geometry incenter circumcircle inequalities inradius Euler vector

Let $ABC$ be a triangle with orthocenter $H$, incenter $I$ and centroid $S$, and let $d$ be the diameter of the circumcircle of triangle $ABC$. Prove the inequality
\[9\cdot HS^2+4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 3d^2,\]
and determine when equality holds.

some ideas on how to solve the problem:

may it be useful the substitution $9HS^2=4\cdot9OS^2=4(9R^2-a^2-b^2-c^2)$? so it is equivalent to show that  $6R^2+AH\cdot AI+BH\cdot BI+CH\cdot CI \geq a^2+b^2+c^2$, where $R$ is the circumradius, and $2R=d$

recall $AH=2Rcos\alpha$ and $AI=\frac{r}{sen\frac{\alpha}{2}}$, so it reduces to
$3R+r\sum\frac{cos\alpha}{sen\frac{\alpha}{2}}\geq2R(\sum sen^2\alpha) \\\Leftrightarrow R\sum cos2\alpha +r\sum\frac{cos\alpha}{sen\frac{\alpha}{2}}\geq0$

any suggestion?  :(

Thanks for reminding me about this inequality. Well, my solution at the exam was 10 pages long, particularly since I had to prove all the lemmata, what is fortunately not necessary on ML, and at some moment I became angry and started bunching something that later turned out to be a rather easy inequality. So here is [i]my solution[/i] after a dozen of simplifications:

Let R be the circumradius and r the inradius of triangle ABC. Then, the diameter d of the circumcircle is obviously d = 2R. Hence, we can transform the inequality in question as follows:

$9\cdot HS^2+4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 3d^2$
$\Longleftrightarrow$ $9\cdot HS^2+4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 3\cdot\left(2R\right)^2$
$\Longleftrightarrow$ $4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 3\cdot\left(2R\right)^2-9\cdot HS^2$.

Now, we can see that $9\cdot HS^2=4\left(9R^2-\left(a^2+b^2+c^2\right)\right)$. In fact, if O is the circumcenter of triangle ABC, then $9\cdot SO^2=9R^2-\left(a^2+b^2+c^2\right)$ according to http://www.mathlinks.ro/Forum/viewtopic.php?t=17684 post #2, but on the other hand, as a consequence of the Euler line, $HS=2\cdot SO$, and thus $9\cdot HS^2=9\cdot\left(2\cdot SO\right)^2=4\left(9\cdot SO^2\right)=4\left(9R^2-\left(a^2+b^2+c^2\right)\right)$.

Now, we will work with vectors. Since the product of the lengths of two vectors is always $\geq$ to their scalar product, we have $AH\cdot AI\geq\overrightarrow{AH}\cdot\overrightarrow{AI}$, what rewrites as

$AH\cdot AI\geq\left(\overrightarrow{AI}+\overrightarrow{IH}\right)\cdot\overrightarrow{AI}=\overrightarrow{AI}^2+\overrightarrow{IH}\cdot\overrightarrow{AI}=AI^2+\overrightarrow{IH}\cdot\overrightarrow{AI}$.

Similarly,

$BH\cdot BI\geq BI^2+\overrightarrow{IH}\cdot\overrightarrow{BI}$;
$CH\cdot CI\geq CI^2+\overrightarrow{IH}\cdot\overrightarrow{CI}$,

and thus, addition of these three inequalities yields

$AH\cdot AI+BH\cdot BI+CH\cdot CI\geq\left(AI^2+BI^2+CI^2\right)+\overrightarrow{IH}\cdot\left(\overrightarrow{AI}+\overrightarrow{BI}+\overrightarrow{CI}\right)$.

Now, since S is the centroid of triangle ABC, we have $\overrightarrow{SI}=\frac{\overrightarrow{AI}+\overrightarrow{BI}+\overrightarrow{CI}}{3}$, so that $\overrightarrow{AI}+\overrightarrow{BI}+\overrightarrow{CI}=3\cdot\overrightarrow{SI}$, and thus this becomes

$AH\cdot AI+BH\cdot BI+CH\cdot CI\geq\left(AI^2+BI^2+CI^2\right)+\overrightarrow{IH}\cdot\left(3\cdot\overrightarrow{SI}\right)$.

Now, in http://www.mathlinks.ro/Forum/viewtopic.php?t=39447 it was shown that $\measuredangle SIH>90^{\circ}$; hence, the vectors $\overrightarrow{IH}$ and $\overrightarrow{IS}$ form an obtuse angle with each other, i. e. their scalar product $\overrightarrow{IH}\cdot\overrightarrow{IS}$ is $\leq 0$. Hence,

$AH\cdot AI+BH\cdot BI+CH\cdot CI\geq\left(AI^2+BI^2+CI^2\right)+\overrightarrow{IH}\cdot\left(3\cdot\overrightarrow{SI}\right)$
$=\left(AI^2+BI^2+CI^2\right)+\overrightarrow{IH}\cdot\left(-3\cdot\overrightarrow{IS}\right)=\left(AI^2+BI^2+CI^2\right)-3\cdot\underbrace{\overrightarrow{IH}\cdot\overrightarrow{IS}}_{\leq 0}$
$\geq AI^2+BI^2+CI^2$.

Now, we will show the inequality

[color=green][b](1)[/b][/color] $AI^2+BI^2+CI^2\geq a^2+b^2+c^2-6R^2$.

Once this inequality will be proven, we will have

$4\left(AH\cdot AI+BH\cdot BI+CH\cdot CI\right)\geq 4\left(AI^2+BI^2+CI^2\right)\geq 4\left(a^2+b^2+c^2-6R^2\right)$
$=4\left(3R^2-\left(9R^2-\left(a^2+b^2+c^2\right)\right)\right)=3\cdot\left(2R\right)^2-4\left(9R^2-\left(a^2+b^2+c^2\right)\right)$
$=3\cdot\left(2R\right)^2-9\cdot HS^2$,

and thus the problem will be solved. Thus, in order to solve the problem, it remains to prove the inequality [color=green][b](1)[/b][/color].

If the incircle of triangle ABC touches the side CA at a point Y, then the triangle AYI is right-angled at Y (since the tangent to a circle at a point is perpendicular to the radius through this point), and its catets are IY = r (the radius of the incircle) and AY = s - a, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC (this follows from the well-known formulas for the distances from the vertices of a triangle to the points where the incircle touches the adjacent sides). Hence, by the Pythagoras theorem, applied to the triangle AYI, we get $AI^2=AY^2+IY^2=\left(s-a\right)^2+r^2$. Thus,

$\left(AI^2+BI^2+CI^2\right)-\left(a^2+b^2+c^2\right)=\sum\limits_{\text{cyclic}} AI^2-\sum\limits_{\text{cyclic}} a^2$
$=\sum\limits_{\text{cyclic}}\left(\left(s-a\right)^2+r^2\right)-\sum\limits_{\text{cyclic}} a^2=3r^2+\sum\limits_{\text{cyclic}}\left(\left(s-a\right)^2-a^2\right)$
$=3r^2+\sum\limits_{\text{cyclic}}\left(\left(s-a\right)-a\right)\left(\left(s-a\right)+a\right)=3r^2+\sum\limits_{\text{cyclic}}\left(s-2a\right)s$
$=3r^2+s\sum\limits_{\text{cyclic}}\left(s-2a\right)=3r^2+s\left(3s-2\left(a+b+c\right)\right)=3r^2+s\left(3s-2\cdot 2s\right)$
$=3r^2-s^2$.

Hence, we can equivalently transform the inequality [color=green][b](1)[/b][/color] as follows:
$AI^2+BI^2+CI^2\geq a^2+b^2+c^2-6R^2$
$\Longleftrightarrow$ $\left(AI^2+BI^2+CI^2\right)-\left(a^2+b^2+c^2\right)\geq -6R^2$
$\Longleftrightarrow$ $3r^2-s^2\geq -6R^2$ $\Longleftrightarrow$ $3r^2+6R^2\geq s^2$.

Now, since $\left(3r^2+6R^2\right)-\dfrac13\left(4R+r\right)^2=\dfrac23\left(R-2r\right)^2\geq 0$, we have $3r^2+6R^2\geq\dfrac13\left(4R+r\right)^2$, so it will be enough to prove that $\dfrac13\left(4R+r\right)^2\geq s^2$. This is equivalent to $\left(4R+r\right)^2\geq 3s^2$, or, after taking the square root, to $4R+r\geq\sqrt3 s$. Now, I am sure this inequality appeared somewhere on ML, but the proof is so easy that it would be probably more appropriate than a link: It is known that $4R+r=r_a+r_b+r_c$, where $r_a$, $r_b$, $r_c$ are the exradii of triangle ABC (a proof was given at http://www.mathlinks.ro/Forum/viewtopic.php?t=50373 , another one in [url=http://www.math.fau.edu/yiu/Geometry.html]Paul Yiu, Notes on Euclidean Geometry[/url], §2.4.1). Hence, the inequality in question becomes $r_a+r_b+r_c\geq\sqrt3 s$. Now use the formulas $r_a=s\tan\frac{A}{2}$, $r_b=s\tan\frac{B}{2}$, $r_c=s\tan\frac{C}{2}$ to rewrite this as $s\tan\frac{A}{2}+s\tan\frac{B}{2}+s\tan\frac{C}{2}\geq\sqrt3 s$, what simplifies to $\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\geq\sqrt3$. Now, this follows from Jensen's inequality, applied to the function $x\to\tan x$, which is convex in the interval [0°; 90°[ (all the angles $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$ lie in this interval):

$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\geq 3\tan\frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3}=3\tan\frac{A+B+C}{6}$
$=3\tan\frac{180^{\circ}}{6}$            (since A + B + C = 180° by the sum of angles in triangle ABC)
$=3\tan 30^{\circ}=3\frac{1}{\sqrt3}=\sqrt3$.

This completes the solution.

  Darij

Very good but big solution darij :) . I am wondering if there is a manner to solve fast this ineq. Like campos theknik.We will see

[u]I am sorry, Darij, why offered you such quikly the your solution ?[/u]

I can affirm that I were certainly that must to use the position of the incenter $I$ w.r.t. the circle with the diameter $HO$.
The main inequality is $\sum AH\cdot AI\ge \sum AI^2$ The strong inequality $\overrightarrow {IH}\cdot \overrightarrow {IG}\le 0$ is the basis of the previous inequality. The proposed inequality is equivalently with the following inequality:\[ \underline {\overline {\left| \ \blacktriangle \ HI^2+\sum AH\cdot AI\ge 3R^2+OI^2\ \blacktriangle\ \right| }}  \]

[quote="Virgil Nicula"]:\[ \underline {\overline {\left| \ \blacktriangle \ HI^2+\sum AH\cdot AI\ge 3R^2+OI^2\ \blacktriangle\ \right| }}  \][/quote]

Which can be proved... how?

I think we can prove that
with any M $MA^2+MB^2+MC^2 \ge a^2+b^2+c^2-6R^2$ quite simple by using Leibnitz fomula:
$\sum MA^2 =3MG^2 + \sum GA^2 \ge \sum GA^2 = \frac{a^2+b^2+c^2}{3}$
and if we use $a^2+b^2+c^2 \le 9R^2$ then we have
$\frac{a^2+b^2+c^2}{3} \ge a^2+b^2+c^2-6R^2$
may be I have some mistakes
best regard


floor function
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calculus
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[quote="shobber"]Let $a_1,a_2,a_3,\ldots,a_n$ be a sequence of non-negative integers, where $n$ is a positive integer. Let
\[ A_n={a_1+a_2+\cdots+a_n\over n}\ .  \]
Prove that
\[ a_1!a_2!\ldots a_n!\ge\left(\lfloor A_n\rfloor !\right)^n  \]
where $\lfloor A_n\rfloor$ is the greatest integer less than or equal to $A_n$, and $a!=1\times 2\times\cdots\times a$ for $a\ge 1$(and $0!=1$). When does equality hold?[/quote]
http://www.kalva.demon.co.uk/apmo/asoln/asol021.html

I don't know why I can't open all the links from this website.

Kalva is not working anymore. Can someone post a solution? I'm stucked. :P

I have tried induction but failed.

There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
It is understandable because $ \log x! \equal{} \log 1 \plus{} \log 2 \plus{} \cdots \log x$ and $ \log x$ is an increasing function.
However I dont see a rigorous proof now.
Also it can be made monotonically increasing(not strictly) by making $ f(x) \equal{} 0\ \forall x\geq 0$ and $ \leq 1$

$ \sum_{i \equal{} 1}^{n}f(a_n)\geq nf(A_n)$ by Jensen's inequality
$ f(A_n)\geq f(\lfloor A_n \rfloor)$ because $ f(x)$ is monotonically increasing.
$ \sum_{i \equal{} 1}^{n}f(a_n) \equal{} \sum_{i \equal{} 1}^{n}\log a_n! \equal{} \log\prod a_n!$
$ nf(\lfloor A_n \rfloor) \equal{} n\log \lfloor A_n \rfloor ! \equal{} \log (\lfloor A_n \rfloor !)^n$
Thus proved. Equality holds for $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n$ or when some of $ a_i$ are $ 1$ and some are $ 0$.

Kalva still lives in our hearts... and someplace else (Webarchive). So here's a link to the solution:

[url]http://web.archive.org/web/20040313130703/www.kalva.demon.co.uk/apmo/asoln/asol021.html[/url]

Great solution in Kalva! But I can't help saying that the equality case given there is not truly an "iff" statement :P

I think those archive is a little bit different from the real kalva. For example, I didn't see tournament of the towns in kalva before (am I blind?) :P

[quote="Akashnil"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
[/quote]
$ f(x)\equal{}\log x!$ is concave!

[quote="Sunjee"][quote="Akashnil"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
[/quote]
$ f(x) \equal{} \log x!$ is concave![/quote]

Honestly, terms convex and concave always seem to be so confusing: in different literature, I used to find diametrically opposite definitions: in some books I've used, what is concerned concave is convex in others: so some use the terms concave up and concave down- now that's a little more unambiguous.   :wink:

[quote="Sunjee"][quote="Akashnil"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
[/quote]
$ f(x) \equal{} \log x!$ is concave![/quote]
For some function defined on reals, I mean it to be convex iff the 2nd derivative is  always non-negetive. And I'm pretty sure there is a convex function $ f(x)$ which has the required property.

Edit: Can we call a sequence convex? In that case the definition would be $ a_{n \minus{} 1} \plus{} a_{n \plus{} 1}\geq 2a_n$. Jensen possibly holds there.

[quote="Akashnil"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
It is understandable because $ \log x! \equal{} \log 1 \plus{} \log 2 \plus{} \cdots \log x$ and $ \log x$ is an increasing function.
However I dont see a rigorous proof now.[/quote]
Correct!  In fact, there is exactly one such function; it is the logarithm of the [url=http://en.wikipedia.org/wiki/Gamma_function]gamma function[/url] (shifted by one).

Now:  I actually think that the right-hand-side [i]divides[/i] the left-hand-side under suitable conditions, but I'm not sure what they are.

I don't understand in kalva web !
Could anyone explain it? Thanks.

Just start with $ a_1\equal{}a_2...\equal{}a$
Then clearly we have equality. Now think about what happens when we increase one of the $ a_i$ by one, and decrease another by $ 1$. Then we are dividing the $ LHS$ by $ a$, and multiplying by $ a\plus{}1$. In this way we see any change preserves the inequality.

[quote="Akashnil"]There is a convex function $ f(x)$ such that for non-negative integers $ x,f(x) \equal{} \log x!$
It is understandable because $ \log x! \equal{} \log 1 \plus{} \log 2 \plus{} \cdots \log x$ and $ \log x$ is an increasing function.
However I dont see a rigorous proof now.
.[/quote]

Since
$ \log (x\plus{}1)! \minus{} \log x! \equal{} log (x\plus{}1) \geq 0$

so $ f(x)\equal{} \log x!$ is increasing!

I came up with a solution. Does it work?

[b]Solution[/b]

Let $ \lfloor A_n \rfloor \equal{} p$. Let $ b_m \equal{} a_m \minus{} p$ for all $ a_m \ge p$ and let $ c_k \equal{} p \minus{} a_k$ for all $ a_k \le p$. Now note that since $ \sum a_i \ge np$, it follows that $ \sum b_m \ge \sum c_k$. Now consider the following product. Define that,

$ P \equal{} \frac{(p\plus{}1)(p\plus{}2) \dots (p\plus{}b_1)(p\plus{}1)(p\plus{}2) \dots (a\plus{}b_m)}{(p\minus{}c_1\plus{}1)(p\minus{}c_1\plus{}2) \dots (p)(p\minus{}c_2\plus{}1)(p\minus{}c_2\plus{}2) \dots (p\minus{}c_k\plus{}1) \dots (p)}$

There are a total of $ \sum b_m$ factors in the numerator each of which is greater than or equal to $ p$ and there are a total of $ \sum c_k$ factors in the denominator each of which is less than or equal to $ p$ (since it is possible that for some $ i$, $ b_i$ or $ c_i$ is $ 0$). Since $ \sum b_m \ge \sum c_k$, it follows that $ P \ge 1$.

Now note that

$ P \equal{} \frac{a_1!}{p!} \cdot \frac{a_2!}{p!} \cdot \dots \cdot \frac{a_n!}{p!}$

Hence since $ P \ge 1$, it follows that $ a_1!a_2! \dots a_n! \ge (p!)^n$.

Equality holds when $ P \equal{} 1$ which can only occur if all $ c_i$ and $ b_i$ are $ 0$ which occurs when $ a_1 \equal{} a_2 \equal{} \dots \equal{} p$.

For all $a_i$ being equal the equality case is evident. Now assume that all $a_i$ are not equal, and without loss of generality let $a_1 \leqslant a_2 \leqslant  ... \leqslant a_n$. Then for some $x$ such that $a_n \geqslant a_1 +x$, the transformation $(a_1, a_2,...,a_n)\mapsto (a_1 +x, a_2, ...,a_n-x)$ decreases the left hand side. Clearly, the right hand side remains invariant under this transformation. The value of the left hand side decreases by every such transformation. This transformation cannot be performed when all the numbers are equal, hence the minimum value of the left hand side is when all $a_i$ are equal. That, as explained above, gives the equality case. Hence we're done.

Notice that by taking $\ln$ of both sides, the inequality becomes: $\sum_{i=1}^{n}\ln {a_i!}\ge\ln {(\lfloor A_n\rfloor)^n}$
Furthermore let $f(x)=\ln {x!}\Longrightarrow f'(x)=\psi(x+1)\Longrightarrow f''(x)=\psi_1(x+1)>0$ thus the function is strictly convex on the non-negative integers.
Thus $\sum_{i=1}^{n}\ln{a_i!}=\sum_{i=1}^{n}f(a_i)\overset{\text{Jensen}}{\ge}n f\bigg(\frac{\sum_{i=1}^{n}a_i}{n}\bigg)=nf(A_n)=n\ln {A_n!}$
Furthermore: $n\ln {A_n!}\ge\ln {(\lfloor A_n\rfloor)^n}\Longleftrightarrow (e^{\ln {A_n!}})^n\ge e^{\ln {(\lfloor A_n\rfloor)^n}}\Longrightarrow (A_n)^n\ge(\lfloor A_n\rfloor)^n\Longrightarrow A_n!\ge\lfloor A_n\rfloor$ which is clearly true $\blacksquare$.
And we are done! :bye:

let a_1>=a_2>=....a_n Now we will prove this by induction
A_1=a_1 so p(1) is true
now let p(n-1) be true and note p(n) is
rewarding
(a_n)!/([A_n-1]!)>=(([A_n])/([A_n-1]))^n
and this is true because
(a_n)-([A_n-1])=n(([A_n])-([A_n-1])) and a_n>=[A_n]
and equality hold iff (a_i)!=(a_j)! for all 1<=(i,j)<=n


limit
geometry
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calculus
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Find:
\[ \lim_{n \to \infty} \frac{\sqrt{n^2+1}+\sqrt{n}}{\sqrt[3]{n^2+n}-\sqrt{n}}  \]

This appears to diverge, as the numerator is $O(n)$ and the denominator is $O(n^{2/3})$.  Perhaps the first term in the numerator is also meant to be a cube root?

I agree with Xevarion's answer, just divide both numerator and denominator by $n$ and it shows that the limit diverges...


A function f is defined on the set D. The set D is the set of rationals being different from 0 and 1. The function f satisfies for all x \in D the following equation:

 f(x) + f(1 - 1/x) = x

Determine f !


If [i]x[/i],[i]y[/i],[i]z[/i] are positive integers and the sum of any two greater than the third,prove tht..\

  $ [1\plus{}(y\minus{}z)/x]^{x}[1\plus{} (z\minus{}x)/y]^{y}[1\plus{}(x\minus{}y)/z]^{z}$ < 1.
 :lol:   :!:

as $ x,y,z$ are sides of a triangle
We can put $ x \equal{} a \plus{} b,y \plus{} b \plus{} c,z \equal{} c \plus{} a$
Plugging this into the inequality we get
$ LHS \equal{} \prod_{cyc}({{1 \plus{} \frac {b \minus{} a}{a \plus{} b}})^{a\plus{}b}} \\
\equal{} \prod_{cyc}{({\frac {2a}{a \plus{} b})^{a\plus{}b}}} \\
\le \frac {2a \plus{} 2b \plus{} 2c}{a \plus{} b \plus{} b \plus{} c \plus{} c \plus{} a} \equal{} 1$
The Inequality applied here is the weighted AM-GM inequality.......


limit
integration
logarithms
real analysis
real analysis unsolved

Prove that

$ \displaystyle\lim_{n\to\infty}\frac{1}{n}\int^n_1 \|\dfrac{n}{x}\|\,dx\equal{}log_3(4/\pi)$

I don't quite see how this is number theory...

What does $ \left \Vert \dfrac{n}{x}\right \Vert$ stand for?  Is that the fractional part of $ \frac {n}{x}$?  If that's the case then
\[ \lim_{n\to \infty}\frac {1}{n}\int_1^n{\left \Vert \dfrac{n}{x}\right \Vert dx} \equal{} 1 \minus{} \gamma \neq \log_3{\frac {4}{\pi}}
\]
It must be something else then?


We choose a subset of $ \{1, 2, \ldots, 2n\plus{}1 \}$ with maximal cardinality so that none of its elements is equal to the sum of others.
How many elements does it have?

I interpreted [quote="mathmanman"]the sum of others[/quote] to mean "the sum of some subset of the others," not "the sum of some pair of the others."  In that case, $ n \plus{} 1$ is too large when $ n \geq 4$ ($ 9 \equal{} 5 \plus{} 3 \plus{} 1$).  mathmanman, which problem was intended?

Indeed, your interpretation was the intended one [b]JBL[/b].

I have moved Tdl's solution for the "the sum of some pair of others" version into the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=1491 . Let's use this particular topic to discuss the "the sum of some subset of the others" version.

  darij

We can take subset $ \{n\plus{}1,n\plus{}2,...,2n\plus{}1\}$ with property described because sum of every subset with at least 2 elements is $ \geq 2n\plus{}3$.

Suppose we can take elements $ a_1,a_2,...,a_{n\plus{}2}$ with required property which are sorted in increasing order. Consider set $ \{a_{n\plus{}2}\minus{}a_1,a_{n\plus{}2}\minus{}a_2,...,a_{n\plus{}2}\minus{}a_{n\plus{}1}\} \cup \{a_1,a_2,...,a_{n\plus{}1}\}$. Every element of this set is $ \leq 2n$ and $ \geq 1$ hence we have $ a_{n\plus{}2}\equal{}a_{i}\plus{}a_{j}$ at least two times for some $ 1\leq i,j\leq n\plus{}1$. It can't be $ i\equal{}j$ both times so we have contradiction, which show that maximal cardinality is $ n\plus{}1$.


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I'm going through calculus currently and I'd also like to get started in some higher level math. I would like to learn some more advanced combinatorics and linear algebra. Any book recommendations?

How advanced do you want your linear algebra?  Are you looking for computationally-based or proof-based?  I like "Linear Algebra" by Friedberg, Insel, and Spence (nice, hard book).  I know they use Axler's "Linear Algebra Done Right" at Harvard but I don't know much about the book.

At this point, I don't really know which one I'd like to go with (the two types you mentioned), because I know nothing about linear algebra and I don't know how this choice will affect me.

As for linear algebra, someone posted [url=http://joshua.smcvt.edu/linearalgebra/]this online book[/url] a while back, which is a fairly comprehensive introductory textbook; though really, mostly any (non-proof based) introductory textbook should do. Princeton uses Hoffman/Kunz, and JRav mentioned another couple of commonly used authors, though the material may be (comparatively) advanced.

Advanced combinatorics is a bit more vague; while there may be comprehensive texts on combinatorics, you'll probably have to refine your search a bit more; perhaps enumerative combinatorics? graph theory?

graph theory.

EDIT: Is proof-based really hard to deal with? Cause, I feel sort of guilty when I just "learn' a theorem without knowing where it came from .

[quote="azjps"]I should have been more clear with what I meant with "non-proof based"; it does not mean that the basic theorems are not proved. The link I provided does include proofs of the theorems and applications. However, there's a difference (and not just for linear algebra) between covering introductory fully-rigorous linear algebra and introductory computation-based linear algebra; it is rather difficult to do the former without any intuition from the latter. For example, a rigorous linear algebra text will use a much more general context (eg, work in any fields and not just the real numbers) and discuss less of the applications, so it will be more difficult to gain hands-on intuition.[/quote]

Oh god. No. Not multiple fields. Not again after writing that script for sagemath.org. Yeah, I think i'll with go with computational-based.

All that means is that you choose scalars that don't have to be real numbers.  It's not bad.  Besides, you'll have to get used to it for abstract algebra, where $ \mathbb{R}$ is only one example.


integration
calculus
analytic geometry
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real analysis unsolved

[quote="jianuovidiu"]Let $p(y(t)) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{(y(t)-m)^2}{\sigma^2}}$ be gaussian distribution of aleator variable $y(t)$, where $m$ is the mean value of $y(t)$. Show that $\int_{m-3\sigma}^{m+3\sigma}p(y(t))dy(t) \geq 0.95$.[/quote]
Substitute $z = ((y(t)-m)/\sigma$ to reduce the problem to the case $m = 0, \, \sigma = 1$. Then consider the [b]square[/b] of the integral:
$\frac{1}{2\pi}\left( \int_{-3}^3 e^{-x^2/2} dx \right) = \frac{1}{2\pi}\int_{-3}^3 \int_{-3}^3 e^{-\frac{x^2 +y^2}{2}} dx dy \ge \frac{1}{2\pi} \iint_{D_3} e^{-\frac{x^2 +y^2}{2}} dx dy$
where $D_3$ is the circle with radius 3 about the origin.  Use polar coordinates to compute the last integral as
$\int_0^3 r e^{-r^2/2}dr = 1 - e^{-9/2}$ and conclude that the integral in question is at least $\sqrt{1-e^{-9/5}} \approx 0.994$. The exact value is more like 0.997.

[i] This is the same trick that one uses to compute $\int_{-\infty}^\infty e^{-x^2/2}dx$.[/i]


Is there any way to determine the largest rectangle that can be inscribed in a triangle? What a bout the largest square?

thanks :)

[hide]
Let $ M\in AB, \ N\in AC$ such that $ MN\parallel BC$. Let $ K,L$ be points on $ BC$ such that $ KLMN$ be a rectangle

Of course the angles $ B,C$ shouldn't be obtuse. Let $ AD$ be the altitude from $ A$

Let $ E\in MN \cap AD$

Let $ \lambda \equal{} \frac {AM}{AB}$. Then

$ AM \equal{} \lambda AB$
$ AN \equal{} \lambda AC$
$ LD \equal{} \lambda BD$
$ DK \equal{} \lambda DC$
$ AE \equal{} \lambda AD$

Then $ DE \equal{} (1 \minus{} \lambda)AD$
$ LK \equal{} LD \plus{} DK \equal{} \lambda(BD \plus{} DC) \equal{} \lambda BC$

$ (KLMN) \equal{} LK \cdot DE \equal{} \lambda(1 \minus{} \lambda)BC\cdot AD \equal{} \lambda(1 \minus{} \lambda)\cdot 2(ABC)$

So we obtain maximum value for $ (KLMN)$ when $ \lambda \equal{} \frac {1}{2}$. This maximum value is equal to $ \frac {(ABC)}{2}$

$ \star$ The rectangle $ KLMN$ is square only if $ KL \equal{} ED \iff \lambda BC \equal{} (1 \minus{} \lambda)AD \iff \lambda \equal{} \frac {AD}{AD \plus{} BC}$

$ \star$ The largest rectangle is a square if and only if $ AD \equal{} BC$

[color=brown]Note
$ (X)$ denotes the area of $ X$[/color][/hide]


[quote="i_like_pie"]How many digits are in the number $25^{21}\cdot2^{48}?$[/quote]

[hide="Solution"]Firstly, you have to orgainze the data (unless you are willing to compute it, which I am not). The number of zeros must be found.

$5^{42}\times 2^{48}= 5^{42}\times 2^{42}\times 2^{6}$. That is equal to: $10^{42}\times 2^{6}= 64\times 10^{42}$. THerefore, there are 42 zeroes. Then, the number of other digits must be found. There are two of these. Therefore, is it $42+2=\boxed{44\text{ digits}}$.[/hide]

[hide="Content Hidden Reveal to Click"] Prime factoring, we have $2^{48}*5^{42}$. We note that this is equal to $10^{42}*2^{6}$, which is 64 with 42 zeroes after it, a total of 44 digits.[/hide]

whats the point of posting the same solution 4 times?
-jorian

Because other people want to try to solve the problem...

And when posting a solution, the poster doesn't know that it's the same solution as the person above them. You might have a different solution that may be helpful to people reading the thread.

[quote="perfect628"]And it's fantastic that you read them without trying the problem...[/quote]

That's why we hide our answers.  :wink:

[quote="jhredsox"]whats the point of posting the same solution 4 times?[/quote]

Because others solve it, and don't look at the hidden answers below them, so they don't know really if it has been posted. Also, more answers can clarify small holes in other's logic.

I personally don't think these are adequate reasons. Solve the problem, then read the other solutions.

#1: You can solve the problem and not post it.  :wink:
#2: The small chance that you have a different solution is overwhelmed by the benefit of reading the previous solutions, and thus not adding to unnecessary clutter.
#3: Reading previous solutions may reveal errors (both on your part and others') , also helping readers of the forum. :P

[hide="solution"]$25^{21}\times 2^{48}= 5^{42}\times 2^{48}= 10^{42}\times 64$

That can be written as

[u]64000...000[/u]
42 zeros

Therefore, there are 44 digits.[/hide]


AMC
AIME
geometry
perpendicular bisector
cyclic quadrilateral

10.  Triangle ABC has AC = BC and ∠ACB = 106 degrees. M is a point inside the triangle such that ∠MAC = 7 degrees and ∠MCA = 23 degrees. Find ∠CMB.

Okay this one took me 50 minutes probably because I'm stupid, but still the solution I came up with wasn't obvious and was kind of nice.

Hmm... I'll take a shot at it.

[hide]Ok... darn. I can't make a diagram but notice that you can extend AM to a point, lets say D on the perpendicular bisector of AB extended from C.

Angle CMD is 30 degrees, and so is angle MCD. So, MD $\cong$ CD. Notice that AD $\cong$ BD, so $\angle$DAB $\cong$ $\angle$DBA. So, $\angle$ADC $\cong$ $\angle$BDC. And BD = BD. So, the two triangles BMD and BCD are congruent. Wow... So, by, I thought I'd never use this on AIME, CPCTC, $\angle$BMD $\cong$ $\angle$BCD, and $\angle$BCD = $53$. (106/2). So, that's $\angle$BMD. Yay.[/hide]

I'm a retarded idiot. It took me 20 minutes to draw perpendicular bisector of AB, another 30 minutes to miss the obvious stuff.
Here's what I did:

Draw perpendicular bisector of AB and let D be the midpoint of AB. Let AM intersect CD at I. $\angle EAI= \angle ECI = 30$. So ACIE is a cyclic quadrilateral, so $\angle CAI= \angle CEI = 7$. BEI= 60-7=53. since MEB=60, MIB=120. MIBE is cyclic quadrilateral. IEB=IMB=53. CMI=30, so CMI+IMB=83=CMB.

[hide]Draw point N inside ABC such that triangle CNM is equilateral.  triangle NCB :cong: MCA by SAS.  angle MNB = 360-150-60 = 150, so triangle MNB :cong: triangle CNB by SAS.  Triangle BCM is isosceles, angle CMB = angle MCB = 83[/hide]

[quote="zabelman"][hide]Angle ceva:
sin(<MBC)/sin(<MBA) = sin(83)/sin(23)*sin(7)/sin(30)
=2sin(83)sin(7)/sin(23)
=2sin(7)cos(7)/sin(23)
=sin(14)/sin(23).
Notice that <MBC=14 and <MBA=23 makes <ABC=37, which it better be, so these angles work. Then <BMC=180-83-14=83.[/hide][/quote]

Angle ceva... that's a nice theorem to know. Added it to my toolbox =).

A restatement of Ceva's theorem with angles instead of lengths:

AD, BE, and CF are concurrent if and only if
sin(<BAD)/sin(<DAC)*sin(<ACF)/sin(<FCB)*sin(<CBE)/sin(<EBA)=1.

Can you prove it?

[quote="zabelman"]A restatement of Ceva's theorem with angles instead of lengths:

AD, BE, and CF are concurrent if and only if
sin(<BAD)/sin(<DAC)*sin(<ACF)/sin(<FCB)*sin(<CBE)/sin(<EBA).

Can you prove it?[/quote]

Er, if that expression equals one.

Isn't this theorem in AoPS 2? I think right after the proof of Ceva with sides.

[quote="mathfanatic"]Er, if that expression equals one.[/quote]

Thanks. Edited.


PLEASE SEE THE $1$ST DIAGRAM [u]NOT[/u] THE $2$ND DIAGRAM.

this is really weird, I maybe wrong though but,
the area of this triangle, using $\frac{1}{2}*b*c*sin\theta$ where $b$ and $c$ are sides which enclose the angle $\theta$ so the area of the triangle is $\frac{1}{2}*r^2*sin360$ which means that for every triangle drawn in this circle the area is constant?????????? how can that be possible????? Ive seen that it can't. where have I gone wrong can somebody help????? :?  :?  :(  :(  :(  :(

$b,c$ must not only enclose the angle $\theta$, but they must also be sides of the triangle. In your diagram, the lengths $r$ either do not enclose an angle of $360^\circ$, or do not form the sides of a triangle (depending on which triangle you look at).

Just as a reminder, no triangle has any angle larger than $180^\circ$.

sorry but I fail to understand you but I have understood my problem,
Can a moderator delete this thread this is useless.


[hide]I believe that $x^{2}={{n+2}\choose{3}}-{n\choose3}$. And I know that $2^{n}$ is the sum of row n, formally written as ${n\choose0}+{n\choose1}+...+{n\choose{n}}$[/hide]

[quote="frost13"]The first 4 rows(5 including zero row) are the powers of 11

1         = 11^0
11       = 11^1
121      = 11^2
1331    = 11^3
14641   = 11^4

After that you run into carrying and it now longer holds.

1  5  10  10  5  1

as 11^5 =

1
    5
    1  0
        1  0
               5
 +                1
---------------
1   6  1  0  5  1[/quote]
Oh. What about my answer, is it valid? Isn't it the same thing your saying, just in more complex terms or no?

bpms, your's was right, just trying to give a clear example everyone could follow....and see why PT stops "displaying" powers of x at the 5th row.

Your explanation is great and quite mathematically concise!!  I did have to think if using n instead of x for powers of 10 was the best way to go, but seems ok.  Can you write the same thing in Summation notation for 11^x(you would need one more variable for the counter).

[quote="jhredsox"][quote="bpms"][quote="jhredsox"]how is $11^{x}$ and $2^{x}$ and $x^{2}$ incorporated into pascals?

......i already know, its just a question
-jorian[/quote]
[hide="$x^{2}$"]$x^{2}=\binom{n+2}{3}-\binom{n}{3}$[/hide]
[hide="$2^{x}$"]$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n-1}+\binom{n}{n}$[/hide]
[hide="$11^{x}$"]I'm never seen this one, but I think it has to do with the binomial theorem. Expanding $(1+10)^{x}$ we get $\binom{x}{0}+10^{1}\binom{x}{1}+10^{2}\binom{x}{2}+...+10^{n}\binom{x}{x}$[/hide][/quote]

i changed what you said, because the hides weren't showing....well here are the answers

for $2^{x}$ its the sum of every row where as $x$ is the number of the row considering the first row is called, $0$

for $x^{2}$ is when you take the sum of adjacent pairs of triangular numbers for $x$ is the nth pair that you come across

frost had an ok definition of $11^{x}$

but here its is, if you have $1$  $5$ $10$ $5$ $1$ (not actual)

then, you would do this

10000
05000
01000 (the 1 drags over like you have 10 hundreds)
00050
00001
------
16051  which is not $11^{x}$ because i was too lazy to use actual pascal numbers :huh:

-jorian[/quote]

I hope you realize your solution is the exact same as frost13's.

ummmmmmmmmmmmmmmm................ya, mine's just more visual for people who can't do anything in math :(  (me) :(

[quote="frost13"]hmmm my post doesn't seem to have included the spacing I put in:-(  I wonder if I can edit that.  Jorian, did you miss a 10 in the middle of yours?[/quote]
It seems that spacing doesn't really show up well in a regular post. However, if you put the spaces in a code box, they [i]will[/i] show up. Here's an example:
[code]This is just         an example of spacing in a code box.[/code]

Thanks Neal!

[code]
 1
   5
   1 0
     1 0
         5
+          1
-----------
 1 6 1 0 5 1[/code]

There's what it should have looked like....after a few previews to get the spacing right :lol:


The incircle $w=C(I,r)$ of the triangle $ABC$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$. Denote the middlepoints $M\in BC$, $N\in CA$, $P\in AB$ of the sides of this triangle. Prove that the lines which pass through the points $D$, $E$, $F$ and are parallelly with the lines $IM$, $IN$, $IP$ respectively are concurrently.

This is not a very nice problem because the solution has too many old results.
[hide="My solution"]
Let $X,Y,Z$ be respectively the tangent points of $A$-excircle and $BC$, $B$-excircle and $CA$, $C$-excircle and $CA$.
It's well-known that:$M,N,P$ respectively are the midpoint of the segments $XD,YE,ZF$.
In order to prove that [color=red]the lines which pass through the points $D$,$E$,$F$  and are parallelly with the lines $IM$,$IN$,$IP$ respectively are concurrently[/color],I will prove that [color=blue]the lines which pass through the points $X$,$Y$,$Z$  and are parallelly with the lines $IM$,$IN$,$IP$  respectively are concurrently[/color]
On other hand,the lines which pass through the points $X$,$Y$,$Z$  and are parallelly with the lines $IM$,$IN$,$IP$ are respectively $AX$,$BY$,$CZ$.So they are concurrently
[/hide]

Here is another solution.
I will also prove that $\overline{IT}=3\overline{GI}$
where $T$ the point where they intersect and $G$ is the centroid of $\triangle ABC$.

I assume that $\triangle ABC$ is not isosceles.

I define the points $X,Y,Z$ as above, in the solution given by James Potter. That is:
$X$ is the tangent point between $A$-excircle and side $a$
$Y$ is the tangent point between $B$-excircle and side $b$
$Z$ is the tangent point between $C$-excircle and side $c$

We know that
$\overline{XM}=\overline{MD}$, and $AX \parallel IM$

The parallel line from $G$ to $AX$ meets the side $BC$ at $G_a$.

$\overline{AG}=3\overline{GM} \Rightarrow \overline{XM}=3\overline{G_aM}$

Since $\overline{XM}=\overline{MD}$ we get $\overline{MD}=3\overline{G_aM}$

similarly,

$\overline{NE}=3\overline{G_bN}$
$\overline{PF}=3\overline{G_cP}$

where
$G_b \in b, GG_b \parallel BY$
$G_c \in c, GG_c \parallel CZ$

The parallel line from $D$ to $IM$ intersects the line $GI$ at the point $Q$ (assuming that $b \neq c$)

$GG_a \parallel AX\Rightarrow GG_a \parallel IM$

$ \begin{array}{l} GG_a \parallel IM \parallel DQ \\ \\ \overline{MD}=3\overline{G_aM} \end{array} \}\Rightarrow \overline{IQ}=3\overline{GI}$

Similarly, we define the points $R,S \in GI ,$ such that $ER \parallel IN, FS \parallel IP$.

then we have

$\overline{IR}=3\overline{GI}$
$\overline{IS}=3\overline{GI}$

$\overline{IQ}=\overline{IR}=\overline{IS}\Rightarrow Q \equiv R \equiv S \equiv T \Rightarrow$

So the three lines are concurrent at a single point $T \in GI$ and in addition we have $\overline{IT}=3\overline{GI}$

[color=blue]--------------------------------------------------------------------------------------------------------------------
[u]P.S.[/u][/color]

We did assume that $ABC$ is not isosceles.

If it is, say $b=c \neq a$ then $I \in AM$ so the line $GI$ is median and also angle bisector.
The asked line (parallel from $D$ to $IM$) is this line $GI$, so we can't find the point $Q$.

But the other equations $\overline{IR}=3\overline{GI}$ and $\overline{IS}=3\overline{GI}$ are true and the lines $ER,FS$ intersect on the line $GI$, at the point $R \equiv S$.

on the other hand, if $\triangle ABC$ is equilateral, then the three lines are the medians, so they are concurrent  at the center of the triangle $G \equiv O \equiv I$.

In any case, the equation $\overline{IT}=3\overline{GI}$ holds.


Hey guys, I am writing a math paper with Latex. But I don't know how to do some stuff...

How do you number the pages starting from the pages, but not starting from the abstract page?
How do you space down the page? Like you know how you press ENTER and the line goes down?
And...how do you "cite" a picture? Like you know how you write "This picture is obtained from http://www.blahblahblah.com." below the picture?

That is all. Thank you for reading and helping me out.

For pages you can begin counting pages from any page you want, just put the command:
[code]\setcounter{page}{1}[/code]


part i. If $f$ is $C^{n}$ on an interval and has $n+1$ distinct zeroes, prove that $f^{(n)}$ has at least one zero on the interval.

part ii. If $f$ is $C^{n}$ on an interval and $f^{(n)}$ never vanishes, then $f$ has at most $n$ zeroes on the interval.

ii) have from i). i) have from Rolle's theorem.  :wink:

the two question have from rolle's theoreme


analytic geometry
conics
ellipse
calculus
derivative
function
geometry

One day, Johnny gave Josie 4 red roses, each with 33 to 47 thorns. When Josie grabbed the 4 red roses with her right hand, she felt an estimate of 2, 3, or 4 thorns. Assuming that Zero is the center of the universe and everything in this universe is math, if each thorn gives Josie 200 units of ouch and each rose gives Josie 500 units of joy (-500 units of ouch), under what conditions will Josie accept another red rose from Johnny?

I have NO idea.

And I heard the stupid answers don't work (Johnny uses a vase).

Maybe Johnny needs to give her differnt kinds of flowers, since she wouldn't accept roses  :)

Silly you, the obvious answer makes the most sense. Let the 4 roses be circles, who lie in the cartesian coordinate plane as such:

O O
 O O

With the circles touching. Then the ellipse of her hand comes into contact with the roses a maximum of 4 times.Now, adding 0 to each of these roses places them in the center of the universe, or in standard position.
If Johnny gave here a fifth rose, she would transfer 2 roses to her other hand, since otherwises she might touch all 33 spikes on the rose by clasping her hand around it. Then, logically, Johnny would infer to place the roses in the center of her 2 hands, where she would hold them in place. Therefore, she would have lost 400 units of happiness, but gained 500, so she would infer that if Johnny was backwardsly inferable, or the derivative of his inferability existed, he would be able to place the rose in the middle of the two hands, as optimal happinness would be achieved. To calculate if Johnny's mind is backwards inferable, we would try and find a function for the area under his mind. Thereby, the requirements would be if Johnny could find a function for te area under his mind. If he can't find it, he is either not backwards inferable, or he doesn't have the mind capacity to let Jodie have optimal happiness. Q.E.D.

P.S. This was an excellent exercise for "pseudo-mathematic" I believe it's called, making everything make sense under false views of reality.

P.P.S. The second row of circles is spposed to have a space before them.


geometry
analytic geometry
inequalities
ratio
geometric transformation
homothety
reflection

Let l be any line passing through the centroid of a triangle ABC. Prove that the area of 2 parts of ABC (divided by l) differ by not more than 1/9 [ABC].

This is an exercise of barycentric coordinates. I try to use barycentric coordinates to solve it without success. Can anyone give me a hand on it?

Let be $a=BC, b=CA, c=AB$ and denote $G$ the centroid of $\triangle ABC$.  WLOG we can suppose that the line $\ell$ meet the sides $BC, AC$ in $X, Y$ respectively. The barycentric coordinates of $A, B, C, G, X, Y$ wrt the reference triangle $\triangle ABC$ are:

$A(1,0,0), B(0,1,0), C(0,0,1), G(1,1,1)$

$X(0,u,v)$, with $u \geq 0, v \geq 0$ and $u \geq v$ since $BX \leq BM$, ($M=$ midpoint of $BC$)

$Y(u,0,u-v)$

$
[XCY] = [ABC] \cdot \left| {\begin{array}{*{20}c}
   0 & u & v  \\
   0 & 0 & 1  \\
   u & 0 & {u - v}  \\
\end{array}} \right| = [ABC] \cdot \frac{{u^2 }}{{\left( {2u - v} \right)\left( {u + v} \right)}}
$

We have:

$\left| {[XCY] - [ABXY]} \right| \le \frac{1}{9}[ABC]{\rm{   }} \Leftrightarrow $

$\left| {[XCY] - [ABC] + [XCY]} \right| \le \frac{1}{9}[ABC]{\rm{   }} \Leftrightarrow $

$\left| {2[XCY] - [ABC]} \right| \le \frac{1}{9}[ABC]{\rm{   }} \Leftrightarrow $

$\left| {\frac{{2u^2 }}{{\left( {2u - v} \right)\left( {u + v} \right)}} - 1} \right| \le \frac{1}{9}{\rm{   }} \Leftrightarrow $

$\frac{{v\left( {u - v} \right)}}{{\left( {2u - v} \right)\left( {u + v} \right)}} \le \frac{1}{9}{\rm{   }} \Leftrightarrow $

${{\frac{{ - \left( {u - 2v} \right)^2 }}{{\left( {2u - v} \right)\left( {u + v} \right)}} \le 0}}$

The theorem is proved since the last inequality is verified for every $u, v$ with $u \geq 0, v \geq 0, u\geq v$.        $\blacksquare$

[quote="siuhochung"]may i know why you can set the coordinates of Y like that? Shouldn't it be $v,0,u-v$?[/quote]

The equations of the line $AC, XG$ , in barycentric coordinates, are

$AC: \qquad y=0$

$GX: \qquad \left| {\begin{array}{*{20}c}
   x & y & 1  \\
   1 & 1 & 1  \\
   0 & u & v  \\
\end{array}} \right| = 0 \qquad \Rightarrow \qquad \left( {u - v} \right)x + vy - uz = 0$

Thus the intersection point of $GX$ and $AC$ is $Y=(u,0,u-v)$.

Leon

I think the solution without barycentric coordinates is more instructive:

In the following, the area of an arbitrary polygon $P_1P_2...P_n$ will be denoted by $\left[P_1P_2...P_n\right]$. Also, let S = [ABC] be the area of triangle ABC. Finally, let G be the centroid of the triangle ABC, so that the line l passes through this point G.

The line l intersects two of the sides of triangle ABC (hereby, I really mean the sides as the segments between the vertices, not as their extensions). Let's WLOG assume that the line l intersects the sides AB and BC. Denote by M and N the respective points of intersection. Then, the line l divides the triangle ABC in two parts: the triangle MBN and the quadrilateral AMNC. Of course, [MBN] + [AMNC] = [ABC] = S. We have to prove that $\left|\left[MBN\right]-\left[AMNC\right]\right|\leq\frac19S$. In order to prove this, we will show that $\frac49S\leq\left[MBN\right]\leq\frac12S$.

[hide="Why this is enough"]
Why is showing $\frac49S\leq\left[MBN\right]\leq\frac12S$ enough to solve the problem? In fact, since

$\left[MBN\right]-\left[AMNC\right]=2\cdot\left[MBN\right]-\left(\left[MBN\right]+\left[AMNC\right]\right)=2\cdot\left[MBN\right]-S$,

this will yield $\left[MBN\right]-\left[AMNC\right]\leq 2\cdot\frac12S-S=0$ on the one hand, so that $\left|\left[MBN\right]-\left[AMNC\right]\right|=\left[AMNC\right]-\left[MBN\right]$, and on the other hand, it will follow that $\left[MBN\right]-\left[AMNC\right]\geq 2\cdot\frac49S-S=-\frac19S$, so that $\left[AMNC\right]-\left[MBN\right]\leq\frac19S$, and thus $\left|\left[MBN\right]-\left[AMNC\right]\right|\leq\frac19S$. This will solve the problem.[/hide]

Thus, in order to complete the solution, it is enough to show that $\frac49S\leq\left[MBN\right]\leq\frac12S$.

Well, let X, Y and Z be the midpoints of the sides BC, CA and AB of triangle ABC; then, the centroid G of triangle ABC is the point of intersection of the medians AX, BY and CZ. Since the centroid of a triangle divides every median in the ratio 2 : 1, we have $\frac{BG}{GY}=2$, so that $BG=\frac23\cdot BY$. Thus, the homothety with center B and factor $\frac23$ maps the point Y to the point G. Let this homothety map the points C and A to some points V and U, respectively; then, from the properties of homotheties, it follows that:

- these points V and U lie on the segments BC and BA and satisfy $BV=\frac23\cdot BC$ and $BU=\frac23\cdot BA$;
- since the points C, A and Y are mapped to the points V, U and G, and the point Y is the midpoint of the segment CA, its image G is the midpoint of the segment VU.
- since our homothety maps the points A and C to the points U and V and the point B to itself (since B is the center of the homothety), it maps the triangle ABC to the triangle UBV, and thus, as the factor of the homothety is $\frac23$, we have $\left[UBV\right]=\left(\frac23\right)^2\cdot\left[ABC\right]=\frac49S$.

Also, it is well-known that a median of a triangle bisects the triangle's area; thus, $\left[ABX\right]=\frac12\cdot\left[ABC\right]=\frac12S$.

(I really don't know why these trivialities took me half a page to explain. :oops: )

Now let's start fire. Since the angles UGA, AGC and VGC form together an angle of 180°, the line l, passing through their common vertex G, must pass through one of these angles. More precisely, it passes through one of the angles UGA and VGC, since it cannot pass through the angle AGC, since then it would not intersect both of the sides AB and BC of triangle ABC (except of the case when it coincides with one of the lines AG and CG, but in this case it passes through one of the angles UGA and VGC as well). We can WLOG assume that the line l passes through the angle UGA. Then, the point M lies inside the segment UA. Of course, since the line l passes through the angle UGA, it must also pass through its vertical angle XGV, and thus the point N lies inside the segment XV.

Now we will show that $MG\geq NG$. In fact, since the centroid of a triangle divides every median in the ratio 2 : 1, we have $\frac{AG}{XG}=2$, so that AG > XG; thus, if X' is the reflection of the point X in the point G, so that X'G = XG, then AG > X'G, and thus the point X' lies inside the segment AG. Hence, the segment UX' intersects the segment GM at an interior point R. Of course, this implies $MG\geq RG$. Now, since UG = VG (because G is the midpoint of the segment VU) and X'G = XG, we have $\frac{UG}{X^{\prime}G}=\frac{VG}{XG}$, so that Thales yields UX' || VX, what, again, yields by Thales $\frac{UG}{RG}=\frac{VG}{NG}$. Since UG = VG, this becomes RG = NG. But $MG\geq RG$; thus, $MG\geq NG$.

Now, since the area of a triangle equals $\frac12$ of the product of two sides and the sine of the angle between them, we have $\left[MGU\right]=\frac12\cdot UG\cdot MG\cdot\sin\measuredangle UGM$ and $\left[NGV\right]=\frac12\cdot VG\cdot NG\cdot\sin\measuredangle VGN$, so that UG = VG, $MG\geq NG$ and < UGM = < VGN yield $\left[MGU\right]\geq\left[NGV\right]$, and thus

$\left[MBN\right]=\left[BUGN\right]+\left[MGU\right]\geq\left[BUGN\right]+\left[NGV\right]=\left[UBV\right]=\frac49S$.

On the other hand, again since the area of a triangle equals $\frac12$ of the product of two sides and the sine of the angle between them, we have $\left[AGM\right]=\frac12\cdot AG\cdot MG\cdot\sin\measuredangle AGM$ and $\left[XGN\right]=\frac12\cdot XG\cdot NG\cdot\sin\measuredangle XGN$, so that AG > XG, $MG\geq NG$ and < AGM = < XGN yield $\left[AGM\right]\geq\left[XGN\right]$, and thus

$\left[MBN\right]=\left[BAGN\right]-\left[AGM\right]\leq\left[BAGN\right]-\left[XGN\right]=\left[ABX\right]=\frac12S$.

So, altogether, we have found that $\frac49S\leq\left[MBN\right]\leq\frac12S$, and the problem is solved.

[b]EDIT:[/b] More can be said: Since [MBN] + [AMNC] = S, we have [AMNC] = S - [MBN], so $\left[MBN\right]\leq\frac12S$ yields $S\geq 2\cdot\left[MBN\right]$ and thus $\left[AMNC\right]=S-\left[MBN\right]\geq 2\cdot\left[MBN\right]-\left[MBN\right]=\left[MBN\right]$. Hence, $\frac{\left[AMNC\right]}{\left[MBN\right]}\geq 1$. On the other hand, from $\frac49S\leq\left[MBN\right]$ we see that $\left[AMNC\right]=S-\left[MBN\right]=\frac94\cdot\frac49S-\left[MBN\right]\leq\frac94\cdot\left[MBN\right]-\left[MBN\right]=\frac54\cdot\left[MBN\right]$, so that $\frac{\left[AMNC\right]}{\left[MBN\right]}\leq\frac54$. Hence, $1\leq\frac{\left[AMNC\right]}{\left[MBN\right]}\leq\frac54$.

  Darij

[quote="levi"]I can suppose $M\in (AB), N\in (AC)$. I note $AM=xc, BN=yb$, where $x,y\in {\left[ \frac 12 ,1\right]},\ [MAN]=xy\cdot S\ $.
Is wellknown the relation $\ \frac {MB}{MA}+\frac{NC}{NA}=1,\ i.e.\ \frac 1x +\frac 1y =3$. Consequently, $[MAN]-min.\Longleftrightarrow xy-min.$
$\Longleftrightarrow \frac 1x\cdot \frac 1y -max.\ (\ \frac 1x +\frac 1y =3-constant\ )\Longleftrightarrow \frac 1x =\frac 1y =\frac32\Longleftrightarrow x=y=\frac 23 $.  Thus,\[\overline{\underline{\left| [MAN]\ge \frac 49 \cdot S\right| }}\ \ (1)\].
I note $\ u=\frac 1x,\ v=\frac 1y\Longrightarrow u,v\in [1,2],\ u+v=3$. Consequently, $[MAN]-max\Longleftrightarrow xy-max.$
$\Longleftrightarrow uv-min.\Longleftrightarrow u(1-u)-min.$, where $u\in [1,2] \Longleftrightarrow u\in \{ 1,2\}$. Thus,
\[\overline{\underline{\left| [MAN]\le \frac 12 \cdot S\right| }}\ \ (2)\] In conclusion, from the relations (1) and (2) results that $|[BCNM]-[MAN]|=|S-2\cdot [MAN]|\in {\left[ 0,\frac 19\right].}$

[b]Remark.[/b] If$\ I\in MN$, Then $\ b\cdot \frac{MB}{MA}+c\cdot \frac{NC}{NA}=a$. You can propose a alike problem !

Soon I will present a generalization of this problem for a any point $P\in (ABC)$
 without use barycentric coordinates.[/quote]
[u][b]Generalization[/b]:[/u] Let $\ \triangle ABC$, a point $P(x,y,z),\ x,y,z\ge o,\ x+y+z=1$ (barycentric coordinates)

and the points $M\in (AB),\ N\in (AC)$ such that $\ P\in MN$. Prove that:

\[\overline{\underline{\left| 4yz\cdot S\le [MAN]\le \frac{max\ \{ y,z \} }{x+max\ \{ y,z \} }\cdot S\right|}}\ \ \ (*)\]
 [b]Remarks.[/b] The point ${\ P\in [ABC]}$, because $x,y,z\ge 0$. In the relation (*) we have equalities iff:

$\ \nwarrow \frac{AM}{yc}=\frac{AN}{zb}=2;\ \nearrow \ M=B\ (for\ y\le z)\ or\ N=C\ (for\ y>z)$.

[u][i]Some particular cases:[/i][/u]

1. $\ P:=G\Longrightarrow |||\nwarrow MN\parallel BC\ \nearrow M=B\vee N=C.$
2. $\ P:=I\Longrightarrow |||\nwarrow AM=AN=\frac{bc}{s}\ \nearrow M=B\vee N=C$. (For $\ \nwarrow $, the points M,N belong to the circle which is tangent to the lines AB,AC and to the circumcircle).


inequalities
function
inequalities proposed

Prove that for any three positive real numbers $ a,b,c, \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \ge \frac{9}{2} \cdot \frac{1}{a\plus{}b\plus{}c}$.

$ a\plus{}b\equal{}x$ , $ b\plus{}c\equal{}y$ ,$ c\plus{}a\equal{}z$ after cauchy-schwarz inequality

We know that ,for all $ x,y,z,m,n,p$ positive reals it's true that:
$ \frac{x^2}{m}\plus{}\frac{y^2}{n}\plus{}\frac{z^2}{p}\ge\frac{(x\plus{}y\plus{}z)^2}{m\plus{}n\plus{}p}$
with this inequality : $ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\ge\frac{(1\plus{}1\plus{}1)^2}{(a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)}\equal{}\frac{9}{2} \frac{1}{a\plus{}b\plus{}c}$

We can also assume $ a\plus{}b\plus{}c\equal{}1$ since the inequality is homogeneous. Now the function $ f(x)\equal{}\frac{1}{1\minus{}x}$ is convex and by Jensen's inequality it is done.

that's how to derive the overrated Nesbitt's Inequality  :lol:


geometry
geometric transformation
reflection

A disabled tanker leaks kerosene (n=1.2) into the Persian Gulf, creating a large slick on the top of water (n=1.3). If you are scuba diving directly under the region of the slick, whose thickness is 460nm, while the Sun is overhead, for which wavelength(s) of visible light is the transmitted intensity strongest?

i'll probably run out of oxigen waiting for a interference pattern to show.but  i decide to take a helicopter and of course since i am in persian gulf i'll stay low.but wait...something is happening i see some colours from the poluted sea...the kerosene must be it so i try to explain myself what's going on.i know the path difference is 2*n1*d+lambda/2 and the intensity is stronghest if the path is k*lambda.since the wavelength of the visible light ranges from 400 to   700 nanometers then i have to solve this system:
2*n1*d=(k-1\2)/lambda and 400 nm<lambda<700 nm
d-thickness and lambda -wavelength

so i see for what k-integer i fiind lambda between 400-700

now i cannot figure what would happen if i would scuba dive.another opinion?

Can you show me why the path length difference is 2*n1*d+lambda/2? What is n1 and d in your formula?

first i want to say sorry because i see now that a scuba diver will also see a interference pattern just that the path difference is only 2*n1*d (n1-refraction index of the kerosene d-thickness of the kerosene on the water)
     but i was correct in the helicopter scene.to demonstrate the formula you must take 2 rays:one that reflects on the kerosene from the air and another one that refracts in the kerosene then reflects on the separation medium of kerosene-water then refracts again in the air(and because of the first reflection on a medium with a higher index you add lambda/2)

I think the path difference is just 2d because the Sun is overhead. Moreover, the shift in phase for two reflecte rays is $\pi$ (off the higher index of refraction). So, the final phase shift can be considered 0. Therefore, we have the formula for strongest transmitted intensity: $2d=m \frac{\lambda}{n_{1}}$. Put it between 400nm and 700nm to solve for m.
  Any idea?

Can you guys take a look at this problem again?


function
probability
integration
real analysis
real analysis unsolved

Let $ \{f_{\gamma}\}_{\gamma\in \Gamma}$ be an arbitrary set of real-valued $ \mathcal{X}$-measurable functions on $ X$ and $ \{a_{\gamma}\}_{\gamma\in \Gamma}$ be real constants and $ Q$ be a probability measure on $ (X, \mathcal{X})$ for which $ \int f_{\gamma}dQ\equal{}a_{\gamma}$ for every $ \gamma\in \Gamma$. Let $ M$ be the subspace of $ L^1(Q)$ spanned by $ 1$ and the $ f_{\gamma}$'s.
Is $ M$ closed? If so give me some hint to prove.
Thanks.

I just wonder, why it should be closed? There is absolutely no restriction on $ f_\gamma$ except integrability w.r.t. $ Q$.


Write the equation for the surface generated by revolving the given around the indicated axis.

4x + 9y^2 = 36 around the y-axis


Hi, we having an arguement at work. The odds of hitting the big game tonight is 1:135,000,000. What would the odds of me winning be if I bought 2 tickets or 10 tickets? Some people think if you buy 10 tickets the odds will go all the way down to 1:13.5 million. I know you couldn't shave off 122 million just buy buying 10 tickets..

If the odds are 1:135,000,000 and you buy 2 tickets your odds of winning are 1:67:500:000. Your odds would be exactly double (half the odds)  since you have 2 out of 135 million combinations. If you bought 10 tickets your odds would indeed be 1-13.5 million to win the lottery. Don't feel bad, but this is a very common mistake that people make. Just think of it as having 130 socks in a drawer with 130 different numbers on them. If you pull 2 socks out of a drawer you have twice the odds (half 65-1) to pull the sock with your number. In the longrun if you pulled out 2 socks you'd pull one of the correct numbers 1 out of every 65 times. This same theory applies no matter how high the number is. If the lottery was 100 trillion to 1, and you bought 100 tickets the odds would go down to 1 trillion to 1. If you have any other questions feel free to post in this forum. Hope I could help.

It still doesn't seem right. If the odds are 135 million to 1. You have 1:135 mil to win on one ticket and 1:135 mil to win on the other ticket. I don't see how the odds would be cut in half.. Can anyone else try to explain this easier?

There is no other way, or easier way to explain this. Ok, I'll try it this way. There are 135 million different combinations. You take 135 million different garbage cans and put 1 ticket in each garbage can. In 1 of the garbage cans there will be 1 winner. So we got that far. Now the next step to take 13,500,000 garbage cans and put 10 tickets in each one. There will still be 1 winning ticket in one of the garbage cans, only now theres 13,500,000 garbage cans. There is 10% less garbage cans because they each have 10 more tickets in it. So yes! If you buy 10 tickets you have a 10x better chance of winning, although the odds are still miniscule. Now to answer you other question about buying 2 tickets think of 67,500,000 garbage cans and put 2 tickets in each one. You have twice the odds of winning (1:67,500,000). Even to break it down even further. If you bought 1000 tickets your odds would increase 1000%. You would have a 1:135,000 to win the jackpot. If you have trouble imagining your odds dropping that low, just do the garbage can formula. If you have 135,000 garbage cans and put 1000 tickets in each, you will win on one of them.

Hope this helps.

Can anyone else help me out and shed some light on this for Toddy. Thanks

Let's say we took every possible lotto ticket and printed them out.  We would have 135, 000, 000 different tickets.  Now, consider putting your name on one of them and put it back into the pile of tickets.  Now, someone randomnly chooses a ticket.  There are 135, 000, 000 different tickets and only one has your name on it.  Thus your probability of winning is 1 in 135, 000, 000.

Now, consider that you put your name on ten of these tickets.  Then someone picks one at random.  Your probability of winning is now 10 in 135, 000, 000.

Of course, you'll probably end up losing the price of 10 lottery tickets.


Prove that $ 3^{2n\plus{}1}\plus{}2^{n\plus{}2}$ is divisible by $ 7$ for all positive integers $ n$.

[quote="AndrewTom"]Prove that $ 3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2}$ is divisible by $ 7$ for all positive integers $ n$.[/quote]
This is quite a simple one...You may want to do it yourself...The key to solving it is given in the hide tag....
[hide]The three key facts to use are \[ 2^3 \equiv 1 (mod \ 7)\] \[ 3^3 \equiv \minus{}1(mod  \ 7\]
\[ n^{m}\equiv (n\pm xk)^{m}(mod\ k)\]
Now, you take the three cases $ n\equal{}3k,3k\plus{}1 or 3k\plus{}2$ to finish the answer....[/hide]

Thanks. I have already done it (by mathematical induction); I thought it might be of interest to others.

[quote="AndrewTom"]Thanks. I have already done it; I thought it might be of interest to others.[/quote]
Then please post your answer(or atleast some path to it if you are feeling lazy :P )....

hello, rewriting your term in $ 9^n\cdot3 + 2^n\cdot4\4\equiv 2^n\cdot3 + 2^n\cdot4\equiv2^n\cdot7\equiv 0 \mod 7$
Sonnhard.


function
logarithms
ratio
arithmetic sequence

Let $a \in \mathbb{R}$ and $f : \mathbb{R} \to \mathbb{R}$ s.t. \[ f(x)f(y)+f(x)+f(y)=f(xy)+a, \ \forall x,y \in \mathbb{R} \ . \]

Determine all simultaneously continuous and bijective functions which satisfy the above condition.

Well i think i solved it...indeed i just see that if you sum 1 to each side you can factorize ( f(x) +1 )( f(y) +1) = f(xy) + 1 then lets say f(x) = g(x) -1

and that makes g(x) g(y) = g(xy) and we can see that g(x) is also bijective and continous and that is a famous funtion which is the identity iff is continuos and bijective. I think so...

im i right?

i only managed to get to the part $g(x) g(y) = g(xy)$. after that i looked at the official solution, cause i didn't know how to continue.

anyway, what's your final answer?

Well I'm not sure what the answer is, but in trying to find it I found that $f(1)=0$, $f(0)=-1$, and for any $x_1\in\mathbf{R}$ the sequence of numbers $\ldots,f(\omega_{-2}x_1),f(\omega_{-1}x_1),f(\omega_0x_1),f(\omega_1x_1),f(\omega_2x_1),\ldots$ form an arithmetic sequence with a common difference of $f(x_1)+1$, where for any $n\in\mathbf{Z}$, $\omega_n$ is defined as the unique real number such that $f(\omega_n)=n$ (by bijectivity). Maybe something interesting could come out of this?

By the way, I have $0$ previous experience with functional equations, so probably nothing I did had any real relevance to the problem.

Edit I: Heh, I just realized that $f(1)=0$ and $f(0)=-1$ follow from the observation about the arithmetic sequence :D

Edit II: ...and now I realized that it also implies $f(-1)=-2$. (sub-edit: apparently it has already been noticed that $a$ must be $0$?)

indeed, $a=0$ (very easy to see)

but to solve $g(x) g(y) = g(xy)$ where $g$ is bijective and continuous is very hard for someone like me , with not so much experience

the key in the soln given is to write $h(x) = \ln g \left( e^x \right)$ then see what happens

dunno how they came up with that. i think it may be some standard procedure, but it's the first time i see this trick

AntonioMainenti, i should try some more. i'm sure that the problem can be done without the use of $h$

I'll just post what I have so far starting from the beginning (given that $a=0$, which is obvious).

Let $f(\omega_n)=n$ for all $n\in\mathbf{Z}$. Then for some $x_1\in\mathbf{R}$, $nf(x_1)+f(x_1)+n=f(\omega_nx_1)$. Hence $f(\omega_{n+1}x_1)-f(\omega_nx_1)=f(x_1)+1$ for all $n\in\mathbf{Z}$. Taking $x=-1,0,1$ in the last equation we find $f(-1)=-2$, $f(0)=-1$, $f(1)=0$. Letting $n=-2$ in the same equation gives \[f(x_1)+1=f(\omega_{-1}x_1)-f(\omega_{-2}x_1)=f(0)-f(-x_1)=-1-f(-x_1)\\\\\Longrightarrow-\{f(x_1)+1\}=f(-x_1)+1.\]Thus $f+1$ is an odd function. The question now, of course, assuming I didn't screw up somewhere, is whether $f+1$ being odd implies that $f$ satisfies the conditions. Probably not, because I'm thinking we should be able to list the possible functions..

Edit: ..and also some of the steps are not reversible at all.

Off-topic edit: perfect ratio, where are you from? I noticed that your AIM handle is my zip code!


algebra
polynomial
calculus
algebra unsolved

solve equation $ \minus{}2x^3\plus{}10x^2\minus{}17x\plus{}8\equal{}2x^2. \sqrt[3]{5x\minus{}x^3}$

[quote="hxd"]solve equation $ \minus{} 2x^3 \plus{} 10x^2 \minus{} 17x \plus{} 8 \equal{} 2x^2. \sqrt [3]{5x \minus{} x^3}$[/quote]

What an interesting olympiad problem! Where is it coming from ?

Raising to power $ 3$ gives : $ \minus{} 8x^9 \plus{} 120x^8 \minus{} 804x^7 \plus{} 3136x^6 \minus{} 7794x^5$ $ \plus{} 12702x^4 \minus{} 13457x^3 \plus{} 8856x^2 \minus{} 3264x \plus{} 512 \equal{} \minus{} 8x^9 \plus{} 40x^7$

$ \iff$ $ 120x^8 \minus{} 844x^7 \plus{} 3136x^6 \minus{} 7794x^5 \plus{} 12702x^4$ $ \minus{} 13457x^3 \plus{} 8856x^2 \minus{} 3264x \plus{} 512 \equal{} 0$

$ \iff$ $ (6x^2 \minus{} 17x \plus{} 8)$ $ (20x^6 \minus{} 84x^5 \plus{} 258x^4 \minus{} 456x^3 \plus{} 481x^2 \minus{} 272x \plus{} 64) \equal{} 0$

And so at least two real roots : $ \frac {17 \minus{} \sqrt {97}}{12}$ and $ \frac {17 \plus{} \sqrt {97}}{12}$

It remains to study real roots of $ 20x^6 \minus{} 84x^5 \plus{} 258x^4 \minus{} 456x^3 \plus{} 481x^2 \minus{} 272x \plus{} 64 \equal{} 0$ and it seems to me that this polynomial has no real roots.

[i][u]Nota [/u]: All hand-made + standard spreadsheet calculus [/i]:)

i have another solution for your nice equation
$ x=0$ is not the root of the equation so the equation is equivalent to
$ \frac{2x^3-10x^2+17x-8}{2x^2}+2-x=\sqrt[3]{x^3-5x}+2-x$
$ \Leftrightarrow\frac{-6x^2+17x-8}{2x^2}=\frac{6x^2-17x+8}{\sqrt[3]{(x^3-5x)^2}+(2-x)^2-\sqrt{x^3-5x}(2-x)}$
${ \Leftrightarrow(6x^2-17x+8)(\frac{1}{\sqrt[3]{(x^3-5x)^2}+(2-x)^2-\sqrt{x^3-5x}(2-x)}+\frac{1}{2x^2}})=0$
notice that the equation${ \frac{1}{\sqrt[3]{(x^3-5x)^2}+(2-x)^2-\sqrt{x^3-5x}(2-x)}+\frac{1}{2x^2}}$=0 has no roots then we have the result as pco


percent
ratio
geometry
geometric transformation
rotation

A wheel with a rubber tire has an outside diameter of $ 25$ in. When the radius has been decreased a quarter of an inch, the number of revolutions in one mile will:

$ \textbf{(A)}\ \text{be increased about }2\% \qquad\textbf{(B)}\ \text{be increased about }1\%$
$ \textbf{(C)}\ \text{be increased about }20\% \qquad\textbf{(D)}\ \text{be increased about }\frac {1}{2}\% \qquad\textbf{(E)}\ \text{remain the same}$

If the diameter of the wheel is $ 25$ inches, its radius is $ \frac{25}{2}$ inches.  Its diameter is therefore $ 25\pi$.  If its radius is decreased by a fourth of an inch, its new radius is $ \frac{49}{4}$ and its new circumference $ \frac{49\pi}{2}$.  Let there be $ n$ inches in a mile.  Our percent change is thus given by:
$ \frac{\frac{n}{\frac{49\pi}{2}}\minus{}\frac{n}{25\pi}}{\frac{n}{25\pi}}\equal{}\frac{50}{49}\minus{}1\equal{}.204...$.  Our answer is thus an increase of $ 2\%$.

[quote="nittanylion"]Its diameter is therefore $ 25\pi$  Let there be $ n$.[/quote]
Replace 'diameter' with 'circumference'.
Its circumeference is therefore $ 25\pi$.

[quote="nittanylion"]If the diameter of the wheel is $ 25$ inches, its radius is $ \frac {25}{2}$ inches.  Its diameter is therefore $ 25\pi$.  If its radius is decreased by a fourth of an inch, its new radius is $ \frac {49}{4}$ and its new circumference $ \frac {49\pi}{2}$.  Let there be $ n$ inches in a mile.  Our percent change is thus given by:
$ \frac {\frac {n}{\frac {49\pi}{2}} \minus{} \frac {n}{25\pi}}{\frac {n}{25\pi}} \equal{} \frac {50}{49} \minus{} 1 \equal{} .204...$.  Our answer is thus an increase of $ 2\%$.[/quote]

I think you calculated the ratio of the diameters, and not of the circumfrences.

$ (12.25^2)\pi\equal{}150.0625$  $ (12.5^2)\pi\equal{}156.25$

$ \frac{150.0625}{156.25}\equal{}.9604$

The circumfrence of the smaller tire is 96.04% of the larger tire. It takes approximately 1.041 rotations of the smaller tire to equal one rotation of the larger one. Shouldn't the answer be about 4%?

My answer is F, none of the above.

zserf, you calculated areas, not circumference.


If $x$ and $y$ satisfy $\frac{1}{x}+\frac{1}{y}\le \frac{1}{2}$, $x>2$ and $y>2$, compute the minimum value of $2x+y$.

[hide="Solution"] $\frac{1}{2}(2x+y) \ge \left( \frac{1}{x}+\frac{1}{y}\right) (2x+y) \ge \left( \sqrt{2}+1 \right)^{2}= 3+2 \sqrt{2}\implies 2x+y \ge \boxed{6+4 \sqrt{2}}$

This minimum is obtained when $x = 2+\sqrt{2}, y = 2 \sqrt{2}+2$. [/hide]


An incline of weight $\ M$ moves rests on a frictionless surface.  On the incline is a block of mass $\ m$ and there is a coefficient of friction $\ \mu$ between the incline and the block.  A horizontal force of magnitude $\ F$ acts horizontally on the incline.  Find the minimum and maximum values of $\ F$ such that the block does not slip.

This is an interesting problem, and it's too bad it's been overlooked. But could you please clarify wheather the incline rests or not (it's much easier if it rests), and is the angle of inclination given?

The angle is given to be $\theta$.  I don't know what you mean by does the incline rest.  It says in the problem that the incline is moving.

Consider in two dimensional view,-a right triangle ABC with right angle at B.The base is BC and the hypotenuse is BC. Then as per your problem , the mass $m$ is placed on the hypotenuse  AC.Assume that the force is applied on face AC.Now the force would required would be maximum if the direction of $F$ has the effect of PUSHing side AC and the force would be minimmum if  $F$ has the effect of PULLing AC.

Let $N$ be the normal reaction force between the mass $m$ and $M$
and let a be the common horizontal acceleration of the masses.
The equations of motion are-

$F+{\mu}N{\cos{\theta}}-N{\sin{\theta}}= Ma$(for minimum force)
 $F-{\mu}N{\cos{\theta}}+N{\sin{\theta}}= Ma$ 9for force to be maximum

The other two equations aarre common for both
$N{\cos{\theta}}+{\mu}N{\sin{\theta}}= mg$[for vertical equilibrium since $M$ and $m$ both should have equal vertical accelerations and the vertical acceleration of M is zero obviously]

$N{\sin{\theta}}-{\mu}N{\cos{\theta}}= ma$[M and m should have equal horizontal accelerations a]

Solving these three equations we get the minimum and maximum values of the required force as -
$F_{min}$ $=$ $(M-m)g$ $\frac{1-{\mu}cot{\theta}}{1+{\mu}tan{\theta}}$ and
$F_{max}$ $=$ $(M+m)g$ $\frac{1-{\mu}cot{\theta}}{1+{\mu}tan{\theta}}$


geometry
circumcircle
analytic geometry
Euler
graphing lines
slope
geometry proposed

Find the coordinates of the center of the circumcenter of the triangle with vertices at (0,0), (a,0) and (b,c) .

If the coordinates are $ \left(x_{0},y_{0}\right)$, it is clear that $ x_{0}\equal{}\frac{a}{2}$.  Then, $ R^{2}\equal{}\frac{a^{2}}{4}\plus{}y_{0}^{2}\equal{}\left(\frac{a}{2}\minus{}b\right)^{2}\plus{}\left(y_{0}\minus{}c\right)^{2}$, or $ y_{0}\equal{}\frac{b^{2}\plus{}c^{2}\minus{}ab}{2c}$, and we are done...

I've just worked on this problem which asked to show the different steps:

Euler was correct that the centroid, orthocenter, and circumcenter are collinear.

Follow these steps :

Draw a “generic” triangle with vertices at (0,0), (a,0) and (b,c).

Find the equations of any two medians.  Each will be in terms of a, b, and c.

Now solve those two equations simultaneously to find the centroid.

In similar fashion, find the orthocenter and circumcenter.

Your results will again be in terms of a, b, and c.

Now show that the slopes from the centroid to each of the other centers are equal.

There, you’ve proved that the three centers are collinear!

Bonus :

try to prove that the orthocenter lies twice as far from the centroid as the circumcenter does.

[color=darkblue]B.t.w. $ \frac{\overline{GH}}{\overline{GO}}\equal{}\frac{\overline{GN}}{\overline{GI}}\equal{}\minus{}2$ (where $ N$ is the [b]Nagel's point[/b]) for [b]Leo Yard[/b] a little theory :

$ \left\{\begin{array}{c}A(x_{1},y_{1})\ ,\ B(x_{2},y_{2})\\\\ d\ :\ \Delta (x,y)\equiv ax\plus{}by\plus{}c\equal{}0\\\\ A\not\in d\ ,\ B\not\in d\ ,\ AB\not\parallel d\\\\ M\in AB\cap d\end{array}\right\|$ $ \implies$ $ \frac{\overline{MA}}{\overline{MB}}\equal{}\frac{\Delta (x_{1},y_{1})}{\Delta (x_{2},y_{2})}\equal{}\frac{ax_{1}\plus{}by_{1}\plus{}c}{ax_{2}\plus{}by_{2}\plus{}c}$.[/color]


Digging up time capsules seems to work...

breathe deeply

take a nap, i took a nap but when i woke up... still no call :(  not a good feeling..

hahah, i didn't even make USAMO. i'm so lucky. wait...no i'm not.

taking a nap is a good idea...convincing yourself that you have a relatively important physics test tomorrow that has been neglected so far is also a good idea.

Play me a game of chess. :P

I need my points back.

I made MOP in 2002 and 2003 but last year I just had a total meltdown and didn't make it. Believe me when I say that I was stunned at how poorly I had done (I knew I'd done poorly, but it was just stunning)

This year I'm a Winner.

So if you didn't make MOP its not the end of the world, regardless of your expectations.

[quote="beta"]Ah I want to calm down a little but I can't, I can't concentrate on homework or do anything else.. help!!!![/quote]

hope homework is all you have, i've got a test tomorrow. . .

i have 3 tests and a journal to do... ew..

Write your own obituary.  :D

Seriously though, I am writing my own obituary. But only because that's my english writing assignment due Thursday.  :)

Still no call for me!! While a lot of the freshmen I know already got it.. sigh it's over :(

haha i didn't do any homework monday or tuesday, then missed half my classes, so i'm still behind now.

if any of you get "f"s, file stress as an illness and see if you get excused :D

Wait did any 9th grader on the East Coast got the call?

*Hopefully* the scores list is soon to come.

By the way I want to thank the MOP/AMC people for all of their hard work.  In all the confusion of today and yesterday, it seems we have forgotten all the important effort they put in to get these graded and then decide who to invite, so thanks very much for all of that.

JB

Ditto to that.  I went a little insane over the last few days.  Let that insanity be a compliment to the AMC! :D

[quote="jackb1115"]*Hopefully* the scores list is soon to come.[/quote]

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-results.html

so can you explain the meaning of this "hidden link"? or are all of us to be left hanging on the edge of our seat?     (or maybe i should know already sorry)

[quote] It's not a hidden link, you just need to refresh the page with your browser.

[/quote]

Haha

Yeah, me too. I totally overestimated... well not by that much, but by a fair amount.

Yea... actually i did overestimate too... but o well.

Since I'm sure you guys are wondering, (from an unofficial but reliable source)

Brian Lawrence, 10th grader, got the 41.

Eric Price with 35

I got a 30 btw, so I guess they liked my stuff for the LHS of #6 better than I did.

Oh well, I overestimated as well.  I think we all do.


There is an expression

1000!/10 to the power n    (only 10 the the power n)

what is the value of n so that the expression has the least integral value.???

Do solve it

[quote="schoolgirl"][quote="t0rajir0u"]In other words, how many zeros are at the end of $1000!$?[/quote]

There are 200 multiples of 5 from 1 to 1000, among these multiples, $25=5^{2}$, $125=5^{3}$, $625=5^{4}$, therefore $1000!$ is divisible by $5^{206}$, hence there are 206 zeros at the end of $1000!$, n = 206.[/quote]

You undercounted. Look at multiples.

[quote="SM4RT"]isnt it

$\text{Total}=\lfloor\frac{1000}{5}\rfloor+\lfloor\frac{1000}{25}\rfloor+\lfloor\frac{1000}{125}\rfloor+\lfloor\frac{1000}{625}\rfloor=200+40+8+1=\boxed{249}$[/quote]

This is also what I got.


probability
factorial
geometry
3D geometry
logarithms
puzzles

How can you make 24 with 10,16,9 and 4 and JUST division, subtraction, multiplication and/or addition?

9*16/(10-4)

How about try:

10, 10, 4, 4
8, 8, 3, 3
1, 3, 4, 6

:D I also have a program that can generate these if you want more.

Hehe nice. I got the last two first, since I'm well acquainted with 1,3,4,6. Spent two weeks at camp trying to figure it out freshmen year. Heck, it's even on my shirt. And knowing that helped with 3,3,8,8. Beta told me 4,4,10,10. ^_^

Are there multiple ways to do any one of them?

When my cousin and I are on a long car ride and one of us has cards handy, we always spend the whole time dealing random combinations and trying to make them equal to 24. Once I argued with her for so long over whether or not we should be allowed to use exponentiation...

Hehe, a lot of people at our school actually play 24 with cards. Keep going until loser runs out. That's usually me.  :oops:

I always get 1,3,4,6 though! :)

How about my personal favorite, 7, 7, 3, 3?

I actually made a list that shows all the solvable cases.

I also like this one
1, 4, 5, 6
There are two solutions.

have alotta time on your hands lol?? :P

irt JBL's 3,3,7,7
[hide]
(3+3/7)*7
yea that ones pretty cool
[/hide]

Hello!  lightrhee
Do you know how many cases are solvable?
If possible, would you post them here also?

Well, every case that he didn't list would be solvable...

Out of $_{16}C_4$=1820, 458 cases are not solvable. However, probability of non-solvable case to occur is not $\frac{458}{1820}$. Some cases are "heavier" then other cases.

Hey! I'm back from Waterloo Seminar.
Some people wanted to see my program for generating the list. My code in Maple is shown below. If anyone finds a bug or a way to improve, please let me know(since I'm a horrible programmer).

[color=red]> restart;
> cmake_24:=proc(am,bm,cm,dm)
> check:=false:
> fn:=[am,bm,cm,dm]:
> fn:=ListTools[Reverse](sort(fn)):
> for a from 1 by 1 to 3 do
> for b from a+1 by 1 to 4 do
> tn:=fn: tn[a]:=0: tn[b]:=0: tn:=ListTools[Reverse](sort(tn)): x:=fn[a]: y:=fn[b]: oo:=[x+y,x-y,x*y,x/y,y/x]:
> for c from 1 by 1 to 5 do
> tn[3]:=oo[c]:
> for d from 1 by 1 to 2 do
> for e from d+1 by 1 to 3 do
> sn:=tn: sn[d]:=0: sn[e]:=0: sn:=ListTools[Reverse](sort(sn)): x:=tn[d]: y:=tn[e]: os:=[x+y,x-y,y-x,x*y,0,0]:
> if y<>0 then os[5]:=x/y end if:
> if x<>0 then os[6]:=y/x end if:
> for f from 1 by 1 to 6 do
> if os[f]>0 then
> sn[2]:=os[f]: x:=sn[1]: y:=sn[2]: ot:=[x+y,x-y,y-x,x*y,0,0]:
> if y<>0 then ot[5]:=x/y end if:
> if x<>0 then ot[6]:=y/x end if:
> for g from 1 by 1 to 6 do
> if ot[g]=24 then check:=true end if:
> end do: end if: end do: end do: end do: end do: end do: end do:
> check
> end proc:

> count:=0:
> for x from 1 by 1 to 13 do
> for y from x by 1 to 13 do
> for z from y by 1 to 13 do
> for w from z by 1 to 13 do
> if cmake_24(x,y,z,w)=false then
> print(x,y,z,w): count:=count+1
> end if:
> end do:
> end do:
> end do:
> end do:[/color]

I'm assuming we're only using *,/,-. + because with stuff like ! things like 1,1,1,1 become possible as (1+1+1+1)! = 24, see?

[quote="opple"][quote="8th Army"]Try 1, 4, 5, 6[/quote]

6/(5/4-1)[/quote]
There is one more solution. Can you find it? :D

Well it's essentially the same but...
4/(1-5/6)

Here are some easy ones  :D ...
6, 6, 6, 6
5, 5, 5, 5
4, 4, 4, 4
3, 3, 3, 3
12, 12, 12, 12

CHALLENGE PROBLEM  :D
1, 1, 1, 24

[quote="K81o7"]CHALLENGE PROBLEM  :D
1, 1, 1, 24[/quote]
1*1*1*24?

[quote="no_one_2000"]lightrhee:  The answer to 1,4,5,6 that I saw is 4/(1-5/6)=24 (which is similar to what opple said).  Is that what you were thinking of?

[hide="I remember 24..."]
I remember playing the 24 game.  I always did the "Platinum" cards (Decimals, Fractions, Exponents, Algebra... though they changed Decimals to Integers recently).  To win (or get anywhere) in the tournaments, you had to memorize every card to the point where you could quickly glance at four numbers and immediately spit out the final solution.  That took a lot of the fun out of it, but at least the cards they have at the final rounds required a little more than pure memorization.

I always liked the four cards they had on the sides of the box.  As I recall, some of them required really weird solutions.

Somebody should find the probability of "eight times three" being a solution to a randomly selected Single Digit card.  I proctored that one year and all I heard was "Eight times three... Eight times three... Eight times three" :P[/hide][/quote]
For 1, 4, 5, 6, there are only two solutions. $\frac4{1-\frac56}$ and $\frac6{\frac54-1}$

Yeah...the challenge problem was a joke, as you probably guessed...
I just found all of the ones below 15 which I could think of which were 4 of 1 number to make 24.
Here are some more...
1, 2, 5, 10
3, 6, 6, 6
2, 4, 6, 10 (find two ways)

[quote="Hamster1800"]here's a challenge one for yall. 2,5,5,5. It's on the "impossible" list but i found a way to solve it.

You can also do it with 2,5,5,6 which is also on the impossible list.

same with 2,9,9,9, but with the rules that I play with, the solution is banned. The other two aren't though.

4,4,6,6 has a solution that is also banned.

I'll stop with these but think about them.[/quote]
Are $25-\frac55$ and $25-6+5$ the ones you have in your mind? Or did you actually find legal solutions(making 25 with 2 and 5 is not allowed in normal 24 game)?
What do you mean by banned? You mean like $(\frac{9-\frac99}2)!$?

Those aren't the solutions I had in mind. Keep trying!

The rules that I play with require:

no factorials

any radical must be associated with a number, for example, to take a cube root, you need a 3.

Logarithms must have a whole number base which counts as a card (however any time i try to apply logarithms, it can be solved anyway lol)

Is it $5^2-\frac55$ and $5^2-6+5$?(powers are also not allowed in normal 24 game)

yep, but I play with powers so i guess i don't play by the rules.

Try the other ones now.


calculus
integration
function
limit
geometry
rectangle
analytic geometry

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and let $\{a_n\}_n$, $\{b_n\}_n$ be sequences of reals such that
\[ \lim_{n\to\infty} \int^1_0 | f(x) - a_nx - b_n | dx = 0 . \]

Prove that:

a) The sequences $\{a_n\}_n$, $\{b_n\}_n$ are convergent;

b) The function $f$ is linear.

Assuming it's not linear, take three non-collinear points $(a,f(a)),(b,f(b)),(c,f(c))$ on its graph. We can then choose rectangles around the three points with sides parallel to the axes, so that there's no line intersecting all three rectangles. If, for example, the rectangle around $a$ has its vertical sides on the lines of $x$-coordinate $a-\varepsilon$ and $a+\varepsilon$, then for $a_nx+b_n$ which does not contain any point in this rectangle, the quantity $\int_{x-\varepsilon}^{x+\varepsilon}|f(x)-a_nx-b_n|dx$ is always larger than some $\delta_a>0$. We can find $\delta_b,\delta_c$ in a similar way, and then we will always have $\int^1_0|f(x)-a_nx-b_n|dx>\min(\delta_a,\delta_b,\delta_c)$, which contradicts the hypothesis.

Assume that $ \int_{0}^{1} |ax+b| \leqslant \varepsilon $ .Then
$ \int_{0}^{1} ax+b \leqslant \int_{0}^{1} |ax+b| \leqslant \varepsilon $
and

$ \int_{0}^{1} ax^{2}+bx \leqslant \int_{0}^{1} |x||ax+b| \leqslant
\int_{0}^{1} |ax+b| \leqslant \varepsilon $.

Therefore $ \frac{1}{2}a + b \leqslant \varepsilon $ and $ \frac{1}{3}a + \frac{1}{2}b
\leqslant \varepsilon $ . Changing $ a $ to $ -a $ and $ b $ to $ -b $ we get that
$ | a + 2b | \leqslant 2 \varepsilon $ and $ | 2a + 3b | \leqslant 6 \varepsilon $.
Therefore $ |a| = | 2(2a+3b) - 3(a+2b) | \leqslant 3 | a + 2b | + 2 | 2a + 3b |
\leqslant 18 \varepsilon $ and $ |b| = | 2(a+2b) - (2a+3b) | \leqslant 2 | a + 2b | +
| 2a + 3b | \leqslant 10 \varepsilon $.

Now if $ \int_{0}^{1} | f(x) - a_{n}x - b_{n} | \leqslant \varepsilon $ and
$ \int_{0}^{1} | f(x) - a_{m}x - b_{m} | \leqslant \varepsilon $
then $ \int_{0}^{1} | (a_{n}-a_{m})x + (b_{n} - b_{m}) | \leqslant 2 \varepsilon $
so $ |a_{n} - a_{m}| \leqslant 36 \varepsilon $ and $ | b_{n} - b_{m} | \leqslant
20 \varepsilon $. So if $ \lim_{n \rightarrow \infty} \int_{0}^{1} | f(x) - a_{n}x - b_{n} |
 = 0 $ then $ \{ a_{n} \} $ and $ \{ b_{n} \} $ are Cauchy sequences, hence they are convergent.
Suppose $ a = \lim_{n \rightarrow \infty} a_{n} $ and
$ b = \lim_{n \rightarrow \infty} b_{n} $. Then we have
$ \int_{0}^{1} | f(x) - ax - b | = 0 $, so because $ f $ is continuous, we get that
$ f(x) = ax + b $.


i was just wondering how die hard u harry fans are. so submit your answer to the poll. i certainly will go at midnight. but its not like its life or death to me. anyways. nuff said

oops.
the poll didnt work
just say yes or no or sumat like that. since im lazy i wont make another poll or anything. but the options open to anyone else. nuff said

Erm, I'm not a [i]fanatic[/i] per se, but I'll probably read it it a couple of hours anyway so why bother.
Oh, btw, search for it on Kazaa. There's a bunch of funny fakes...

-Isaac

What exactly is the yes/no question? But yep I definitely love harry potter. Will be reading the 5th book as soon as I can get my hands on it, but then I'll probably read it far too fast and then I finish :(

oops again. the question was are u gonna go get harry potter 5 at midnight when it comes out. but the poll didnt work. nuff said

No...I live in a small town, we'll probably get it in a few months once the wagons come 'round  8-) . Anyways, yesterday on CNBC they said at the Amazon warehouse there were 8 ft barricades and armed guards [b]just[/b] for the Ord of Phoenix shipment.

HAHAHAHA... just read on Yahoo... 7680 copies were stolen... lol.
2 1/2 inches thick and 896 pages... can't wait...

-Isaac

I'm not getting it because I don't especially like Harry Potter.

I'm probably going to get it the day after.

Well I'm not going at midnight, but I definitely want to get it. :D

even though its a few days after the book came out. call the store first to make sure they have it. cause at my borders they said that the next shipment would be on tuesday. which is today. if any of you guys went to a local book store i commend you. cause not that many people think of those places. they automatically go to borders or barnes and nobel. or something like that.

Have any of you read the 5th book yet?  I finished it two days ago and I was IMMENSELY disappointed.  I was waiting for an action packed and suspenseful book... I got a elongated and boring book with about only 200 pages of action.  The first 600 pages were total gibberish.  Rowling like purposely elongated it to make it 896 pages!  It was generally a good story... if you can over look a half-a-dozen hundred pages of pointless stuff that could be deleted from the story to make it better.  If it was only shorter... like 500 pages or even 400 pages, it would have been much more interesting... did I mention that it ended really abruptly too?... not to mention how HARRY SCREAMS AT THE READER EVERY OTHER PAGE!  Thats about my criticism... I'm writing a review for the book... not done yet...

-Taylor

it was pretty good. but i agree wit dragonslayerxp. i got rather annoyed with how many "..."s they had and how many "-"s they had. every other line. sheesh. she totally changed her writing style. the last 100 pages were the best. it took a whole 200 pages just to get to hogwash. c'mon give me a break. goblet of fire was much better.

Teen Angst sums it up fairly well though... but I think its still an understatement.  I doubt teen angst causes you to go around yelling your head off at insignificant things every other line.  I also agree with syntax error about how the last 100 pages were the best... well, actually I would say the last 100 pages that were 20 pages from the end.  The last 20 pages I thought were dull.  She talks about how Harry was mad and didn't want to go to the feast.  Then she spends about 2 pages talking about how he looks at Cho and how she's going out with another guy... I thought that was lame.

A very disappointed reader,

Taylor

now when i think about it, i dont mind the book with pages that could be cut out. this is so cause weve waited (well, most of us) 3 years for this book. and i dont really mind as long as it continues the series along.....

[color=indigo]I got the book the day it came out, started reading it a week later and finished in a week.[/color]

it took you a whole week to read it. wow

[color=violet][i]LESS[/i] than a week.  I was busy visiting my cousin and didn't have much time to read.  I could read it in a few days, except I didn't have enough time.

P.S. Mom says [/color][color=indigo]THIS COLOR[/color][color=violet] is hard to read.[/color]

sure is, but if you highlight it, it doesnt matter. highlighting is funnnn zilla

woo, syntax is gonna love spoiler text huh?
I just read Jean Aule's book [u]The Clan of the Cave Bear[/u] in a few days. It was a lot better than I thought it would be.

-Isaac

Read that sometime last year - it was interesting.

Herriot? Alfred Wight? Heard of him...have yet to read though. How are his books?

Alfred Wight? No clue who he is. James Herriot was a veterinarian in Britain around WW2. He wrote a series of really good books about his daily life. They're really funny  :lol:. Try them sometime...

-Isaac

James Herriot is the pen name of James Alfred Wight, if I remember correctly.

James Herriot ugh. whatever floats your boat i guess. he was a little redundant with his titles...


Assume that $ S\equal{}(a_1, a_2, \cdots, a_n)$ consists of $ 0$ and $ 1$ and is the longest sequence of number, which satisfies the following condition: Every two sections of successive $ 5$ terms in the sequence of numbers $ S$ are different, i.e., for arbitrary $ 1\le i<j\le n\minus{}4$, $ (a_i, a_{i\plus{}1}, a_{i\plus{}2}, a_{i\plus{}3}, a_{i\plus{}4})$ and $ (a_j, a_{j\plus{}1}, a_{j\plus{}2}, a_{j\plus{}3}, a_{j\plus{}4})$ are different. Prove that the first four terms and the last four terms in the sequence are the same.

Suppose, for the sake of contradiction, that $ (a_1,a_2,a_3,a_4)\ne(a_{n\minus{}3},a_{n\minus{}2},a_{n\minus{}1},a_{n})$.  Since $ n$ is maximal, the 5-sequences $ (a_{n\minus{}3},a_{n\minus{}2},a_{n\minus{}1},a_{n},0),(a_{n\minus{}3},a_{n\minus{}2},a_{n\minus{}1},a_{n},1)$ have both occured.  Since $ (a_1,a_2,a_3,a_4)\ne(a_{n\minus{}3},a_{n\minus{}2},a_{n\minus{}1},a_{n})$, there must be some element before these two 5-sequences.  By definition, it cannot be $ a_{n\minus{}4}$.  Thus, $ 1\minus{}a_{n\minus{}4}$ must be the element in $ S$ that is right before both 5-sequences.  But then the sequence $ (1\minus{}a_{n\minus{}4},a_{n\minus{}3},a_{n\minus{}2},a_{n\minus{}1},a_{n})$ occurs twice.  Contradiction!


Could some one please help me with this question

A sheet of cardboard measures 20 cm × 30 cm. a square measuring x
cm × x cm is cut from each corner so that an open box can be formed.
a)	Express the length and the width of the base of the box in terms of x.
b)	If area of the base of the box is 264cm², form an equation (factorized) involving the area and x.
c)	Solve the equation to find x.
d)	What are the dimensions of the box when the area if the base is 264cm²?

Looks like a homework question. Anyway:

(a) length=20-2x; width=30-2x

(b) and (c):

(20-2x)(30-2x)=264
4(x-10)(x-15)=264
x^2-25x+84=0
(x-21)(x-4)=0
x=4,21

if x is 21, the length and width is negative and therefore impossible.

therefore x=4

(d) dimensions: 12cm*22cm*4cm


how do you do this problem

on a regular cylinder of radius 2 and height 5, what is the shortest along the fase of the cylinder from point A on the top base of the cylinder to point B diametrically opposite point A on the bottom base.

Point a is on the left corner of the cylinder and point b is on the corner diagnol from that.
[b]Note: i couldn't get a picture of the cylinder in here. [/b]

please include explanation

Unfold the cylinder so that the points A and B are on a rectangle measuring $5\times 4\pi$, and the circumference of the cylinder forms one side of the rectangle.  Since A and B are opposite on the cylinder, B is on the middle of one side (assuming that A is in an opposite corner).  Then the shortest distance, by the Pythagorean theorem, is $\sqrt{25+4\pi^2}$.

I believe that you mean that you have to travel only on the cylinder and not theough it...am I right?

yep samath, you got the right answer to. i checked :starwars:


Let p be a prime number such that $\large p=1(mod 4)$. Evaluate:
$\large \sum ([\frac{2k^2}{p}]-2[\frac{k^2}{p}])$ with k from 1 to (p-1).

It's trivial.Note that there are (p-1)/4 quadratic residure in the interval[p-1/2;p-1]

It's similiar with P5a  of VietnamTST 2005


trigonometry
number theory unsolved
number theory

Find all pairs $(n,r)$ with $n$ a positive integer and $r$ a real such that $2x^2+2x+1$ divides $(x+1)^n - r$.

You need $-\frac{1\pm i}{2}$ to be roots of $(x+1)^n - r$, which leads to $r = \left(\frac{1\pm i}{2}\right)^{\!n} = \frac{1}{2^{n/2}} \;\!e^{\pm i n\pi/4}$. This forces $n = 4k$ and $r = \frac{(-1)^k}{2^{2k}}$, since $r$ has to be real. That's all, I mean...

$2x^2+2x+1=0$ has two roots $\frac{-1}{2}\pm\frac{i}{2}$ which means that $r=(\frac{1+i}{2})^n=2^{\frac{-n}{2}}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^n=2^\frac{-n}{2}e^{\frac{n{\pi}i}{4}}$.Also $(\frac{1+i}{2})^n-r=(\frac{1-i}{2})^n-r=0 \Rightarrow (1-i)^n=(1+i)^n \Rightarrow e^{\frac{-n{\pi}i}{4}}=e^{\frac{n{\pi}i}{4}} \Rightarrow \sin\frac{{\pi}n}{4}=0 \Rightarrow \frac{{\pi}n}{4}=k\pi$
 for some $k \in \mathbb{Z} \Rightarrow n=4k$.


function
algebra unsolved
algebra
USAMO
1993

For each integer $\, n \geq 2, \,$ determine, with proof, which of the two positive real numbers $\, a \,$ and $\, b \,$ satisfying \[ a^n = a + 1, \hspace{.3in} b^{2n} = b + 3a  \] is larger.

Does this wierd way work, as it is very different from Kalva's answer

[hide]Lemma: a>1

If a=1 then the equation becomes 1=2, a contradiction.  If a<1, then a^n<a and a^n-a cannot equal 1, a contradition.

a=sqrt^n(a+1)
b=sqrt^(2n)(b+3a)

denote the symbol ? as not knowing which side is greater

sqrt^n(a+1)?sqrt^(2n)(b+3a)
(a+1)^2?b+3a
a^2+2a+1?b+3a
a^2-a+1?b

So a>b if a^2-a+1>b.  Let a^2-a+1=c and c>a

a<c?sqrt^(2n)(b+3a)
(a+1)^2<c^(2n)?b+3a
(a+-a+1)<c^(2n)-3a?b

Let d=c^(2n)-3a where d>c

c<d?sqrt(2n)(b+3a)
c^2n-3a<d^2n-3a?b

Now we can define a function for the left side as F(m)=F(m-1)^(2n)-a
This functions grows to infinity as m grows to infinity.  The function does not converge as m grows to infinity because F(m-1) grows as m grows and the difference between successive terms increases.

$\infty>b$ so a>b

[/hide]
[hide][/hide][/hide]

I get this strange feeling I'm missing something, since this seemed really easy...  Oh well, I'll post my solution anyway.

[hide]
We have that $a^{n}= a+1 > 1$, so we have that $a > 1$.  The second equation tells us that $b^{2n}= b+3a > 3+b > 3$.  This tells us that $b>1$.  Now we can show that $b^{2n}-b$ is increasing at least when $b > 1$.  Suppose that $x$ and $y$ are real numbers such that $x>y>1$.  This means that $x^{2n-1}-1>y^{2n-1}-1 \implies x^{2n}-x > x(y^{2n-1}-1) > y^{2n}-y$ (true since $x^{2n-1}-1,y^{2n-1}-1 > 0$).  Suppose that $b \ge a > 1$.  Then we have \[b^{2n}-b-3a \ge a^{2n}-a-3a = (a+1)^{2}-4a = (a-1)^{2}> 0\] So we have $b^{2n}> b+3a$ if $b \ge a$, which is impossible.  Thus we have $a > b$ and we're done.[/hide]

We square and rearrange the first equation and also rearrange the second.
\[ a^{2n}\minus{}a\equal{}a^2\plus{}a\plus{}1\]
\[ b^{2n}\minus{}b\equal{}3a\]
It is trivial that $a>1$, $b>1$, thus
\[ a^2\plus{}a\plus{}1>3a\]
\[ a^{2n}\minus{}a>b^{2n}\minus{}b\]
where we substituted to achieve the result.  This writes $(a-b)(a^{2n-1} + a^{2n-2}b + \cdots + ab^{2n-2} + b^{2n-1} - 1) > 0$, and as the second factor is positive, it follows $a>b$.

@limac, I know this is a really late post, but you can show that:
$f(x)=x^(2n)-x$ is an increasing function when x>1. Wouldn't that do the trick.
Are there any other ways of proving this?

Let assume for sake of contradiction that $b>a$, and as above $a,b>1$ so, $b^{2n}>a^{2n}$, $b+3a>a^2+2a+1$, $b+a>a^2+1$, since we assume that $b>a$, $b/a>1$, so $2>b/a+1>a+1/a$, so $2>a+1/a$, contradiction, since $a>1$

I claim that $a>b$ for all positive integers $n$.

First note that from the equality $a^n=a+1$ we have that raising $a$ to an integer power increases its value, implying that $a>1$.  Similarly, we may deduce $b>1$.  Now note that we can write $a^{2n}=a^2+2a+1$, so\[a^{2n}-b^{2n}=(a^2+2a+1)-(b+3a)=(a^2-a+1)-b.\]In order to prove our claim, it suffices to show that $a^2-a+1>b$ for all $n$.  If we can do this, then we have $a^{2n}-b^{2n}>0$, so $a^{2n}>b^{2n}\implies a>b$.

Suppose the contrary, that $b\geq a^2-a+1$.  As $b$ is positive, increasing $b$ increases the discrepancy between $b^{2n}$ and $b$.  Therefore, if we can show that $b^{2n}>b+3a$ for only the case $b=a^2-a+1$, then our desired follows.

Note that from $a>1$ we have $a-1>0$, and squaring this gives $a^2-2a+1>0\implies a^2-a+1>a$.  Now raising both sides to the $(2n)^{\text{th}}$ power gives\[(a^2-a+1)^{2n}>(a^n)^2=(a+1)^2=(a^2-a+1)+3a\implies b^{2n}>b+3a,\] contradiction.  Therefore $a^2-a+1>b$ for all positive integers $n$, and we are done.  $\blacksquare$

Note that $a^n=a+1>1\Rightarrow a>1$ and, thus, $b^{2n}=b+3a>3\Rightarrow b>1$.  Now noting that $(a-1)^2>0\Rightarrow a^2+1>2a$,
\begin{align*}
a^{2n}-a&=a^2+a+1\\
&>3a\\
&=b^{2n}-b
\end{align*}
Now, note that $f(x)=x^{2n}-x$ is increasing for all $x>1$ (seen easily by differentiating).  Thus, we may conclude that $f(a)>f(b)\Rightarrow a>b$.$\blacksquare$

Another solution using Calculus:
First its easy to notice using the expressions that $a>1,b>1$
Consider the function $f(x)=x^{2n}-x-3a$ for $x\in [1,\infty)$ clearly $x=b$ is a root of it. We see that $f'(x)>0$ for all $x\in [1,\infty)$, thus $f$ is strictly increasing on this interval. We know that such a function can intersect $x-$axis at most once thus $b$ is the only possible root. Now we have $f(a)=(a-1)^2>0$ and $f(1)<0$ thus using Intermediate Value Theorem  $1<b<a$.


geometry
trigonometry
calculus
calculus computations

1)A ladder 10ft long rests against a vertical wall.If the bottom of the ladder slides away from the wal at a speed of 2ft/s,how fast is the angle between the top of the ladder and the wall changing when the anglie is pi(3.14...)/4 rad?

2)Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rads/s.Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length Pi(3.14..)/3.

show steps plz i really need help :(

1)if we denote the angle between the top of the ladder and the wall by $\theta$ then we have $\sin \theta = \frac{x}{10}$ where $x$ is the distance from the bottom of the ladder to the wall. Use the chain rule $\frac{d (\, \sin \theta)}{d \theta} \frac{d \theta}{dt} = \frac{d (\, \sin \theta)}{dt} = \frac{1}{10} \frac{dx}{dt}$
now you know what to do with the latter equation?  :)

2) denote the angle between the given sides by $\theta$ then the area of the triangle is $A= \frac{1}{2} (4)(5) \sin \theta$
$\frac{dA}{dt}=10 \cos \theta \frac{d \theta}{dt}$   :)


Show that the polynomial $x^{8} +98 x^{4}+1$ can be expressed as the product of two nonconstant polynomials with integer coefficients.

That is strange, when I make your multiplication I get $x^8+34x^4+1$...


Every square of a 8x8 chessboard is numbered from 1 to 64 in the usual way (from left to right and from top to bottom). An array of knights is called peaceful if no pair of knights attack each other. For each peaceful array the sum of the numbers written on the corresponding squares is computed. Find the largest such sum.

[hide="I think..."]
We can put 32 knights on all black or all white fields so:
2+4+6+...+64=32*33=1056
[/hide]

Placing 32 knights on all squares of the same color is an obvious candidate for a maximum. Actually, an easy pigeon-hole argument ensures that 32 is the maximum number of knights that can be placed peacefully. What is hard to prove is that the above configuration is actually a maximum. I have an exhaustive solution based on analizing cases on a 4x4 chessboard, but I'm looking for a somehow easier solution.


Let $ABCD$ be a convex quadrilateral such that: $\angle ABC=\angle ADC=135^\circ$ and $AC^{2}.BD^{2}=2AB.BC.CD.DA$
Prove that: $AC\perp BD$

Let $\alpha=m(\widehat{AC,\,BD})$ and $[XYZ]$ be the area of triangle $XYZ$
We have:
\begin{eqnarray*}[ABCD]^{2}&=&\frac{1}{4}(AC\cdot BD\cdot\sin\alpha)^{2}\le\frac{1}{4}\cdot AC^{2}\cdot BD^{2}\\ &=&\frac{1}{2}AB\cdot BC\cdot CD\cdot DA\\ &=&(BA\cdot BC\cdot\sin 135^\circ)(DA\cdot DC\cdot \sin 135^\circ)\\ &=&2[ABC]\cdot 2[ADC]\\ &\le&\left([ABC]+[ADC]\right)^{2}=[ABCD]^{2}\end{eqnarray*}
Hence: $\sin\alpha=1\Longrightarrow \alpha=90^\circ\Longrightarrow AC\perp BD$