follows that $\angle FAB = \frac{\alpha}{3}$, $\angle FBA = \frac{\beta}{3}$.[/quote]

At first, Yetti, thanks a lot for doing this work. The above quoted passage from your solution of the problem actually contains a nice proof of Morley's theorem, so you have done even more than was asked for.

The only complaint I have to your solution is that "Hosenberger" should be "Honsberger" throughout the text. :mrgreen:

Now, to Pestich's post:

[quote="pestich"]Mathworld (ref. to Well's book) claims that the envelope
   of Simson lines is a deltoid (oriented as Morley's triangle) and it touches
   the sides of the original triangle in points isotomic to feet of the altitudes.

   Is it obvious?[/quote]

What? That the envelope is a deltoid? By no means. A proof can be found in

[url=http://racefyn.insde.es/Publicaciones/racsam/art%C3%ADculos/racsam%2095_1/Guzm%C3%A1n.pdf]M. de Guzmán, [i]The envelope of the Wallace-Simson lines of a triangle. A simple proof of the Steiner theorem on the deltoid.[/i], Rev. R. Acad. Cien. Serie A. Mat. Geometría y Topología 95 (2001) 1, p. 57-64[/url].

I would call the proof anything but simple, but it is probably much simpler than all proofs of this fact that were known before.

[quote="pestich"]What triangle center is the radical point of the 3 circles
   constructed on these isotomic points as diameters?[/quote]

You mean the three circles whose diameters are the segments joining the feet of the altitudes of triangle ABC with the reflections of these feet in the midpoints of the respective sides of triangle ABC ? The radical center of these three circles is X(185), the Nagel point of the orthic triangle of triangle ABC. This is, by the way, quite easy to prove synthetically.

[quote="pestich"]   It also seems that the 3 distances from the midpoints of the original triangle sides
   to the nearby touching points of Simson lines with the 9 PC  (which seem to be
   equal to the distances from same midpoints  to the intersections of sides with
   Simson lines tangent to 9 PC)
   are    linked so that one of the distances is equal to the sum of the other two.[/quote]

If I correctly understand your assertion, then, unfortunately, it is not true. However, since the points where the Simson lines touch the nine-point circle form an equilateral triangle, for any point P on the nine-point circle, the distance from P to one of these points equals the sum of the distances from P to the other two.

To the post by Remike:

[quote="Remike"]One more solution is in the article of L. Emelyanov and T. Emelyanova "The Morley theorem. 100 years later."[/quote]

Where can I find this article? Has it already appeared? I would be very delighted to see it, since I really like Emelyanov's articles!

[quote="Remike"]In this article the authors give two open questions:

1. It is easy to prove that the Morley triangle and the original triangle are perspective. What is the center of the perspective?

2. What is the center of homothety of the triangle PQR (which vertices have Simson lines touching to the nine point circle) and the Morley triangle?[/quote]

Well, what do you mean by "What is the center"? Actually, none of these two centers is one of [i]the[/i] four famous triangle centers ;) . The center of perspective of the Morley triangle and the original triangle is X(357) in Clark Kimberling's ETC; the center of homothety of the triangle PQR and the Morley triangle is, if I'm not mistaken, not in the ETC yet.

Last but not least, I wanted to say that I also have a solution to the initial problem of this thread. It is different from both Yetti's and Remike's solutions, although, maybe from the advanced viewpoint all three solutions are more or less the same. Whether I will post it here... well... that depends from weather ;) .

  darij

[quote="darij grinberg"]
At first, Yetti, thanks a lot for doing this work. The above quoted passage from your solution of the problem actually contains a nice proof of Morley's theorem, so you have done even more than was asked for.

The only complaint I have to your solution is that "Hosenberger" should be "Honsberger" throughout the text. :D
[/quote]
I have seen this proof at the Cut-The-Knot, with a comment that direct proofs are too complicated. I also fixed Honsberger's name. :blush: Thanks.

[quote="Remike"]
Nice solution. I had some similar ideas concerning Simson lines passing through the center of nine point circle. And my solution also uses the same well-known facts about the homothety with center in the orthocenter that maps the circumcircle to the nine point circle. But anyway, I suppose that our solutions are different. So, I will post my solution soon.
[/quote]
Thanks. Please, give me some time to go through the details of your solution.

yetti

For other deepenings on Morley's theorem I suggest the following site http://shk.ans.hive.no/

Leon

Reading your proof, I realized that I did not really show that only 3 Simson lines are tangent to the 9-point circle (or only 3 Simson lines pass through the center of the 9-point circle, which is the same thing). It seemed clear to me that at most 3 such Simson lines can exist, because when the pole moves from one vertex of the triangle $\triangle ABC$ to another, the corresponding Simson line moves from one altitude to another, but I should have said something. Otherwise, the ideas of both proofs are similar, even though I did not see that the homothety of the circumcircle and the 9-point circle WRT centroid $G$ can be used as well. Thanks for posting your solution.

You may consider fixing some typing errors:

[quote="Remike"](3) Let $A'B'C'$ be the Morley triangle of a triangle $ABC$. Then $\angle{C'B'A}=60^{\circ}+\angle{C}/2$.
...
[b]Lemma.[/b] Let $P \in \Omega$. Then the Simson line of $P$ passes through the center of $\omega$ if and only if $PO \perp QB$, where $Q$ is a point at $\Omega$ such that $PQ \perp AC$.
...
We have proved earlier that the central symmetry with center $O$ maps $P_1'P_2'P_3'$ to $P_1P_2P_3$.[/quote]

In lemma 2, the angle $\angle{C'B'A}=60^{\circ}+\angle{C}/3$.
In lemma 7, the Simson line of $P$ passes through the center of $\omega$ if and only if $PO \parallel QB$, as you proved.
Finally, the homothety center of the triangles $P_1'P_2'P_3'$ and $P_1P_2P_3$ is the centroid $G$ of the triangle $\triangle ABC$, as you showed.

[quote="yetti"]even though I did not see that the homothety of the circumcircle and the 9-point circle WRT centroid $G$ can be used as well. Thanks for posting your solution.[/quote]
What do you mean by "centroid"? Is it a common point of all three altitudes (that is the point that I called orthocenter)?

[quote="yetti"]You may consider fixing some typing errors:

[quote="Remike"](3) Let $A'B'C'$ be the Morley triangle of a triangle $ABC$. Then $\angle{C'B'A}=60^{\circ}+\angle{C}/2$.
...
[b]Lemma.[/b] Let $P \in \Omega$. Then the Simson line of $P$ passes through the center of $\omega$ if and only if $PO \perp QB$, where $Q$ is a point at $\Omega$ such that $PQ \perp AC$.
...
We have proved earlier that the central symmetry with center $O$ maps $P_1'P_2'P_3'$ to $P_1P_2P_3$.[/quote]

In lemma 2, the angle $\angle{C'B'A}=60^{\circ}+\angle{C}/3$.
In lemma 7, the Simson line of $P$ passes through the center of $\omega$ if and only if $PO \parallel QB$, as you proved.[/quote]
I have edited my post. Thank you.

[quote="yetti"]Finally, the homothety center of the triangles $P_1'P_2'P_3'$ and $P_1P_2P_3$ is the centroid $G$ of the triangle $\triangle ABC$, as you showed.[/quote]
If I understand you correctly then it is not true. I meant here that $P_1$, $P_2$ and $P_3$ are the points for which the Simson lines pass through the center of $\omega$; and $P_1'$, $P_2'$ and $P_3'$ are the points for which the Simson lines tangent $\omega$. I showed that $P_i$ is diametrically opposite to $P_i'$. So, the central symmetry with center $O$ maps $P_1P_2P_3$ to $P_1'P_2'P_3'$.

The intersection of medians is called centroid in English geometry textbooks and usually denoted $G$.

When reading the problem, I thought that $P, Q, R$ were the the tangency points of the Simson lines with the 9-point circle, not their poles on the circumcircle. Similarly when reading your proof, for the points $P_1', P_2', P_3'$ (originally denoted $P, Q, R$) are mentioned at the end, when everything becomes clear. But of course, you denoted the tangency points $X_1, X_2, X_3$. So, the triangles $\triangle P_1P_2P_3 \sim \triangle P_1'P_2'P_3'$ are centrally similar with homothety center at the circumcenter $O$ and coefficient $-1$, while the triangles $\triangle P_1P_2P_3 \sim \triangle X_1X_2X_3$ are centrally similar with homothety center at the centroid $G$ and coefficient $-\frac 1 2$.

[quote="darij grinberg"][quote="Remike"]One more solution is in the article of L. Emelyanov and T. Emelyanova "The Morley theorem. 100 years later."[/quote]

Where can I find this article? Has it already appeared? I would be very delighted to see it, since I really like Emelyanov's articles![/quote]
Well, I don't know where you can find it. It was published in Russian journal "Mathematics in school" ("Математика в школе"), № 9, 2004. If you know Russian, you can see here a content of this journal:
http://www.schoolpress.ru/shop/aposoft.z/StoreFront-Main .
And here the abstract of this paper is:
http://www.schoolpress.ru/shop/SPStoreFront/Share/public/jornal/matematika/short_text/9_2004/08_9_04_62.html .

By the way, some time ago another article of L. Emelyanov "Geometrical proof of the Ponsele theorem" ("Геометриче?кое доказатель?тво теоремы Пон?еле") was published in this journal (№ 3, 2004). Here the abstract is: http://www.schoolpress.ru/shop/SPStoreFront/Share/public/jornal/matematika/short_text/3_2004/08_3_04_73.html .

[quote="darij grinberg"][quote="Remike"]In this article the authors give two open questions:

1. It is easy to prove that the Morley triangle and the original triangle are perspective. What is the center of the perspective?

2. What is the center of homothety of the triangle PQR (which vertices have Simson lines touching to the nine point circle) and the Morley triangle?[/quote]

Well, what do you mean by "What is the center"? Actually, none of these two centers is one of [i]the[/i] four famous triangle centers ;) . The center of perspective of the Morley triangle and the original triangle is X(357) in Clark Kimberling's ETC; the center of homothety of the triangle PQR and the Morley triangle is, if I'm not mistaken, not in the ETC yet.[/quote]
Well, I can quote from the original text of the article (in Russian), but I'm afraid it is out of law. But I suppose I may retell it in English. So, here my translation of the final part of the article is.

"It seems that we can finish our consideration of the Morley theorem. However, we want to think that this material will have become a beginning of an investigation of the properties of the Morley triangle, because every fundamental theorem generates more questions than answers. For instance, it is easy to prove that the original triangle and its Morley triangle are perspective, that is the lines AD, BE and CF have a common point. What is that point? Is it a well-known point for the triangles ABC and DEF? Does it lie at any well-known line? And there is one more natural question: what is the center of homothety of the triangles PQR and DEF?

Dream, ask questions, put forward ideas --- modern computer provides an opportunity to verify many and many geometrical conjectures. Fortunately, it (computer) can't prove these conjectures, since we need geometrical intuition and mind to prove it. But it can help us to develop both these things."

[quote="yetti"]The intersection of medians is called centroid in English geometry textbooks and usually denoted $G$.[/quote]
Yes, in Russian too.

[quote="yetti"]When reading the problem, I thought that $P, Q, R$ were the the tangency points of the Simson lines with the 9-point circle, not their poles on the circumcircle. Similarly when reading your proof, for the points $P_1', P_2', P_3'$ (originally denoted $P, Q, R$) are mentioned at the end, when everything becomes clear. But of course, you denoted the tangency points $X_1, X_2, X_3$. So, the triangles $\triangle P_1P_2P_3 \sim \triangle P_1'P_2'P_3'$ are centrally similar with homothety center at the circumcenter $O$ and coefficient $-1$, while the triangles $\triangle P_1P_2P_3 \sim \triangle X_1X_2X_3$ are centrally similar with homothety center at the centroid $G$ and coefficient $-\frac 1 2$.[/quote]
I understand you now. There are two homotheties that maps $\Omega$ to $\omega$: with positive coefficient and with negative coefficient. You consider a homothety with negative coefficient and with center at $G$. And I consider a homothety with positive coefficient and with center at $H$. You claim that the triangles $P_1P_2P_3$ and $X_1X_2X_3$ are homothetic with coefficient $-1/2$ and center at $G$. Well, it's true. And I claim that the triangles $P_1'P_2'P_3'$ and $X_1X_2X_3$ are homothetic with coefficient $1/2$ and center at $H$. Well, it's also true.


Simplify and express your answer as a common fraction.

\[ \dfrac{\frac{1}{1\plus{}\frac{2}{1\plus{}3}}}{\frac{3}{2\plus{}\frac{1}{1\plus{}1}}}\]


Find the inverse of the $n$x$n$ "lower" triangular matrix. Prove that the matrix you give is indeed the inverse.

By lower triangular matrix I mean the matrix with a 1 in every entry below or on the main diagonal, and a 0 in every other entry.

perform the following row operations on the matrix:

$R_{n}\mapsto R_{n}-R_{n-1}$,
$R_{n-1}\mapsto R_{n-1}-R_{n-2}$,
...
$R_{2}\mapsto R_{2}-R_{1}$.

the result is $I_{n}$. performing these same operations on $I_{n}$ now gives the inverse of the original matrix: the result is a main diagonal of 1's, the diagonal below that of all -1's, and everything else 0.


Prove that the geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity.

If algebraic and geometric multiplicities are defined via the Jordan form of a matrix, as some authors do, your argument is obvious.

Here is another treatment:
Let $ k$ be the dimension of the eigenspace $ V_{\lambda}$ corresponding to the eigenvalue $ \lambda$. Then extend $ V_\lambda$ to a basis $ \mathcal B$ for your vector space $ V$. Clearly the matrix of your operator, say $ \tau$, with respect to $ \mathcal B$ has the form:
$ \begin{pmatrix}\lambda I_k & A\\ 0& B\end{pmatrix}$.
But, $ c_{\tau}(x)=\det(xI-[\tau]_{\mathcal B})=(x-\lambda)^k\det(xI_{n-k}-B).$ Here $ n$ is the dimension of $ V$. Hence, the result follows.


geometry
circumcircle
geometric transformation
reflection
analytic geometry
trigonometry
parameterization

Let $ \Gamma$ be the circle with radius $ r$ and center $ O$ and $ \Gamma'$ be the circle with radius $ r'$ and center $ O'$. Let $ T$ be a variable point of  $ \Gamma$ and the tangent line $ t$ passing by $ T$  intersect the circle $ \Gamma'$ at two points $ A$ and $ B$. Find the locus of the circumcenter of the triangle $ BOA$ when $ T$ varies on $ \Gamma$. I know, making a figure, that the locus is a circle of center $ O'$ with a radius depending of $ OO'$, $ r$ and $ r'$. But how to prove it? Thanks in advance.

First, invert the figure in the circle $ \Gamma$ with center $ O.$ Tangent of $ \Gamma$ at $ T$ goes to a circle $ \mathcal T^*$ with diameter $ OT.$ Arbitrary circle $ \Gamma'$ with center $ O'$ (not entirely inside of $ \Gamma$) goes to a circle $ \Gamma'^*$, intersecting the circle $ \mathcal T^*$ at $ A^*, B^*.$ Circumcircle $ \mathcal C$ of the $ \triangle OAB$ goes to the line $ A^*B^*$ and its circumcenter $ C$ into the reflection $ C^*$ of $ O$ in $ A^*B^*.$ Let $ E \in OO'$ be the center of $ \Gamma'^*$ and $ s$ its radius. Denote $ d \equal{} OO',\ e \equal{} OE.$ Since $ O$ is the similarity center of $ \Gamma'^*$ and $ \Gamma'$ with similarity cefficient $ k \equal{} \frac {r^2}{d^2 \minus{} r'^2}$ (power of inversion divided by power of the inversion center to the original circle), we have $ e \equal{} kd, s \equal{} kr'.$

Let $ O$ be the coordinate origin and $ OO'$ the x-axis. Equations of the circle $ \mathcal T^*$ with fixed diameter $ r \equal{} OT$ and passing through the origin $ O$ is

$ \mathcal T^*: \ \ x^2 \minus{} xr \cos \theta \plus{} y^2 \minus{} yr \sin \theta \equal{} 0,$

while equation of the circle $ \Gamma'^*$ centered on the x-axis at $ E \equal{} (e, 0)$ and with radius $ s$ is

$ \Gamma'^*: \ \ x^2 \minus{} 2ex \plus{} e^2 \plus{} y^2 \equal{} s^2.$

Subtracting equations of the 2 circles from each other gives equation of their radical axis $ A^*B^*$:

$ A^*B^*: \ \ F(x, y, \theta ) \equal{} x(r \cos \theta \minus{} 2e) \plus{} yr \sin \theta \plus{} e^2 \minus{} s^2 \equal{} 0.$

Transforming the last equation to polar coordinates $ (\varrho, \phi)$ by $ x \equal{} \varrho \cos \phi,\ y \equal{} \varrho \sin \phi,$

$ A^*B^*: \ \ \Phi(\varrho, \phi, \theta ) \equal{} \varrho r \cos (\phi \minus{} \theta) \minus{} 2 \varrho e \cos \phi \plus{} e^2 \minus{} s^2 \equal{} 0.$

This is equation of a family of lines depending on the parameter $ \theta.$ Equation of their envelope $ \mathcal K$ (if any) is obtained by eliminating this parameter from the equations $ \Phi(\varrho, \phi, \theta ) \equal{} 0$ and

$ \frac {\partial \Phi}{\partial \theta} \equal{} \varrho r \sin (\phi \minus{} \theta) \equal{} 0.$

The 2nd equation yields $ \phi \minus{} \theta \equal{} n \pi,\ \cos(\phi \minus{} \theta) \equal{} \pm 1.$ Substituting this to the polar equation $ \Phi(\varrho, \phi, \theta) \equal{} 0$ of the line $ A^*B^*,$ the envelope equation comes out as

$ \mathcal K: \ \ \pm \varrho r \minus{} 2 \varrho e \cos \phi \plus{} e^2 \minus{} s^2 \equal{} 0,$ or

$ \varrho \equal{} \frac {e^2 \minus{} s^2}{r} \cdot \frac {1}{\frac {2e}{r} \cos \phi \pm 1} \equal{} \frac {L}{\epsilon \cos \phi \pm 1}.$

This is polar equation of a conic with focus $ O,$ major axis line $ \phi \equal{} 0$ (i.e., the line $ OO'$), excentricity $ \epsilon \equal{} \frac {2e}{r}$ and semilatus rectum $ |L| \equal{} \frac {|e^2 \minus{} s^2|}{r}.$ The correct sign, obtained by demanding $ \varrho > 0,$ depends on radii $ r, r'$ and center distance $ d \equal{} OO'$ of the circles $ \Gamma, \Gamma'.$ For an ellipse $ (\epsilon < 1$), the correct sign is the same as the sign of $ e^2 \minus{} s^2 \equal{} k^2(d^2 \minus{} r'^2)$ or $ d \minus{} r'$ ($ O$ outside or inside of $ \Gamma'$), for a hyperbola ($ \epsilon > 1$) one sign holds for each branch.

Since $ A^*B^*$ is tangent to this conic, the foot $ F$ of perpendicular from the focus $ O$ to $ A^*B^*$ lies on its pedal circle $ \mathcal P$ centered at $ P \in OO'$ on the conic major axis. Since $ F$ is midpoint of $ OC^*,$ the image $ C^*$ of the circumcenter $ C$ lies on a circle $ \mathcal Q.$ centrally similar to $ \mathcal P$ with similarity center $ O$ and coefficient 2, also centered on $ OO'.$ As a result, $ C$ lies on the inversion image $ \mathcal Q^*$ of $ \mathcal Q,$ also centered on $ OO'.$ The center $ Q$ of $ \mathcal Q$ is the other conic focus

Assume $ \epsilon > 1$ (hyperbola) and $ d > r'$ ($ O$ outside $ \Gamma'$). Substituting $ \phi \equal{} 0$ into the hyperbola equation, $ q \equal{} OQ \equal{} \frac {L}{\epsilon \plus{} 1} \plus{} \frac {L}{\epsilon \minus{} 1} \equal{} \frac {2\epsilon L}{\epsilon^2 \minus{} 1}.$ Radius of $ \mathcal Q$ is equal to the conic major axis length, $ t \equal{} \frac {OQ}{\epsilon} \equal{} \frac {2L}{\epsilon^2 \minus{} 1}$ for a hyperbola. Again, $ O$ is the similarity center of $ \mathcal Q^*, \mathcal Q$ with similarity coefficient $ m \equal{} \frac {r^2}{q^2 \minus{} t^2} \equal{} r^2 \cdot \frac {\epsilon^2 \minus{} 1}{4L^2},$ so that distance of the center of $ \mathcal Q^*$ from $ O$ is

$ mq \equal{} r^2 \cdot \frac {\epsilon^2 \minus{} 1}{4L^2} \cdot \frac {2\epsilon L}{\epsilon^2 \minus{} 1} \equal{} r^2 \cdot \frac {\epsilon}{2L} \equal{} r^2 \cdot \frac {2e}{r} \cdot \frac {r}{2(e^2 \minus{} s^2)} \equal{} r^2 \cdot \frac {kd}{k^2(d^2 \minus{} r'^2)} \equal{} d.$

Since $ d \equal{} OO',$ the center of $ \mathcal Q^*$ containing the circumcenter $ C$ is identical with the center $ O'$ of $ \Gamma'.$


Let $P(x)$ be a polynomial with integer coefficents. And let $Q(x) = \prod \limits^n_{k=1} P(x^k) + 1.$ Determine whether $Q(x)$ has integer roots dependently on $n$.

Is it right that we have to define all natural $n$ s.t. there is $P(x)\in Z[x]$ that $Q(x)$ has integer root?


AMC
USA(J)MO
USAMO
AIME
inequalities
pigeonhole principle
AMC 10

lol i clicked 42 just for the heck of it.

Well that depends on who's doing the expecting. If you ask Delong, I'm getting a 42. If you ask me, I'm getting just enough to make red mop, or 14+ if both #1 and #4 are really easy (read: not geometry).

I'm hoping for a decent improvement, so I said 15-21 (I got 10 last year.)  Given the fact that I had a 42-point increase between my first AMC-10 and my second, a 5-point increase between my first AIME and my second, and a 20-point increase between my first AMC-12 and my second, I'm thinking that that might be a reasonable goal.

[quote="bubala"][quote="Ignite168"]I'm going to be taking the USAMO unofficially, I'm so mad I didn't qualify because of 1 error, I think I could easily get like a 14 if I made it  :mad:[/quote]

Take it from the guy who scored a 3 twice in a row Ignite, even if you think you could solve two problems, getting a 14 is far more difficult than you realize.

--the bubala[/quote]

He was joking anyway; he qualified.

does he ever not joke?

hmm...i put 15-21 b/c i hope that that will be enough for blue MOP (at least, the upper end of that range) which is a goal that i have worked EXTREMELY hard for.

plus i scored 7 last year (2 points from red MOP)

[quote="Treething"]uhhh

hopefully enough to get into red mop

meaning i'll be spending my 4.5 hours bashing the living crap out of inequalities with my multivariable calc knowledge and gross misapplications of pigeonhole, CRT, and muirhead

be prepared, usamo graders[/quote]

Just a tip. If you actually do find yourself writing a proof with Lagrange Multipliers, make sure that you do all the arithmetic (solving the n equations for n variables) in your proof, or else you will get very few points.

Zero, since I didn't make it. I'll take it unofficially just to see how I do. Zero, probably, but then again I can''t really know whether I'd get a 0 or a 1 on any of the problems.

[quote="Treething"]uhhh

hopefully enough to get into red mop

meaning i'll be spending my 4.5 hours bashing the living crap out of inequalities with my multivariable calc knowledge and gross misapplications of pigeonhole, CRT, and muirhead

be prepared, usamo graders[/quote]

If there are inequalities...hm

And tim's facebook says you don't know the CRT

42. obviously. since Delong says so.   :lol:
In reality, hoping for 15-21 so that i at least improved....

[quote="undefined117"]mine is maybe 1 problem right.. so 7/42 :blush:[/quote]

As an 8th grader with your AMC10 and AIME scores, solving one problem fully would be very *very* impressive. Unless you've taken practice tests before and *can* solve one problem on each one...?

Anyways my goal is a non-negative score, although I'd be happy to get a negative score also.

Hm I said 13-31 in the last topic. I guess expected score is then $\frac{13+31}{2}= 22$ assuming a linear probability distribution or symmetric about the median or whatever. I'll be happy if I do in fact get a $22$.

[quote="probability1.01"]Hmmm... I clicked 8-14. Since I voted 0-9 in another topic, I guess my score will be 8 or 9.[/quote]

dude, come on.. there are only 6 problems. Even you can't pull off an 8  :P

Good luck to everyone taking it tomorrow

I'd say 1-7.  I don't think I'll draw blanks on every single question and be able to write nothing.  If I get lucky, I'll see a question that I can solve, and maybe get a 7 on that problem.


When I move my mouse over the private messages it says that 1 unread message.

But I've read all messages.

What is this? :?

Looks like a bug :? . What style are you using? Also please tell me if refreshing the window / log in/out clears it ...

This is bit weird.

I had AoPS as my board style and when I changed to MathLinks, I have "no new private messages."

? :?

[quote="Silverfalcon"]This is bit weird.[/quote]I agree :) ... it might have been a cookie issue, or a private message that has been sent to you by someone, and then deleted during the time you spoted the link and when you clicked it.

Does returning to AoPS style cause the same problem? ...

Thanks Valentin. :D

Back to ages where I used MathLink Board Style :D  :D  :D


superior algebra
superior algebra unsolved

Let $I$ be an ideal of the ring $R$ and $f$ be nonidentity permutation of the set $\{\ 1,\dots ,k\}$.suppose that for every $0\neq a\in R,aI\neq 0,Ia\neq 0$ hold;for any elements of $x_1,x_2,\dots,x_k\in I$,$x_1x_2\dots=x_{f1}\dots x_{fk}$holdsprove that $R$ is commutative.

This one has been lying here for quite some time, huh? :)

If $f1=1$, the problem is reduced to smaller $k$'s, so we can assume this is not the case. We replace $x_1$ with $ax_1$ for an arbitrary $a\in R$, and we get $ax_1\ldots x_k=x_{f1}\ldots ax_{ff^{-1}1}\ldots x_{fk}$ on the one hand, and $ax_1\ldots x_k=ax_{f1}\ldots x_{fk}$ on the other hand. Equating those, we get $x_{f1}\ldots ax_{ff^{-1}1}\ldots x_{fk}=ax_{f1}\ldots x_{fk}$. In this relation, we can eliminate $x_{ff^{-1}1},\ldots,x_{fk}$ one by one because of the fact that $tI=0\rightarrow t=0$, and we're left with a relation of the type $ax_1\ldots x_u=x_1\ldots x_ua,\ \forall x_i\in I,\ \forall a\in R$ for some $1\le u\le k-1$. If we take $a$ from $I$, this is a relation just like the one we started with, except that the permutation $f$ is a cycle here. This means that from now on we can assume $f$ to be a cycle.

Now, if $f$ is a cycle and $k\ge 3$, then either $fk\ne 1$ or $f^2k\ne 1$, and if we pull the same trick as before, with the arbitrary $a$ and the elimination of $x_{ff^{-1}1},\ldots,x_{fk}$, we'll eliminate at least two $x_i$'s. This leaves us with an analogous problem, but for smaller $k$, and we can use induction. This means that it suffices to prove the whole thing for $k=2$ (and $f$ being a transposition).

We have $x_1x_2=x_2x_1,\ \forall x_i\in I$. This means that for all $a\in R$, we have $x_2ax_1=x_1x_2a=x_2x_1a$, so $I(ax_1-x_1a)=0,\ \forall x_1\in I,\ \forall a\in R$. We thus get $I\subset Z(R)$ (the center of $R$). Finally, for $a,b\in R,\ x_1\in I$, we have $ax_1b=x_1ab=x_1ba\Rightarrow I(ab-ba)=0$, so $ab=ba$, and we're done.

From $ ax_1...x_u\equal{}x_1...x_ua$ you can finish it faster:
$ abx_1...x_u \equal{} a(bx_1...x_u) \equal{} bx_1...x_ua \equal{} bax_1...x_u$, so $ (ab\minus{}ba)$ is killed by $ I^u$ which means (by hypothesis) that $ ab\minus{}ba\equal{}0$


AoPSwiki
articles
Alcumus
videos
Pascal\u0027s Triangle

There are Alcumus videos on those things.  I find them very useful.  Now, everybody can access all the videos, so watch them now before you can't!

[quote="Cliu0301"]Pascal's Triangle and the Hockey Stick Theorem are very useful in mathematics, so I heard.

I want to learn it but the AOPSWiki article confused me.

Can you guys help?[/quote]

Try intro. to counting&probability book (AOPS), that gives you a great understanding in pascal's triangle. You get to derive the proofs, and all.
Its great fun!


A shape that includes the point (3,8) is reflected across the line y = x + 2. At what point does the point (3,8) land?

A store puts everything on sale for 20% off. If the sales tax is 8%, what percent of the original marked price is the final cost including tax? Round answers to the nearest tenth.

CHALLENGE: SIG FIG MANIA

Answer the following in significant figures.

1. 2.0 x 3.625
2. 0.0000003 / 1.0

How many sig figs do each of these numbers have?

1. 1.204
2. 0.200000
3. 0.036230
4. 100.0

hmm, these look suspiciously like homework problems as opposed to challenge problems (who gives sig figs in a challenge problem?)

SIF FIG

sounds like a sciences class and the problems are too easy to not be hw

[quote="soulzmischief"]lol, nice try buddy

u n00b[/quote]

goodness, person. i bet you're in math counts and you're in my science class. you're either T-guy, Ma-guy, L-guy, or Mi-guy. Most likely you're T-guy.

it has been a day, and we did sig figs three weeks ago.

answers: (6,5), 86.4, 7.3, .0000003, 4, 6, 5, 4

T-guy, shut up. and dont even THINK about talking to me during school

[quote="pakagawa"]What's bad about science?

What's a significant figure anyway?

(I grew up learning in foreign languages; I might know it already)[/quote]

The number of significant figures a number has tells you how accurate that number is.  The number of sig. figures in 2.0021 is 5.   I ll tell you more if you want but you might already know it so I ll just wait.  It's not very hard to learn so anyone can just google it and learn it quickly.

[quote="soulzmischief"]lol, nice try buddy

u n00b[/quote]

are u from hopkins?

hey buddy its yuchen.. ur mathcounts tutor

have a nice day... and i'll make sure thursday is pretty painful for you

:)

[quote="soulzmischief"]... and i'll make sure thursday is pretty painful for you

:)[/quote]

i cant wait.  har har har

btw

im sure that you know we're past sig figs and all that velocity ****. except for the fact that theres a test this friday. im ready to die.

and you know all the science hw we get through your sister.


a)If  $ a,b,c > 0$ such that  $ abc \equal{} 1$  then
\[ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {9}{2(a \plus{} b \plus{} c)}
\]
b) If $ a,b,c>0$ such that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}1$, then \[ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {9}{2}\]

[quote="Marius Mainea"]Let $ a,b,c > 0$.  Prove that:

 a) $ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\leq\frac {a \plus{} b \plus{} c}{2}$
[/quote]
Maybe
$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}{2}$ $ ?$ :wink:

[quote="arqady"][quote="Marius Mainea"]
[/quote]
Maybe
$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}{2}$ $ ?$ :wink:[/quote]

 Yes ,you are right. :|

[quote="Marius Mainea"]Let $ a,b,c>0$. Prove that:
a)$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}{2}$[/quote]
Similar problem posted previously.
Use $ H\ddot{o}lder$ on $ L.H.S.$. Followed by some generalizations. :wink:

[quote="arqady"][quote="Sunkern_sunflora"][quote="Marius Mainea"]Let $ a,b,c > 0$. Prove that:
a)$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}\geq\frac {a \plus{} b \plus{} c}{2}$[/quote]
Similar problem posted previously.
Use $ H\ddot{o}lder$ on $ L.H.S.$. [/quote]
How we can use Holder? I don't see it. :maybe:  Explain please. Thank you![/quote]
It 's here
http://www.mathlinks.ro/viewtopic.php?t=216009&sid=eee3567d29fe4d8663712a426107f404
Hi arqady; Can U post your proof ?


Let a,b,c are positive real numbers such that $ \left\{ \begin{array}{l}0 < a < b \\b^2  \minus{} 4ac \le 0 \\
 \end{array} \right.$

Find the min value of $ S \equal{} \frac{{a \plus{} b \plus{} c}}{{b \minus{} a}}$

hello, prove that $ \frac{a\plus{}b\plus{}c}{b\minus{}a}\geq3$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, prove that $ \frac {a \plus{} b \plus{} c}{b \minus{} a}\geq3$.
Sonnhard.[/quote]

Hello!
This is problem 5 in Gifted school examination of my city. Can u solve this problem detailly? I need a detail solution! :)
Thanks so much! ;)

L_Euler.

The answer of [b]Dr Sonnhard Graubner[/b] is right. Now I will solve it in detail.
Consider $ P(x) \equal{} ax^2 \minus{} bx \plus{} c$. Because $ b^2 \minus{} 4ac\le 0$ and $ a > 0$ therefore $ P(x)\ge 0$ for all real number $ x$. Put $ x \equal{} 2$, we will get $ 4a \minus{} 2b \plus{} c\ge 0$, which is also means that $ a \plus{} b \plus{} c\ge 3(b \minus{} a)$, or $ \frac {a \plus{} b \plus{} c}{b \minus{} a}\ge 3$. Then you will get the result :lol:

equality holds if $ a\equal{}1,b\equal{}4,c\equal{}4$


I was recently doing a problem that related to quadratic equations.

I am asked to complete the square and solve.

Solve: $x^{2}-2x-9 = 0$

My answers:
$x$ = $\sqrt{10}-1$
$x$ = $-\sqrt{10}-1$

However, those were said to be incorrect.

The correct answers are

$x = 1+\sqrt{10}$
and
$x = 1-\sqrt{10}$

What did I do wrong?

--------

1) $x^{2}-2x = 9$

2) $\frac{1}{2x}*-2x$

3) $x^{2}-2x+1 = 9+1$

4) $(x-1)(x-1) = 10$

5) $\sqrt{x-1}^{2}= \sqrt{10}$

6) $x-1 = \sqrt{10}$

From there, I started solving.

You are mostly correct, but I I have a few comments:

$\sqrt[]{{x-1}^{2}}= \pm \sqrt[]{10}$, not just $\sqrt[]{10}$

If $x-1 = \pm \sqrt[]{10}$, then $x = 1 \pm \sqrt[]{10}$, giving the correct answers.  Maybe you just made a mistake in solving for x in the last step?

[hide="hint"]in step 5,just remeber that
$\sqrt{(x-1)^{2}}=|x-1|$
so
$x-1=\sqrt{10}$
and
$x-1=-\sqrt{10}$[/hide]

Why is the $1$ put before the $\sqrt{10}$?
That is what I do not understand.

If I am isolating $x$, then I need to add the one to both sides...

Ah. I subtracted instead of added...

*Points finger to head*
*Bang*


geometry
3D geometry
geometric transformation
rotation

There are six uncoloured wooden cubes of equal size and three different colours. On each of the cubes exactly three sides shall be completely covered with one colour. On each of the cubes all three colours shall be used.

Is it possible, to colour the six cubes as mentioned above, such that each cube can be distinguished from each of the other ones ?

Hint: Two coloured cubes are not to be distinguished if they can be transformed into each other by means of rotations and/or translations.

Yes; paint as follows, letting the three colors be red, blue, and green for simplicity.  The locations of the colors don't even matter...once again, this seems a bit easy: (spoiler)

[hide]3 red, 2 blue, 1 green

3 red, 2 green, 1 blue

3 blue, 2 red, 1 green

3 blue, 2 green, 1 red

3 green, 2 red, 1 blue

3 green, 2 blue, 1 red[/hide]


A stone is dropped into a well and the report of the stone striking the bottom is heard $ 7.7$ seconds after it is dropped. Assume that the stone falls $ 16t^2$ feet in $ t$ seconds and that the velocity of sound is $ 1120$ feet per second. The depth of the well is:

$ \textbf{(A)}\ 784 \text{ ft.} \qquad\textbf{(B)}\ 342 \text{ ft.} \qquad\textbf{(C)}\ 1568 \text{ ft.} \qquad\textbf{(D)}\ 156.8 \text{ ft.} \qquad\textbf{(E)}\ \text{none of these}$

...is AHSME equiv to AMC or AIME?

[hide="solution"]
$ d \equal{} 16t_1^2 \qquad(1)$
$ d \equal{} 1120t_2 \qquad(2)$

$ t_1 \plus{} t_2 \equal{} \frac {d}{1120} \plus{} \frac {\sqrt {d}}{4} \equal{} 7.7$

Let $ d \equal{} a^2$

$ a^2 \plus{} 280a \minus{} 8624 \equal{} 0 \Leftrightarrow (a \minus{} 28)(a \plus{} 308)$
$ d \equal{} 28^2 \equal{} 784$

Answer $ \rightarrow \boxed{A}$

Are you allowed calculators on this exam?[/hide]

AHSME was the equivalent of AMC and AIME together until about 1984, when it was "downgraded" to only AMC because of the introduction of the AIME.
And the first pocket calculator was invented in 1967, so I don't think calculators were allowed.  :D

[quote="mathwizarddude"]We just have to solve for $ t$ in $ 1120(7.7 \minus{} t) \equal{} 16t^2$ and then plug that into $ 16t^2$ to get the answer. As a side note, the acceleration of the stone is taken to be $ 32$ feet per second.

By the way, how many problems are there on the AHSME?

EDIT: Sorry, I messed up and have corrected it; I have no idea why I integrated instead of differentiating...[/quote]
[quote="facis"]mathwizard, the physics equation is $ \delta x \equal{} 1/2at^2 \plus{} v_0 t$ so the acceleration is $ 32$ feet per second, which is commonly used as the Earth's acceleration due to gravity.[/quote]For the sake of accuracy, acceleration is feet per second [b]squared[/b] or $ \text{ft}/\text{sec}^2$

oops... stupidity at its best, I guess.  ft/s^2 sounds better... And while I'm at it, I meant to use a $ \Delta$ not a $ \delta$.


function
limit
real analysis
real analysis unsolved

Let $f: X\rightarrow Y$ be a continuous function and let $E$ be dense in $X$.  Prove that $f\left(E\right)$ is dense in $f\left(X\right)$.

Fix $y\in f(Y)$. Exist $x\in X$ such that $f(x)=y$ and sequence $\{e_{n}\}\subset E$ such that $\lim{e_{n}}=x$. Final $\{f(e_{n})\}\subset f(E)$ and $\lim{f(e_{n})}=f(x)=y$.

If we're in a general topological space, note that $f^{-1}(\overline{f(E)})$ is a closed subset containing $E$, hence containing $\overline{E}= X$, so $f(X) \subset \overline{f(E)}$.


topology
real analysis
real analysis unsolved

Family {$V_ {\alpha}$} of open subsets of $X$ is called [i]basis[/i] for $X$, if for arbitrary $x \in X$ and arbitrary $G \subset X$ such that $x \in G$ there exists $V_ {\alpha}$ such that $x \in V _{\alpha} \subset G$. In other words: every open subset of $X$ is sum of some sets of {$V_ {\alpha}$}.

Prove that every metric space which contains countable dense subset has countable basis.

Take the open balls with radii $\frac 1n$ centered at the countably many points in the dense countable set.

:first: perfect solution  :) - it is THAT easy though I like it


calculus
integration
geometry
trigonometry
calculus computations

[quote="kunny"]In the $ x \minus{} y$ plane, find the area of the region bounded by the parameterized curve as follows.

$ \left\{\begin{array}{ll} x \equal{} \cos 2t & \quad \\
y \equal{} t\sin t & \quad \end{array} \right.\ (0\leq t\leq 2\pi)$[/quote]

The curve is not simple, it intersects itself at the point (1,0). So break it up into

$ C_1: \vec{r}(t)\equal{}< \cos 2t, t\sin t>,\mbox{ for }0\leq t\leq \pi$ (Negatively orientated)

$ C_2: \vec{r}(t)\equal{}< \cos 2t, t\sin t>,\mbox{ for }\pi\leq t\leq 2\pi$ (Positively orientated)

Green's Theorem gives the area bounded by the curve as

\[ A\equal{}\minus{}\oint_{\minus{}C_1\cup C_2} y\, dx \equal{} \minus{}2\int_{0}^{\pi}t\sin t \sin 2t\, dt\plus{}2\int_{\pi}^{2\pi}t\sin t \sin 2t\, dt\]

replace $ t$ with $ t\minus{}\pi$ in the second integral to get

\[ A\equal{} \minus{}2\int_{0}^{\pi}t\sin t \sin 2t\, dt\minus{}2\int_{0}^{\pi}(t\minus{}\pi)\sin t \sin 2t\, dt\]
\[ \equal{} \minus{}4\int_{0}^{\pi}t\sin t \sin 2t\, dt\plus{}2\pi\int_{0}^{\pi}\sin t \sin 2t\, dt\]
\[ \equal{} 2\int_{0}^{\pi}(\pi \minus{}2t)\sin t \sin 2t\, dt\equal{} 4\int_{0}^{\pi}(\pi \minus{}2t)\sin^2 t \cos t\, dt\]

now by parts with $ u\equal{}\pi\minus{}2t$

\[ \equal{} \left[ \frac{4}{3}(\pi \minus{}2t)\sin^3 t\right]_{t\equal{}0}^{\pi}\plus{}\frac{8}{3} \int_{0}^{\pi}\sin^3 t\, dt\]
\[ \equal{}\frac{8}{3} \int_{0}^{\pi}\left(\sin t\minus{}\sin t\cos^2 t\right)\, dt\]
\[ \equal{}\frac{8}{3}\left[\minus{}\cos t\plus{}\frac{1}{3}\cos^3 t \right]_{t\equal{}0}^{\pi}\equal{}\boxed{\frac{32}{9}}\]

That's correct answer.

This problem seems to be difficult for applicants for Tokyo University, because Japanese high school students don't study Green's theorem.


Discuss our plans to win the Spirit Award and win the Team Round... MUAHAHAHAHA  :rotfl:  :rotfl:  :rotfl:

You just lost the Spirit Award. Thank you, have a good nats trip.

The four MO team people please post your heights here.

shhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh............................
we don't want to hint at our plan....

Although asking for heights doesn't give away much...

Just letting you guys know that I am going to be 10-15 minutes late for Mathcounts Practice at Mewto's house tomorrow night...

Okay. I'll have little RomanianGenius go tell mom.

Something tells me you guys will win the Spirit Award...

Unless the CA team puts on some epic rickroll...*COUGH MILLER4MATH COUGH*

Although there has been a horrendous amount of spam and useless posts/threads in this forum, I would like to say congratulations to everyone on the Missouri national MC team, being a former Missouri competitor myself.  As everyone knows, Missouri's best rank was second as a team; since all four of you are from Ladue, please, [b]please[/b] make Missouri proud (as well as St. Louis) by at least getting first  :D .   After all, you all live right next to each other, so there's no reason to not get enough practice in, and overall, do your best, congratulations, and make sure at least half the team gets into the Countdown Round.

Thanks Michael, we will try to make you proud by being the first team to be kicked out of National MC.

Is that good?

I don't think that's what he meant by first...

...and I hope you were joking.

[size=200]EPIC WIN....KANSAS GOT FIRST ONE YEAR AT NATS

MISSOURI = EPIC UBER PHAILURE

KANSAS = EPIC UBER WIN[/size]

So that makes the mean place over your 25 years of MC what, 57x24+1 /25=54.76

No Mewto. We definitely beat the Virgin Islands and Guam at least 3 times.

Oh. Well then. It would appear my maths is flawed.

Perhaps your math is just too beastly for mankind to grasp.

SO BEASTLY MMM

This would explain why I am currently failing pre-algebra...

[quote="Mewto55555"][quote="ZzZzZzZzZzZz"][quote="Mewto55555"]Oh dang I totally just remembered. My first ever official coutndown was when ZzZzZ...Z wiped the floor with me and Sam at chapter. He buzzed in like .5 sec and the fail judge guy read the whole problem before calling on him to give the answer  :huh:[/quote]

And then I believe I lost to Shuyang...  Strange how being up against Ladue kids gives you motivation to do well, but then with no more strong competition, the mind just goes blank...[/quote]

Yeah and then you lost first round at state too, and then the same thing happened at nationals  :rotfl:[/quote]

Awful times... Awful times...


Let a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ that follows the conditions:
a) $f(0) = 0$ and $f(2) = 2$.
b) For all $a \in \mathbb{R} - \{0\}$ the tangent line to $f$ on $P(a,f(a))$ meets x-axis on $A$ and y-axis on $B$ such that $A$ is the line-bissector of $\overline{BP}$.

Find $f(3)$.

Solve the differetial equation $f'(x)=\frac{2}{x}f(x)$ and $f(2)=2,$ yielding $f(x)=\frac{1}{2}x^2.$
Thus $f(3)=\frac{9}{2}.$


Be a,b,c  roots of $P(x)x^3+px^2+qx+r=0$ . If $S_n=a^n+b^n+c^n$, n is integer and n>3, being
 $K=S_n+pS{}_n{}_-{}_1+qS{}_n{}_-{}_2$ Find K

what are you asking?  $K$ in terms of $S$?  if thats the case figure out $S_1, S_2, S_3$ using relationships btwn roots and coefficients.

then write $x^3=-px^2-qx-r$
so $x^4=-px^3-qx^2-rx$.

so then $S_4=-pS_3-qS_2-rS_1$  and continue like this so you have

$S_n=-pS_{n-1}-qS_{n-2}-rS_{n-3}$, so $S_n+pS_{n-1}+qS_{n-2}=-rS_{n-3}=K$.


Is 1.9999999999999999999... recurring a rational number?

[hide="working"]let
x=1.999...
10x=19.9999
9x=18
x=2
2 is rational
Therefore as x=2, x is rational[/hide]

[hide="ambiguity of the solution"]In this case we proved that 2 is rational and when x is 2, x is rational, but in the original assumption, we said x=1.999...
so in a way we did not prove that 1.999... is rational at the end.
if this is the case, we can never prove that it is rational
therefore it is not rational (?)

reasoning seems flawless but logic doesn't seem to tie up for either cases???[/hide]

It is very well known that $ 0.9999999...\equal{}1$.
It is not smaller than 1 by some infinitesimal value; this is a common misconception.

[hide="Proof"]
Let $ x\equal{}0.99999...$
Then $ 10x\equal{}9.99999...$
And $ 10x\minus{}x\equal{}9x\equal{}9$.
Dividing both sides by $ 9$, we get $ x\equal{}1$.

Therefore, $ 0.99999...\equal{}1$
[/hide]

Therefore, $ 1.99999...\equal{}2$, which is obviously rational.

There exists the following remarkable result:

$ 0.aaaaa...\equal{}\frac{a}{9}$, where $ a$ $ \in$ $ \{ 0,1,2,3,4,5,6,7,8,9 \}$.

I don't know if I'd call it "remarkable" :wink:

I think you can generalize it for all $ n\in\mathbb Z$.
If $ n \equal{} \overline{a_0a_1\cdots a_m}$, and we want the fractional form of the recurring decimal expansion $ 0.\overline{a_0a_1\cdots a_m}$, we let $ x$ be that value, multiply $ x$ by $ 10^{m \plus{} 1}$, and subtract $ x$ away to get: $ x(10^{m\plus{}1}\minus{}1)\equal{}n$

And therefore, the fractional form is $ \frac {n}{10^{m \plus{} 1} \minus{} 1}$.

If $ n \equal{} a$, $ m \equal{} 0$, and the fraction is $ \frac {a}{9}$, as expected.

I want to address a different step in your logic.

We can never prove that it is rational
Therefore, it is not rational.

The above two steps rely on the false assumption that if something is true, it can be proven. See [url=http://en.wikipedia.org/wiki/Godel%27s_incompleteness_theorem]Godel's Incompleteness Theorem[/url].

In other words, the provability of a statement does not imply anything about that statement's truth.


What is the smallest possible value of $\left|12^{m}-5^{n}\right|$, where $m$ and $n$ are positive integers?

The value can't be zero, by the fundamental theorem of arithmetic. Also, note that $|12^{m}-5^{n}|$ can't be divisible by $2,3$ or $5$, hence the expression can't be equal to $2,3,4,5,6$. Assume it can be equal to $1$. Then either $12^{m}-1=5^{n}$ or $5^{n}-1=12^{m}$. The first case is impossible as LHS is divisible by $12-1=11$ and RHS is not. The second case, for modulo $6$, gives $(-1)^{n}-1\equiv 0\pmod{6}\implies n=2k$. Hence $25^{k}-1=12^{m}$. Since the unit digit of LHS is $4$, that gives $m=4l+2$, hence $25^{k}-1=12^{4l+2}$. Now RHS is divisible by $9$, hence $25^{k}\equiv(-2)^{k}\equiv 1\pmod{9}$, which gives $k=3s$. Therefore $25^{3s}-1=12^{4l+2}$. However, the LHS is divisible by $25^{3}-1=(25-1)(625+25+1)=24\cdot 651=2^{3}\cdot 3^{2}\cdot 7\cdot 31$, and RHS isn't divisible by either $7$ or $31$. Therefore the initial expression can't be equal to $1$. It can be equal to $7$ for $m=n=1$, thus we conclude that $7$ is its minimal value.


Let $p$ be a prime number. Show that $\sigma\in S_n$ is of order $p$ if and only if $\sigma$ can be written as products of commuting $p$-cycles.


geometry
rectangle
perimeter
symmetry
geometry solved

A rectangle has a perimeter of 15cm. What is it`s maximum volume if the rectangle starts spinning around one of it`s sides shaping a cylinder?

I assume one side is: x
I assume one side is: x
The other side is : (15-2x)/2
The other side is : (15-2x)/2

A cylinders volume is : π*r^2*h
Where π=3.14159265359
r: is half of one side as one whole side becomes the cylinders diametre when it starts spinning.
and height is a full opposite site.

I assume  it starts spinning on the side (15-2x)/2 which becomes the cylinders diametre and x its hight:
But the forumla for a cylinders volume needs the radius which is: ((15-2x/2)/2) = (15-2x)/4

The formula for the cylinders volume then is: π*x((15-2x)/4)^2

the Above calculation results in a volume for the cylinder as :
(225πx) - (60πx^2)+(4πx^3) /16

As we want to know the maximum volume we derive(typo) it:

V`=37,699x^2 - 376,991x + 706,858
This is a simple 2nd degree equation resulting in x1=(7,5) x2=2,5

Now to get the maximum volume we put 2,5 into the Volume formula since this is a maximum:

(225π*2,5) - (60π*2,5^2) - (4π2,5^3) / 16 = 49,08

BUT HERE is my problem: The answer ought to be 196 and therefor (225π*2,5) - (60π*2,5^2) - (4π2,5^3) / [b]4[/b]
is the correct volume formula for the cylinder.

But since one side of the original rectangle is (15-2x/2) then the raidus for the cylinder is ((15-2x/2)/2)

My fault accurs as I get the forumla for the cylinder to be split by 16, it should be 4, can anyone see where in the transitiation/calculation from rectangle to cylinder and coverting diametre to a radius I do wrong since the above calculation is correct except the volume is to be split by 4 and not as I get 16.

Thanks in advance, Stephan

[quote="gossen19"]r: is half of one side as one whole side becomes the cylinders diametre when it starts spinning...
[/quote]
No, cylinder radius is one whole side. The rectangle is not spinning around one of its symmetry axes, but around one of its sides.


http://www.youtube.com/watch?v=VjgidAICoQI
What are your opinions?

Neal Adams is a comic book artist, not a scientist.

http://en.wikipedia.org/wiki/Expanding_earth_theory

A takedown of Neal Adams is available [url=http://scienceblogs.com/goodmath/2006/12/wacky_physics_it_must_be_right_1.php]here[/url].


I just took the SAT 1.
hope to get a 800 in math. does any one remember a mathquestion where a stupid mistake could be done?

in the past one i got 1 question wrong.... i always do silly mistakes.. :?

To be honest, I didn't see any today that may have caused stupid mistakes? I mean...I suppose there was that one question with the chart and it asked for the number of books sold in a certain week...(this better not count as revealing questions) I mean you could've read the chart wrong...but otherwise...eh...

[quote="manuel"]true,
i think the answer to this oen was 46,000 if i'm not mistaken...
i think i was careful to read the chart well (i am the kind of person that could have read it wrong)[/quote]

Different problem.

i almost missed the $m^2-n^2=12$, find values of $(m-n)$ question. then i realized 1 didn't work.

[quote="Rushil"]What about  SAT II, how'd it go for ya!!![/quote]

Math IIC was a piece of cake.

Physics... was another story. I left 4 blank, which might kill my 800. I hope not, though, there's just too much material to study and formulas to memorize. The problems were easier than the AP test, but they don't give you formulas which is stupid; I forgot some of the waves/sound/optics ones.

[quote="manuel"]what answer did u selected inthis one..
2 right?[/quote]
 :(  now i realize only 1 is right. yes i did put 2 right. there goes the 800.

oh... so u got 790? that's good too,
but wait i don't get it, what were the possible answers to this question?

btw, i wrote the best essay i've done in my life!!! so i hope toget a good writing score too

Edit: Oops. Forgot that you can't reveal questions.  ;)

That retarded essay! Stopped in mid-sentence on my third paragraph!

[quote="Forsaken"]Edit: Oops. Forgot that you can't reveal questions. [/quote]
don't worry. we won't tell anyone.  ;)

and i wonder how they will like an essay that isn't divided into paragraphs. i just couldn't find a spot where my ideas transitioned where paragraphs were needed. o well.

[quote="paladin8"]

Physics... was another story. I left 4 blank, which might kill my 800. I hope not, though, there's just too much material to study and formulas to memorize. The problems were easier than the AP test, but they don't give you formulas which is stupid; I forgot some of the waves/sound/optics ones.[/quote]

it has a nice curve. i got an 800 after guessing on 4 or 5, and im sure i missed a few more than that.

Ugh. Essay.  :wallbash_red: I don't GET IT. I wrote for the entire 25 minutes, right? And all I got was a page and a half in normal writing. But the girl next to me? (I glanced over after we were closing the tests up) She wrote a FULL TWO PAGES... and her writing was [size=59]tiny[/size]. HOW? How do people [i]do that????[/i]

I wrote two full pages because i did an essay in the same topic for school...that's good luck :D

writing two pages shouldn't be that difficult. i spent the first 10 minutes trying to decide which side to argue and then come up with details. so i wrote a page and a quarter in around 15 minutes. 2 pages isn't out of the question at all.

If I didn't even start my conclusion, what's the max score I can get? If I had finished,
I think I would have gotten a 10, so how much would that be without a conclusion?

Man, I just realized that I have at least 2 wrong on math and 3 wrong on CR. I hope I can pull a 2200.  :(

Lol unless you guys are seniors, there's really no need to worry. Plenty of more opportunities.

As for the essay, note that some people just memorize a format and plug in the essay to an outline they've already decided on. Lol. It's very difficult to just see a question and start writing madly for 25 minutes unless you do that, especially if you're trying to get a full 2 pages at reasonable size.

Essay scoring depends on the length of the essay as well. So unless you've written more than 3/2 pages, an 11-12 is out of the question and a 9-10 is pushing it. Otherwise, it just matters on how coherent your essay is and whether or not you've used correct syntax, good vocab, and supported your thesis. My two, unprofessional cents on that matter...

I am a senior.  :(

At least I know for sure that the 3 paragraphs I wrote this time are better than the ones I wrote on the March test. (Got a 6/12)  :blush:

[quote="Rushil"]What about  SAT II, how'd it go for ya!!![/quote]

math II C >> easy(answered every question) > should get 800

physics>> easy >> left 2 blank>> should get 800

chemistry>> Super easy>> answered every question except one which i think was WRONG!! the choices were incorrect, that would be q 57 i think >> should get 800.

thats it. :)

Which question are you talking about in Chem (Sat II) that  you think is wrong. PM me!!!

For people that took the Physics test, out of curiousity did you guys use a review book? If so, which one(s)? And were they any good?

Same with chem.

i took AP physics B, C and AP chem last year. and got 5's in all of them, so i didn't have anyproblem reviewing stuff,
but i went to the library and checked out  the Kaplans SAT review for chem (which wasn't that good!), and also i looked at the SAT physics book by princeton hall the night before the test, that was pretty good, bettar than the Kaplan one.

oh i forgot to mention that the Kaplan book that i said, was for 03-04. so it was a bid too old, but i don't like Kaplan anyways.

~amir

for chem (which i took a couple years ago), i took the AP test in may (got a 5) and then took the SAT II in June. I was totally out of practice, didn't even go over the stuff til two days before the SAT II, and I still got a 760. i used princeton review, which isnt really the best review book, but i like the way theirs is organized. (you have to be very careful when going over math-type problems, because they most definitely WILL make mistakes).

[quote="amirhtlusa"][quote="paladin8"]

Physics... was another story. I left 4 blank, which might kill my 800. I hope not, though, there's just too much material to study and formulas to memorize. The problems were easier than the AP test, but they don't give you formulas which is stupid; I forgot some of the waves/sound/optics ones.[/quote]

well in the AP test they don't give the formulas up to the FR section either.

but SAT physics is way easier than AP test (not that AP test is hard, cause its not!). for answering the MC problem all they ask you in the multiple choice problems is to know the basic knowledge of the subject, thats it.  :)[/quote]

True; in general the questions were a lot easier. But the AP test is graded on a ridiculous curve, so if you forget like 5 formulas it doesn't hurt your chances of pulling an easy 5. I don't know about the SAT, though. The curve is supposed to be generous, but how generous is the question.

Kaplan = bad review books.

I would suggets PR like most people, but I also like 5 Steps to a 5 for APs. Or barrons.


calculus
integration
geometry
derivative
function
geometric series
calculus computations

Find the area under the curve $ f(x)\equal{}4x^3 \plus{} 3x^2 \plus{} 2x \plus{}1$ from $ x \equal{} 0$ to $ x \equal{} 2$.

You need to tell us the value of $ f$ and its first derivative at some point otherwise the answer is not unique.

Because of the constants and stuff that are canceled out when we find the derivative? I'll just change the problem. Sorry about that, I'm still learning all this calculus stuff.

Do you know how to find the antiderivative of $ f$? If so, evaluate it at $ x \equal{} 2$ and $ x \equal{} 0$ and then subtract its value at $ x \equal{} 0$ from its value at $ x \equal{} 2$.

[hide="Solution"]
From the rule $ f(x)\equal{}x^{n}\implies f'(x)\equal{}nx^{n\minus{}1}$, and noticing the pattern in the function $ f$ given ($ 4x^3$, etc), the antiderivative is pretty obvious.

$ F(x)\equal{}\int f(x)\; dx\equal{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}C$

By the fundamental theorem of calculus, the answer is $ F(2)\minus{}F(0)\equal{}\frac{2^{5}\minus{}1}{2\minus{}1}\minus{}1\equal{}\boxed{30\text{ square units}}$.

(That last step used the sum of a finite geometric series to skip a little calculation).
[/hide]


trigonometry
induction
LaTeX
linear algebra
matrix
geometry
ARML

My hope is that we can get more interest in proofs by doing a proof marathon.  The power question score for SC needs to come up.  Anyways, here is the first item that needs to be proven.  This is a quite famous identity.  I will refrain from naming it in hopes that you guys will try to prove it on your own.  Prove:  \[{ ( \ cos\theta + i\sin}\theta )^n = \cos n\theta + i\sin n\theta , n\in {\rm Z}  \]

[hide="HINT"] Show that the identity is true for positive integers first by using mathematical induction[/hide]

[hide="Hint"]
Have you even tried induction yet?[/hide]

i don't know if this attempt is strong enough to fully prove it, but it's proof by induction

show it works for n=1:
$\cos\theta+\i\sin\theta=\cos\theta+\i\sin\theta$
assume it works for some n=k:
$(\cos\theta+\i\sin\theta)^k=\cos(k\theta)+\i\sin(k\theta)$
show it works for n=k+1:
$(\cos\theta+\i\sin\theta)^{k+1}=\cos((k+1)\theta)+\i\sin((k+1)\theta)$
$(\cos(k\theta)+\i\sin(k\theta))(\cos\theta+\i\sin\theta)=$
$\cos(k\theta)*\cos\theta+i\sin(k\theta)*\cos\theta+i\sin(\theta)*\cos(k\theta)-\sin(k\theta*\sin\theta=$
combining real and imaginary parts and using trig identies:
$\cos(k\theta+\theta)+i\sin( k\theta+\theta)=$
$\cos((k+1)\theta)+i\sin((k+1)\theta)$
which is what you would get by pluging in k+1 into the equation. this satisfies positive integers.
to show negative integers, joe said to use  (cis theta)^n where n is negative (i.e. n=-k for positive k) is the same as 1/(cis theta)^k but i don't feel like doing that right now. maybe shobhit will help me out. i'm $\text{\LaTeX}$ed out :(
and if there are any errors in the proof, deal with it and assume i meant to put the correct thing  :P

I thought I told you *not* to use \i and just put i?  I don't even know what those \i 's are :?

Wow...you posted exactly what i said over IM.  If I knew that was going to be posted here I would have said it more clearly ;)

$\text{Problem 2}$

Let $\{F_i\}_{i=1}^\infty$ denote the Fibonacci sequence.

Prove that \[ \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_n & F_n\\ F_n & F_{n-1}\end{pmatrix}  \]

[quote="joml88"]I thought I told you *not* to use \i and just put i?  I don't even know what those \i 's are :? [/quote]
well, some of them came out ok  :blush:

and problem 2 is pretty cool. i did it recently so i'll let someone else tackle it (cough*shobhit*cough)

Guys,
Furious made a big "no no" in the world of proofs.  You cannot start off by equating things you want to prove.  You need to start with one side until it looks like the other side of the equation.  Now, that is not to say that you can't figure out how to write the proof by equating things you want to show, but when you finally write the proof up, you need to work from one side of an equation until you can get to the other.  Otherwise, you are assuming that the claim is already true when it may not be, e.g. Prove 1 = 2.

He wrote what he wanted to prove first and then worked entirely on the LHS showing that it is equivalent to the RHS.  So I think his proof is OK.  It could use a lot of work on neatness/clarity/etc though.

Yeah,
I should have looked more closely.  To be honest, I did not read the whole thing.

[quote="furious"]and problem 2 is pretty cool. i did it recently so i'll let someone else tackle it (cough*shobhit*cough)[/quote]
[quote="furious"] but i don't feel like doing that right now. maybe shobhit will help me out.[/quote]
For now, I will watch you guys solve these proofs and learn from your solutions. That is atleast how I improve in certain areas of math, I watch before I attempt! Give me about a month, and in April, a month before ARML, I'll start doing these ;)

But if you insist, here's how i would do the negative numbers part, let $x=-y$, so therefore, we have:

$(cis\theta)^x=(cis\theta)^{-y}=((cis\theta)^y)^{-1}$, if we assume demoivres theorem is true, we have $cisx\theta=cis{-y\theta}$ but since $x=-y$, then that holds true for negative numbers...is that good enough?

[quote="Iversonfan2005"]
For now, I will watch you guys solve these proofs and learn from your solutions. That is atleast how I improve in certain areas of math, I watch before I attempt! Give me about a month, and in April, a month before ARML, I'll start doing these ;)[/quote]

You should still start trying now.  It's easier for us to tell you when you've done something wrong.  Additionally, you learn much more from doing it yourself.  Watching someone else prove something is far different from actually doing it yourself.

Joe,
I do not think the Fibonacci matrix equation is correct.  Take a look, and you should see.

Hey guys, Jim here.  Mr. R. is finally forcing me to deal with computers.  (AHHHH!)

Here is the proof for problem 2 (note correction):
P(n) = $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix}$

We will argue by induction that this holds true for integers $n>1$
First, we show that is is true for $n=2$:

P(2) = $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^2 = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} F_3 & F_2 \\ F_2 &{ F_1}\end{pmatrix}$

Now, assume this is true for $n=k$ i.e.

P(k) = $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1}\end{pmatrix}$

Now we plug in $k+1$

P(k+1) = $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{k} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1}\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_{k+1} + F_{k+1} & F_k \\ F_k + F_{k-1} & F_k\end{pmatrix} = \begin{pmatrix} F_{(k+1)+1} & F_{k+1} \\ F_{k+1} & F_{k}\end{pmatrix}$

So $P(k) \Rightarrow P(k+1)$ and hence, by PMI, the proposition is true for all integers $n>1$

Here is Problem 3, sticking with matrices:

Prove:

$\begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} {2^n} & 2^n-1 \\ 0 & 1 \end{pmatrix}$

for all positive integer values of n.

Also, determine whether or not this holds true for $n=-1$ :ninja:

[hide="Solution"]
We begin with the base case, $n=1$:

\[ \begin{bmatrix} 2 & 1 \\ 0 & 1\end{bmatrix}^1 = \begin{bmatrix} 2^1 & 2^1-1 \\ 0 & 1\end{bmatrix} \]

which obviously holds.

Now, we assume that the assertion is true for $n=k$:

\[ \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}^{k} = \begin{bmatrix} 2^k & 2^k-1\\ 0 & 1\end{bmatrix}. \]

We proceed to show that it holds for $n=k+1$:

\[ \begin{eqnarray*} \begin{bmatrix} 2 & 1 \\ 0 & 1\end{bmatrix}^{k+1} &=& \begin{bmatrix} 2 & 1 \\ 0 & 1\end{bmatrix}^k\begin{bmatrix} 2 & 1 \\ 0 & 1\end{bmatrix}\\ &=& \begin{bmatrix} 2^k & 2^k-1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 0 & 1\end{bmatrix}\\ &=& \begin{bmatrix} 2^{k+1} & 2^{k+1}-1\\ 0 & 1\end{bmatrix} \]

So the assertion is true by the Principle of Mathematical Induction (PMI).[/hide]

Let's try something other than induction....

$\text{Problem 4}$

Prove that $\sqrt{11}$ is irrational.

Let's say $\sqrt{11}$ is rational, then it can be written in the form $11=\frac{a^2}{b^2}$, so therefore $b^2 \cdot 11=a^2$, because an even divided by an even results in an even number, both $a$ and $b$ are odd. But no odd square divides into another odd square unless they have they can be written with the same base. But in this case, this is impossible since the exponent will always be an even number, therefore, $\sqrt{11}$ is irrational.....maybe.... :? $\text{QED?}$

An even divided by an even is always even?  How about 6/2 or 40/8?  That's a really good attempt for not having seen a problem like this before.  Try again though, cause you're thinking in the right direction.

[quote="joml88"]An even divided by an even is always even?  How about 6/2 or 40/8?  [/quote]

wow,  :|  :huh:

ok, just gotta think some more

EDIT: John, why did u delete your post? i'm not gonna post the answer now cuz i'll just copy you word for word

Sorry, I posted that before I saw all this extra stuff, so I was going to get rid of it and let you think more. Seeing as I usually post things really early in the morning, I'm not used to posting at the same time as other people. Anyway...

Suppose $\sqrt{11}=\frac ab$ in simplest terms. Then $11b^2=a^2$ and $11|a$. Let $a=11c$; then $b^2=11c^2$ and $11|b$. Therefore, $\frac ab$ is not in lowest terms, which is a contradiction. QED

$\text{Problem 5}$
Prove that e is irrational, assuming $e=1+\frac 1{1!}+\frac 1{2!}+\frac 1{3!}+...$


This question appears in the section on transcendence bases in Hungerford's Algebra.

If $F = K(u_{1},\ldots,u_{n})$ is a finitely generated extension of $K$ and $E$ is an intermediate field, then $E$ is a finitely generated extension of $K$.

do you know german? :-D

http://www.matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=49649

Sorry but I'm afraid i don't know german.


[img]http://i165.photobucket.com/albums/u46/137456/polemath.png[/img]

Two poles with diameters 10 and 30 are bound together by a metal band. How could I find the length of the metal band?

Knowing that tangents are perpendicular to the radiuses of the circles, I tried drawing trapezoids and such, but to no success. =(

[img]http://ipicture.ru/uploads/080604/a4q1JcZ7Yi.png[/img]
We have $ AD\equal{}5$, $ CE\equal{}15$. Draw perpendicular from $ A$ to $ EC$. THen $ ADEF$ is rectangle and $ AD\equal{}EF\equal{}5$, so $ FC\equal{}15\minus{}5\equal{}10$. $ A,B,C$ are colinear and $ AC\equal{}5\plus{}15\equal{}20$. Then by Pythagora $ DE\equal{}AF\equal{}\sqrt{AC^2\minus{}CF^2}\equal{}\sqrt{400\minus{}100}\equal{}10\sqrt{3}$

Ah, I can't believe I never thought of that. From that triangle I was able to find the angles to also find what fraction of the circumfrences were being touched by the band and added it to 2(DE) to find the answer. Thanks!


I have this idea.

Why don't we create a separate page in AoPs wiki for each of the books listed [url=http://www.artofproblemsolving.com/Wiki/index.php/Math_books]here[/url] and then one that separate page, we can let users give their in-depth review of it if they own the book and then give it a rating and maybe other stuff. Over time, more and more reviews will accumulate.

I think this will be very helpful to those people looking to buy certain books especially if the reviews are from AoPSers. (and it will be very organized).

If you make this I have no objection [i]as long as[/i] you make the page for the book and then make a separate page for reviews.

For example, the main article could be:

Challenging Problems in Geometry

and the review page would be:

Challenging Problems in Geometry/Reviews

A link to the second would be on the first, of course.  Using this format should keep things neater.

I fifth that! (I think) :lol:
I may be able to review some of the well-known books (AOPS vol.1 and 2, ACOPS, geometry revisited, etc) once I finsih some of them.


Determine all positive integers $k$ such that \[\frac{d(n^{2})}{d(n)}= k\] for some $n \in \mathbb{N}$.

Let $ n\equal{}\prod p_{i}^{a_{i}}$, so $ d(n)\equal{}\prod (a_{i}\plus{}1)$ and $ d(n^{2})\equal{}\prod (2a_{i}\plus{}1)$. To have $ d(n^{2})\equal{}kd(n)$, obviously $ k$ must be odd.

We will prove that for any odd $ k$ it works. Obviously $ k\equal{}1$ works, so let $ w$ be the smallest odd positive integer for which it doesn't, and write $ w\equal{}2^{m}k\minus{}1$, with $ k<w$ an odd positive integer. So, $ k\equal{}\frac{d(n_{0}^{2})}{d(n_{0})}\equal{}\frac{\prod (2a_{i}\plus{}1)}{\prod (a_{i}\plus{}1)}$. Defining $ b_{i}\equal{}2^{i}((2^{m}\minus{}1)k\minus{}1)$, we get $ \frac{\prod (2b_{i}\plus{}1)}{\prod (b_{i}\plus{}1)}\equal{}\frac{2^{m}k\minus{}1}{k}$, so $ \frac{2^{m}k\minus{}1}{k}\equal{}\frac{d(n_{1}^{2})}{d(n_{1})}$.

We can pick $ n_{1}$ such that $ \gcd(n_{0},n_{1})\equal{}1$ (since we're just constructing 'an' $ n_{1}$), thus $ \frac{d((n_{0}n_{1})^{2})}{d(n_{0}n_{1})}\equal{}\frac{d(n_{0}^{2})}{d(n_{0})}\frac{d(n_{1}^{2})}{d(n_{1})}\equal{}2^{m}k\minus{}1\equal{}w$ and we have constructed a valid $ n$ for $ w$, contradiction. So it is true for all odd positive integers.


algebra
polynomial
calculus
integration
real analysis
real analysis unsolved

algebraic complex z if and only if
there are integers $a_{i}$  with $i=1,n$ not all 0
$\sum_{i=0}^{n}a_{i}\cdot z^{n-i}=0$

how do i show that the set of algebraic numbers is a countable one?

Let $P_{n}$ be the set of all polynomials of degree at most $n$ with integral coefficients. Then the map $c_{n}x^{n}+\dots+c_{1}x+c_{0}\mapsto (c_{n}, \dots, c_{1}, c_{0})$ is an injection from $P_{n}$ to $\mathbb{Z}^{n+1}$. Hence each $P_{n}$ is countable. It follows that the set of all polynomials with integral coefficients is countable. Each polynomial has finitely many roots, so the set of algebraic numbers is countable.


Let be $ a,b\in \mathbb{N}$ such that $ \sqrt{a}\plus{}\frac{1}{\sqrt{b}}\in \mathbb{N}$ . Prove that $ a,b$ are perfect squares .

Let $ \sqrt{a}\plus{}\frac{1}{\sqrt{b}}\equal{}x$
Then $ (x\minus{}\frac{1}{\sqrt{b}})^2\equal{}x^2\minus{}\frac{2x}{\sqrt{b}}\plus{}\frac{1}{b}\equal{}a$

So $ \sqrt{b}\equal{}\frac{q}{p}$ is rational. It's easy to see that $ \sqrt{b}$ is an integer.

Then from the original question $ \sqrt{a}$ is also rational $ \Leftrightarrow \sqrt{a}$ is an integer also, done.


trigonometry
Euler
geometry unsolved
geometry

Given the triangle $ ABC$ and $ DEF$ is its intouch triangle. Let $ D^'$ be the reflected image of the point $ D$ through the line $ EF$ ( $ ED^' = ED$ and $ FD^' = FD$). Let $ A_0$ be the intersection of the line  $ AD^'$ and the line $ BC$. $ B_0$ and $ C_0$ are determined similarly. Prove that $ A_0$, $ B_0$ and $ C_0$ are collinear.

[hide="Hint "]Prove that the points $ A_0,I,C_0$ are collinear and your problem become obvious :wink: .
[/hide]

Here is complete proof,but it is very ugly proof,with using trigonometry:
Let $ \angle BAC \equal{} 2\alpha,\angle ABC \equal{} 2\beta,\angle BCA \equal{} 2\gamma$.Then it is easy to understand that:
$ \frac {\sin\angle C_0CA}{\sin\angle C_0CB} \equal{} \frac {FE}{ED}\cdot\frac {\sin\angle(\beta \minus{} \gamma)}{\sin\angle(\alpha \minus{} \gamma)} \equal{} \frac {\cos\alpha}{cos\beta}\cdot\frac {\sin\angle(\beta \minus{} \gamma)}{\sin\angle(\alpha \minus{} \gamma)}$.
Hence:
$ \frac {BC_0}{C_0A} \equal{} \frac {BC_0}{C_0C}\cdot\frac {C_0C}{C_0A} \equal{} \frac {\sin\alpha\cdot\sin\angle(\alpha \minus{} \gamma)}{\sin\beta\cdot\sin\angle(\beta \minus{} \gamma)}$.
Similarly:
$ \frac {AB_0}{B_0C} \equal{} \frac {\sin\gamma\cdot\sin\angle(\beta \minus{} \gamma)}{\sin\alpha\cdot\sin\angle(\beta \minus{} \alpha)}$
$ \frac {CA_0}{A_0B} \equal{} \frac {\sin\beta\cdot\sin\angle(\beta \minus{} \alpha)}{\sin\gamma\cdot\sin\angle(\alpha \minus{} \gamma)}$
Now if we apply Menelaus theorem to $ \triangle ABC$ and transversal $ \overline {A_0B_0C_0}$,we'll obtain the result. :)
Do you need in more details?
Best regards.

1). $ DD'$, $ EE'$ and $ FF'$ altitudes in $ \triangle{DEF}\Longrightarrow$ point $ \{H_1\} \equal{} DD'\cap EE' \cap FF'$ is ortocentre in $ \triangle{DEF}$.
2). $ IH_1$ euler line of $ \triangle DEF$ and $ A_0, B_0, C_0 \in IH_1.$

My hint can be proved using relation(see my trig. proof) $ \frac {BC_0}{C_0A}$ and barycentric coordinates. :wink:

[quote="mihai miculita"]
2). $ IH_1$ euler line of $ \triangle DEF$ and $ A_0, B_0, C_0 \in IH_1.$[/quote]

... and I remember proving this at http://www.mathlinks.ro/Forum/viewtopic.php?t=20666 ...

  darij


induction
number theory proposed
number theory

Let us consider a simle graph with vertex set $ V$. All ordered pair $ (a,b)$ of integers with $ gcd(a,b) \equal{} 1$, are elements of V.
$ (a,b)$ is connected to $ (a,b \plus{} kab)$ by an edge and to $ (a \plus{} kab,b)$ by another edge for all integer k.
Prove that for all $ (a,b)\in V$, there exists a path fromm $ (1,1)$ to $ (a,b)$.

[hide="hint"]
If $ a,b$ are negative it is always joined to a point with positive arguments.
use induction on $ |a|\plus{}|b|$.
use the fact that $ (a,b)\equal{}(a,a\minus{}b)\equal{}(b,a\minus{}b)\equal{}1$ to deduce there exist $ l,m$ such that $ la\plus{}m(a\minus{}b)\equal{}1...$.
prove that $ (a,b)$is connected to $ (a,a\minus{}b) or (b\minus{}a,b)$
as $ a>b \text{ or }b>a$.[/hide]

Actually the problem is not hard. As you say, Bezout's Identity application will do it here consdiering that ample information is given in the question that $ (a,b)$ is connected to $ (a, b\plus{}kab)$ and $ (a\plus{}kab, b)$ for all $ k \in \mathbb{Z}$.So this with this along with Bezout's identity will give you insight into constructing the path from $ (1,1)$ to $ (a,b)$.

How to construct a path from (a,b) to (a-b,b) using bezout's theorem..??


Consider the system
$ x_1'\equal{}x_1\plus{}x_2$, $ x_2'\equal{}x_2\plus{}x_3$, $ x_3'\equal{}x_3$

Find the solution satisfying $ x_1(0)\equal{}2, x_2(0)\equal{}3, x_3(0)\equal{}4$

[quote="countdownmath"]Consider the system
$ x_1' \equal{} x_1 \plus{} x_2$, $ x_2' \equal{} x_2 \plus{} x_3$, $ x_3' \equal{} x_3$

Find the solution satisfying $ x_1(0) \equal{} 2, x_2(0) \equal{} 3, x_3(0) \equal{} 4$[/quote]

We find $ x_3$. Then we find $ x_2$. Then we find $ x_1$.

We have $ x_3 \equal{} 4e^t$, $ x_2 \equal{} e^t(3 \plus{} 4t)$, $ x_1 \equal{} e^t(2 \plus{} 3t \plus{} 2t^2)$.


Sorry about the wait.  I will be busy with school from now on so if I am a little slow with things don't worry too much.

Round 5 Matchups

HashBrownie    vs.   theone853

white_horse_king88    vs.    nr1337

tandjsnell   vs.  kool_dudy

RC-7th   vs.  mlee06

Problems are due by Saturday.  I should send them out today.

Results so far:

Hashbrownie    ties   theone853  4-4

nr1337 vs.   white_horse_king88  3-TBD

kool_dudy   vs.   tandjsnell    2-TBD

mlee06  def.  RC-7th  2-0(RC-7th forfeited due to vacation)

Those who still need to take it are:
white_horse_king88
tandjsnell

I will get the tiebreakers out to hashbrownie and theone853 as soon as I put them together  ;) .

Just a reminder:  I need to get the answers from white_horse_king88 and tandjsnell by tomorrow or else they will be disqualified.

[edit] I am also going to send out the next round of winner's bracket questions since they aren't really affected by what happens in the losers bracket.

Sorry - My internet was down for about a week. I will take it today.

joml, Are the loser's bracket going to have a round off or are both winners and losers have simultaneous rounds?

I ll have the rounds going at the same time.

AGH! That was very disappointing. That set was much harder and I wasn't prepared when I took it. I spent so much time on number 2, and I still didn't get the answer. At least I have hopes of tieing nr1337.

I just got my first set of tiebreakers.  (Were they suppose to be for me or were they sent to the wrong person?)

I'm not sure if I did well on it but I answered all questions.

Those were for you kool_dudy.

Updated results

Hashbrownie   def   theone853  (after tiebreakers)

white_horse_king88  ties  nr1337 3-3

tandjsnell    ties  kool_dudy  2-2

mlee06   def     RC-7th  forfeit

Man you guys really like tying don't you?   Well at least we have good matchups this week.

Oh...hehe...just makes you work harder joml....

Have you got my tiebreaker score?

Just if you guys were wondering, I was beaten in the tiebreakers 5 to 4.5... Good luck in future rounds, Hasbrownie!

Whew, I get a break for a couple weeks! Yay!  :lol:
It couldn't have come at a better time either, since I'm not having any computer access at all beginning Monday Night EST to at least Friday or Saturday.

I just got home so here is the score of one match.

white_horse_king88 def  nr1337 (tie 5-4)

kool_dudy   def  tandjsnell (tie 2-0)


[quote="rogue"]Two players replace in turn the hearts in the formula
\[ M=\frac{1\cdot\heartsuit+2\cdot\heartsuit+\ldots+1001\cdot\heartsuit}{1+2+\ldots+1001}  \]
starting from the leftmost heart and ending with the rightmost one
by numbers $1, 2, \ldots, 1001$ in some order (each number can be
used only once). When all the hearts are replaced by numbers, the
players evaluate the integer nearest to $M$ and the player who
replaced one of the hearts by this integer wins. Which of the
players has a winning strategy?[/quote]
the first player, $1001=2n+1$
$M= \frac{1\cdot\ (n+1)+2\cdot\ a_1+3\cdot\ ((2n+1)-a_1)+4\cdot\ a_2+5\cdot\ ((2n+1)-a_2)+\ldots+2n\cdot\ a_n+(2n+1)\cdot\ ((2n+1)-a_n)} {1+2+\ldots+(2n+1)}$
$((n+1)-\frac{1}{2})$ < $M$< $((n+1)+\frac{1}{2})$


function
real analysis
real analysis unsolved

Let $ f(x)$ an integrable function in each limited inteval of R and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ x\in{R}$$ y\in{R}$.Prove that:$ f(x)\equal{}f(1)x$

No calculus. $ f$ integrable $ \rightarrow$ bounded, using only this it's easy and well-known.

[quote="lorincz"]$ f$ integrable $ \rightarrow$ bounded.[/quote]
That's using the Riemann theory of integration. What if we mean (locally) Lebesgue integrable? (Which we could weaken to measurable.)

if f is integrable, then pts of discontinuity of f is of measure 0....so you can find $ x\in\mathbb{R}$ such that f is cont at x. and a additive function continuous at 1 pt can be proved to be continuous on the whole of R. the formula is easy,do it first for rationals and then by density of Q and continuity conclude it holds for all reals.

@ dr kent- i havent done lebesgue, so..sorry if i said something stupid


Did you guys hear anything about the team this year? Have you got calls from AAPT or something? I'm on spring break right now and I can't get any information.

Not necessarily - they didn't know mailing addresses (or emails) for a lot of people, so I would contact your school teacher?

edit: oh I said that because even though I turned in the forms, they said they still didn't have any info on me, so I guess they reach you however they can reach you.

[quote="genericme"]AAPT sends out letters and packets of forms via overnight mail.  I think finalists should already have received their letters, but I can't speak for everyone.

Edit: I say that because I've received my letter.[/quote]
Congrats!! But how do they know your address? Do they just send them to the school?

[quote="simon_math"]Congrats!! But how do they know your address? Do they just send them to the school?[/quote]

Thanks.  I wrote my address on a form that I e-mailed to AAPT - they mailed me the materials directly.  Heck, I don't think my school even knows about it.  I'll have to tell my physics teacher.

I'm kinda confused.  :wacko:  But how did they tell you that they wanted your address? Did they have your email or something?

AAPT e-mailed my teacher some forms that all semifinalists from my school were to fill out "in case we were chosen as finalists" (there being two semifinalists from my school), and then he e-mailed them to us.

Oh, I see. Thank you. My teacher didn't tell me anything about the form you just mentioned. BTW, when did your teacher get it? Was it this week or some time earlier?

My teacher received the same form through email April 3rd or 4th.  It said on the form that it must be returned by April 1st, so they revised the deadline to the 6th online.

Seeing as how I haven't been contacted about being a finalist, I don't think that form has any special significance.  Given the date we received it with relation to the date it was supposed to have been turned in, I'm guessing it's just a slip-up AAPT made where they needed more information about the semifinalists.  Remember, semifinalists are supposed to get subscription to Physics Today and some other stuff, so they need your address info regardless of whether you made the team.

Thanks. I am not an official semifinalist because my QF paper arrived too late for them to grade, but 2 days before the deadline of SF test administration, AAPT called my teacher saying I could get a chance to take the SF test. Probably that's the reason why I didn't get the form. Also, we are on spring break this week, so I wouldn't know it even if my teacher has got the form.

So, genericme, how did you do on the semifinal?


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Any idea of a local diffeomorphism $ f: \mathbb{R}^2 \longrightarrow S^2$?

Just project onto a portion of the sphere, any way you feel like.