Perfect Squares
number theory
Additive Number Theory
combinatorics
IMO
IMO 1992

For each positive integer $\,n,\;S(n)\,$ is defined to be the greatest integer such that, for every positive integer $\,k\leq S(n),\;n^{2}\,$ can be written as the sum of $\,k\,$ positive squares.

[b]a.)[/b] Prove that $\,S(n)\leq n^{2}-14\,$ for each $\,n\geq 4$.
[b]b.)[/b] Find an integer $\,n\,$ such that $\,S(n)=n^{2}-14$.
[b]c.)[/b] Prove that there are infintely many integers $\,n\,$ such that $S(n)=n^{2}-14.$

Here's a sketch:
Suppose $ n^2$ can be expressed as the sum of $ n^2\minus{}t$ squares, where $ m^2 \leq m \leq (m\plus{}1)^2\minus{}2$. Then the largest square that can occur in such a representation is $ m^2$, so if the square $ j^2$ occurs $ a_{j}$ times in the representation, then we seek non-negative solutions to the system:
$ \sum_{1 \leq j \leq m} j^2 a_{j}  \equal{} n^2$ ... (1), and $ \sum_{1 \leq j \leq m} a_{j} \equal{} n^2 \minus{} t$ ... (2).
Subtracting (2) from (1): $ \sum_{2 \leq j \leq m} (j^2\minus{}1) a_{j}  \equal{} t$ ... (3), where $ \sum_{2 \leq j \leq m} a_{j} \leq n^2 \minus{} t$, or alternatively:  $ t \plus{} \sum_{2 \leq j \leq m} a_{j}  \leq n^2$ ... (4).  Clearly, we can find non-negative integer solutions to (3) and (4), iff we can express $ n^2$ as a sum of $ n^2\minus{}t$ squares.
(a) To show that $ S(n) \leq n^2\minus{}14$, it suffices to prove $ n^2\minus{}13$ is impossible, i.e. $ t\equal{}13$ doesn't work, i.e. the equation $ 3a_{2} \plus{} 8a_{3} \equal{} 13$ has no solutions, which is clearly true.

Now to find solutions to (3) and (4), for a specific value of $ t$, where $ m^2 \leq t \leq (m\plus{}1)^2 \minus{} 2$: let $ a_{m} \equal{} 1$, and let's try to express the remaining value, $ t\minus{}m^2$, as multiples of $ 3$ and $ 8$. Since any number that is at least $ 13$ can be expressed as $ 3a_{2} \plus{} 8a_{3}$ for some values of $ a_{2}$ and $ a_{3}$, this is possible when $ t \minus{} m^2 \geq 14$. If $ t \minus{} m^2 \leq 13$, let's instead let $ a_{m\minus{}1}\equal{}1$, and express the remainder $ t \minus{} (m\minus{}1)^2$ in the form $ 3a_{2} \plus{} 8a_{3}$. We need to find out for what values of $ t$ this construction works and satisfies the equation (4): $ t \plus{} \sum_{2 \leq j \leq m} a_{j} \leq n^2$. Let's do some computations: For any $ x \geq 14$, we can express $ x \equal{} 3a_{2} \plus{} 8a_{3}$, where $ a_{2} \leq 7$, so then $ a_{2} \plus{} a_{3} \leq 7 \plus{} \frac {x\minus{}21}{8} \equal{} \frac {x\plus{}35}{8}$.
If $ t \minus{} (n\minus{}1)^2 \leq 13$, and $ m^2 \leq t \leq (m\plus{}1)^2 \minus{} 2$, we let $ a_{m\minus{}1} \equal{} 1$, and then we can find suitable $ a_{2}, a_{3}$ so that $ a_{2} \plus{} a_{3} \leq  \frac {t \minus{} (m\minus{}1)^2 \plus{} 35}{8} \leq \frac {(m\plus{}1)^2 \minus{} 2 \minus{} (m\minus{}1)^2 \plus{} 35)}{8} \equal{} \frac {4m\plus{}33}{8} \leq \frac {4n\plus{}33}{8}$.

Then, $ \sum_{2 \leq j \leq m} a_{j} \equal{} 1 \plus{} a_{2} \plus{} a_{3} < \frac {4n\plus{}41}{8}$, and so $ t\plus{} \sum_{2 \leq j \leq m} a_{j} \leq (n\minus{}1)^2 \plus{} 13 \plus{} \frac {4n\plus{}41}{8} < n^2$ for $ n \geq 13$.
That means, for $ 14 \leq t \leq (n\minus{}1)^2 \plus{} 13$, we have taken care of all cases. Let's see what we can do with the same strategy for $ (n\minus{}1)^2 \plus{} 13 < t < n^2$: here let $ a_{n\minus{}1} \equal{} 1$, and we can express $ t \minus{} (n\minus{}1)^2 \equal{} 3a_{2} \plus{} 8a_{3}$, so that $ \sum_{2 \leq j \leq n} a_{j} \equal{} 1\plus{}  a_{2} \plus{} a_{3} < 1 \plus{} \frac { t \minus{} (n\minus{}1)^2 \plus{} 35 }{8} < 1 \plus{} \frac { n^2 \minus{} (n\minus{}1)^2 \plus{} 35}{8} < \frac {n}{4} \plus{} 6$.
That means, when $ t < n^2 \minus{} \frac {n}{4} \minus{} 6$, then $ t \plus{} \sum_{2 \leq j \leq n\minus{}1} a_{j} < n^2$, so we have taken care of these cases as well. So what remains is: $ t > n^2 \minus{} \frac {n}{4} \minus{} 6$, so we need to express $ n^2$ as $ k$ squares for $ k \leq \frac {n}{4} \plus{} 6$.

(b) A little experimentation suggests $ n\equal{}13$ works. In light of previous results, we only need to check that $ 13^2$ can be expressed as sum of $ k$ squares for $ k \leq \frac {n}{4} \plus{} 6 \equal{} 8$ squares. We can compute these constructions: $ 169 \equal{} 144 \plus{} 25 \equal{} 144 \plus{} 9 \plus{} 16 \equal{} 121 \plus{} 16 \plus{} 16 \plus{} 16 \equal{} 100 \plus{} 36 \plus{} 16 \plus{} 16 \plus{} 1 \equal{} 144 \plus{} 9 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \equal{} 144 \plus{} 9 \plus{} 9 \plus{} 4 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 144 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 1$. Thus $ n\equal{}13$ does work.

(c) Suppose $ N \equal{} k$ works. Then if $ k$ is odd, $ N \equal{} \frac {k^2 \plus{} 1}{2}$ works, since: $ ( \frac {k^2 \plus{} 1}{2} )^2 \equal{} k^2 \plus{} ( \frac {k^2 \minus{} 1}{2} )^2$, and we can express $ k^2$ as the sum of $ r$ non-zero squares for $ r \equal{} 0, 1, 2 ... k^2 \minus{} 14$. Thus, by tagging along $ ( \frac {k^2 \minus{} 1}{2} )^2$ to the representation of $ k^2$ as $ r$ squares, for $ r < k^2 \minus{} 14$, we can express $ (\frac {k^2 \plus{} 1}{2})^2$ as the sum of $ r$ squares, for $ r \equal{} 0, 1, 2 ... k^2 \minus{} 13$. By using above results, we only need to make sure that it is the sum of $ r$ squares for $ r < \frac {k^2 \plus{} 1}{8} \plus{} 6$, which follows from what we have just done. Using similar method, if $ k$ is even $ N \equal{} \frac {k^2  \plus{} 4}{4}$ works, since $ (\frac {k^2\plus{}4}{4})^2 \equal{} (\frac {k^2\minus{}4}{4})^2 \plus{} k^2$. Thus starting from $ n\equal{}13$, we can keep constructing larger values of $ n$ which work ($ \frac {13^2 \plus{} 1}{2}$ and so on).

$\bullet$ Representing $n^2$ as a sum of $n^2-13$ squares is equivalent to represent $13$ as sum of numbers of the form $x^2-1$, and that's impossible.

$\bullet$ Define $\mathcal{S}:=\{n\in \mathbb{N}:S(n)=n^2-14\}$. Let's prove that $13 \in \mathcal{S}$: first, consider the following identities

$13^2=8^2+8^2+4^2+4^2+3^2$
$13^2=8^2+8^2+4^2+4^2+2^2+2^2+1^2$
$13^2=8^2+8^2+4^2+3^2+3^2+2^2+1^2+1^2+1^2$

Now, if a representation of $n^2$ has $k$ positive squares and one of them is even, then $n^2$ can be represented also by $k+3$ squares, indeed $(2m)^2=m^2+m^2+m^2+m^2$. We have only to show that $13^2$ is really representable as sum of $2,3,4,6$ positive squares, and that's true indeed
$13^2=12^2+5^2=12^2+4^2+3^2=11^2+4^2+4^2+4^2$ and $13^2=12^2+3^2+2^2+2^2+2^2+2^2$.

$\bullet$ We claim that if $n \in \mathcal{S}$ and $n\ge 13$ then also $2n \in \mathcal{S}$. We notice that if $n^2=\sum_{i=1}^r{x_i^2}$ then also $(2n)^2=\sum_{i=1}^r{(2x_i)^2}$. Replacing $(2x_i)^2$ with $4\cdot x_i^2$. This gives a representation of $(2n)^2$ as sum of $k$ squares for any $k\le 4n^2-62$. Further, we observe that each number $m\ge 14$ can be written as sum of $k\ge m$ numbers of the form $x^2-1$. Therefore if $k\le 4n^2-14$, it follows that $4n^2-k$ is a sum of $k$ numbers of the form $x^2-1$, and consequently $4n^2$ is a sum of $k$ squares.

This is a terribly worded solution, but I believe it works...

[b] a [/b]

I prove that $S(n)=n^2-13$ is absurd proving the rest of this part by definition of $S(n)$.
Start with the set with $n^2-13$ elements $\{1^2, 1^2, \cdots, 1^2\}$.  We are going to have to add some to some elements to get to the desired value.  Changing $1^2$ to $4^2$ gives us a change of $15$, so we are above our upper bound.  Changing $1^2$ to $3^2$ gives us a desire of adding in $5$ more.  However, changing another $1$ to $3$ doesn't work and another $1$ to $2$ is going to give us $3$ more, which done twice doesn't work.

Therefore, no such operation works which will increase $n^2-13$ to $n^2$, and we are done.  $\Box$

[b] b [/b]

Okay so dinoboy told me $13$ worked and gave me some ideas on how to prove it, let's use that here...

Okay so $13^2=13^2$, $13^2=5^2+12^2$, $13^2=12^2+3^2+4^2$, $13^2=8^2+10^2+4^2+1^2$

Now, use the following factorization dinoboy found:  $13^2=8^2+8^2+4^2+4^2+3^2$  Using this, we get $13^2=8^2+8^2+4^2+5^2$.

I define any number $k$ such that $n^2$ can be expressed as the sum of to be a brilliant number base $n$. Therefore, we are going to get $8^2=64*1^2$ so we get $64$ there, and we have two of those so $128$.  Then $4^2+4^2=32$ so we get $32$ more.  Therefore, we get $161$, and the value we desired was $155$ so therefore this is more (the max bound doesn't mean larger values can't work, remember the definition).  This is for the first value of $13^2$.

Therefore, we have proved that $2\pmod{3}$ always gives a brilliant number base $n$.

Now, we prove $1\pmod{3}$ always works.  To do this, note that $13^2=8^2+8^2+4^2+5^2$.  We are going to get $128+16+1=145$ brilliant numbers base $n$.  We desire to have $169-14=154$ values brilliant numbers base $n$, so we have to prove the next few.  $n^2-14$ works is as follows: have a set of $n^2-14$ ones, then change a $1$ to a $3$ giving us+8 and change 2 $1$'s to a $2$ giving us the desired.  For remaining values, change a $1$ to a $2$ proving it works.

For proving $0\pmod{3}$, note that $13^2=8^2+8^2+4^2+3^2+3^2+2^2+1^2+1^2+1^2$  This has $9$ numbers in it, so we have to prove for $6$ it works too, which is done by $13^2=12^2+3^2+4^2$ and use $12^2=6^2+6^2+6^2+6^2$.

From this sum, we get $128+16+4+5=153$ from using basic decomposition.  Therefore, we are happy because we wanted up to $154$, and we are finally done.

So $\boxed{13}$ works yay.

[b] c [/b]

I prove $n$ is a number that works implies $2n$ works also.

I prove that $4n^2$ is expressible as the sum of squares for sets with sizes of $4n^2-56$ to $n^2-15$.  Also sets with size $n^2-14$ to $1$ work because $n$ works.

We prove that $4n^2-56$ works, then $4n^2-57$ and $4n^2-58$ also work.  $n^2-14$ is $\{2,2,2,2,\cdots, 4,4,6)\}$ with $n^2-14$ elements ($n^2-17$ $2$'s), $n^2-15$ is going to be $\{2,2,2,2,\cdots, 8\}$ with $n^2-15$ elements $(n^2-16)$ $2's$ (this means that when you square these numbers it has the desired number of elements I will continue using this notation ), and $n^2-16$ is $\{2,2,2,2,2,\cdots, 6,6)\}$ with $n^2-16$ elements and $n^2-18$ $2's$ Therefore $4n^2-56$ is going to be a set with $8$ $2's$, $4$ $3$'s, and $4n^2-68$ $1$'s which has $4n^2-56$ elements, $4n^2-60$ is going to be a set with $4n^2-64$ 1's and an $4$, and $4n^2-64$ is going to be a set with $8$ $3$'s and $4n^2-72$ $12$'s (essentially 4 times of everything in the above sets) We desire to turn these into the respective sets with $n^2-14, n^2-15, n^2-16$ element sets with sum $4n^2$, decreasing it by $3$ each time proving it works for $0,1,2$ $\pmod{3}$.  We decrease the number of elements by changing every group of $4$ into one group. Therefore, changing a $2$ to a $1$ changes it by $3$, then do this enough time to be able to switch a $4$ to a $2$ without missing any numbers, and continue doing this and we are done $\Box$.

We therefore have to prove it works for $4n^2-55$ to $4n^2-14$.

For $4n^2-14$, we are going to use $\{1,1,1,1,\cdots, 2,2,3\}$ with $4n^2-14$ elements.  The operation decreasing set size by $3$ works here ($1$ to $4$ then remove three $1$'s), so we just prove that $4n^2-14, 4n^2-15, 4n^2-16$ work and use this decreasing function thing (and use $n\ge 13$ so we have enough $1$'s).

For $4n^2-15$, we get $\{1,1,1,1,\cdots, 4\}$, and for $4n^2-16$ we get $\{1,1,1,1,\cdots, 3,3\}$ therefore we are done $\Box$

[b]d)[/b] Prove that if $S(n)\ge n^2/4$ then $n\in\mathcal{S}$.

[quote=HarvardMit]not a good problem tbh[/quote]

not a good opinion tbh

Another solution for part (c): First we observe that
[b]Lemma.[/b] Suppose $n\geq 6$ and $11\leq k\leq n^2-14$. Then, $n^2$ can be written as a sum of $k$ positive squares.
[i]Proof.[/i] We wish to find positive integers $a_1,a_2,\ldots,a_{\ell}$ ($\ell\leq k$) such that
$$ (a_1^2 - 1) + (a_2^2 - 1) + \cdots + (a_{\ell}^2-1) = n^2 - k. $$
If $11\leq k\leq n^2-17$, then we can find an integer $0\leq r\leq 7$ such that $n^2-k-3r\equiv 3\pmod{8}$ and $n^2-k-3r\geq -4$, so if we take $a_1 = a_2 = \cdots = a_r = 2$, then the equation becomes
$$ (a_{r+1}^2 - 1) + \cdots + (a_{\ell}^2-1) = n^2 - k - 3r, $$
Now we have $n^2-k-3r+4\geq 0$. By Lagrange's four-square theorem, we can find non-negative integers $a_{r+1},\ldots,a_{r+4}$ such that
$$ (a_{r+1}^2 - 1) + \cdots + (a_{r+4}^2-1) = n^2 - k - 3r, $$
but these four numbers must be non-zero because $n^2-k-3r\equiv 3\pmod{8}$, so we are done with the cases $11\leq k\leq n^2-17$. For the case $k = n^2-16$, take $a_1 = a_2 = 3$; for the case $k = n^2-15$, take $a_1 = 4$; for the case $n=k^2-14$, take $a_1 = 2$, $a_2 = 3$. [b]Q.E.D.[/b]

In view of the lemma, it suffices to find integers $n\geq 6$ so that $n^2$ can be written as a sum of $k$ positive squares for all $k\leq 10$, but this is easy: define a sequence $\{a_i\}$ of odd positive integers and a sequence $\{b_i\}$ of even positive integers as follows: $a_1 = 3$, and, for $i=1,2,3,\ldots$, $b_i = \frac{a_i^2-1}{2}$ and $a_{i+1} = b_i+1$. It is easy to check that $a_{i+1}^2 = a_i^2 + b_i^2$, and hence for $i\geq 10$,
$$ a_i^2 = a_{i-1}^2 + b_{i-1}^2 = a_{i-2}^2+b_{i-2}^2 + b_{i-1}^2 = \cdots = a_{i-9}^2+b_{i-9}^2 + \cdots + b_{i-1}^2, $$
so $n$ be any of the numbers $a_{10},a_{11},a_{12},\ldots$.

Here's $c.$

If $$n^2=x_1^2+\dots +x_r^2,$$ it follows that $$(2n)^2=(2x_1)^2+\dots+(2x_r)^2.$$ Replacing $(2x_i)^2$ with $x^2_i+x^2_i+x^2_i+x^2_i$ as long as it is possible we can obtain representations of $(2n)^2$ consisting of $r, r+3,\dots, 4r$ squares. This gives representations of $(2n)^3$ into $k$ squares for any $k\leq 4n^2-62.$

Also notice that every number $m\geq 14$ can be expressed as a sum of $k\geq m$ numbers of the form $x^2-1: x\in \mathbb{N},$ which is clear to see. It follows that if $k\leq 4n^2-14,$ then $4n^2-k$ is a sum of $k$ numbers of the form $x^2-1,$ which implies that $4n^2$ is a sum of $k$ squares.

So when $s(n)=n^2-14,$ for any $n\geq 13,$ we have $s(2n)=(2n)^2-14.$ And Q.E.D.


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angle bisector

Let $ABC$ be a triangle with sides of length $a$, $b$ and $c$. Let the bisector of the angle $C$ cut $AB$ in $D$. Prove that the length of $CD$ is \[ \frac{2ab\cos \frac{C}{2}}{a+b}. \]

[quote="4everwise"][hide="Using Triangle Area"]Let $CD=d.$ Note that $[\triangle ABC]=[\triangle ACD]+[\triangle BCD].$ This can be rewritten as $\frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .$

Because $\sin C=2\sin \frac C2 \cos \frac C2$ the expression can be written as $2ab\cos \frac C2=d(a+b).$ Dividing by $a+b,$ $CD=\frac{2ab\cos \frac C2}{a+b}.$[/hide][/quote]

that is pretty nice  :) , for some reason, i never seem to see using that formula...i did this problem the long way when going through aops...

[quote="madatmath"]the result is direct from Apollonius theorem,
 using the property that an angle bisector divides the opposite side in
 the same ratio as the adjacent dsides.[/quote]Can you explain how apollonius works? I never quite got that. :(

[quote="madatmath"]the result is direct from Apollonius theorem,
 using the property that an angle bisector divides the opposite side in
 the same ratio as the adjacent dsides.[/quote]

i believe that you are thinking of stewart's theorem...apollonius is for relating the [i]median[/i] of a triangle to the other sides

oh...and i found a purely geometric proof of the form without the cosine...look at the most current post in my blog  ;)

Yes, sorry, i did mean Stewart's theorem. i confuse them too often.
 Apply Stewart's , then use half angle formula for cosine.

[hide="An elementary proof."]Denote: $P\in AB$ so that $\widehat {ACP}\equiv\widehat {BCP}\ ;$ $M\in (AC)$ so that $PM\parallel BC\ .$
We prove easily that $MC=MP$ and $\boxed {\ l_c\equiv CP=2\cdot MP\cos \frac C2\ }\ .$
$\frac{PA}{b}=\frac{PB}{a}=\frac{c}{a+b}\Longrightarrow \boxed {\ PA=\frac{bc}{a+b}\ }\ .$
$\frac{MP}{BC}=\frac{AP}{AB}\Longrightarrow MP=\frac ac\cdot AP=\frac ac\cdot \frac{bc}{a+b}=\frac{ab}{a+b}\Longrightarrow \boxed {\ MP=\frac{ab}{a+b}\ }\ .$
Therefore, $\boxed {\ l_c=\frac{2ab}{a+b}\cdot \cos\frac C2\ }\ .$

[b]Remark.[/b] $\boxed {\ \frac{1}{MP}=\frac 1a+\frac 1b\ }\ .$[/hide]

Search?

Exact problem was posted in same forum...fairly recently...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=show&t=82787[/url]

i've learned recently that using area instead of pure side length formulas makes these types of problems mounds easier..

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1969Problem5

Vo Duc Dien

Let $a, b, c, d$ be the lengths of $BC, AC, AB, CD$ respectively. Solving for $d^2$ in Stewart's theorem gives us $$d^2 = ab \left(1 - \frac{c^2}{(a+b)^2}\right)$$ Law of cosines states that $c^2 = a^2 + b^2 - 2ab \cos{C}$. Substituting this into the above equation yields
\begin{align*}
d^2 &= ab \left(1 - \frac{a^2 + b^2 + 2ab - a^2 - b^2 + 2ab \cos{C}}{(a+b)^2}\right) \\
&= 2 a^2 b^2 \left(\frac{1 + \cos{C}}{(a+b)^2}\right) \\
&\Longleftrightarrow d = ab \sqrt{2} \left(\frac{\sqrt{1 + \cos{C}}}{a+b} \right)
\end{align*}
Using the identity $\cos{\frac{C}{2}} = \sqrt{\frac{1 + \cos{C}}{2}} \Longleftrightarrow 2\cos{\frac{C}{2}} = \sqrt{2}\sqrt{1 + \cos{C}}$ and substituting yields $$d = \frac{2ab \cos{\frac{C}{2}}}{a + b}$$as desired. $\square$
Perhaps I used an unnecessary number of identities here. :P


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Let $ABC$ be a triangle and let $AU$ be the internal bisector of $\angle A$ with $U$ on $BC$. Let $V$ be on $AB$ and $W$ be on $AC$ such that $V, U, W$ are collinear. Let $I$ be the incentre of $\triangle ABC$ and $I_A$ be the excentre on the internal bisector of $\angle A$. If $IV$ and $I_AW$ meet at $P$ and $IW$ and $I_AV$ meet at $Q$, show that $PQ \perp AU$.

I have a proof with vectors but I was wondering if there's a more classical proof.

A is the external homothety center of the circles $(I), (I_A),$ U is their internal homothety center. Hence U divides the segment $II_A$ in the ratio $\frac{r}{r_a}$ internally, while A divides this segment in the same ratio externally.

$\frac{UI}{UI_A} = \frac{AI}{AI_A} = \frac{r}{r_a},\ \ \ \frac{UI}{UI_A} \cdot \frac{AI_A}{AI} = 1$

Using a central projection, project the line AU to itself and the normal to $AU$ at A to infinity. Since $\angle BAU = \angle CAU,$ the lines AB, AC are projected to lines $A'B',\ A'C'$ parallel to $A'U' \equiv AU$ and equidistant from AU. Since the cross-ratio $\frac{UI}{UI_A} \cdot \frac{AI_A}{AI} = \frac{U'I'}{U'I_A'} \cdot \frac{A'I_A'}{A'I'} = 1$ is preserved and since $A'I_A' = A'I'$ for finite $I_A', I'$ and $A'$ at infinity, it follows that $\frac{U'I'}{U'I_A'} = 1,$ i.e., U' is the midpoint of $I'I_A'.$ Since U' is also the midpoint of V'W', the diagonals $V'W',\ I'I_A'$ of the quadrilateral $I'V'I_A'W'$ cut each other in half and it is a parallelogram. Hence both $P' \equiv I'V' \cap I_A'W'$ and $Q' \equiv I'W' \cap I_A'V'$ are at infinity, i.e., on the image of the normal to AU at A. Consequently, the original points P, Q lie on this normal and $PQ \perp AU.$

[hide="Here is a proof with harmonical division, three similar problems and a generalization."][color=darkblue]The division $(A,I,U,I_a)$ is harmonically, i.e. $\frac{IA}{IU}=\frac{I_aA}{I_aU}\ .$ Thus, the pencil $V(A,I,U,I_a)$ is harmonically. Denote the intersection $R\in AB\cap I_aW\ .$ Therefore, and the division $(R,P,W,I_a)$ is harmonically. But the ray $[AI_a$ is the external bisector of the angle $\widehat {WAR}\ .$ We came to the conclusion that the ray $[AP$ is the internal bisector of the same angle $\widehat {WAR}\ ,$ i.e. $AP\perp AI_a\ .$
We can prove similarly that $AQ\perp AI_a\ ,$ i.e. $A\in PQ$ and $PQ\perp AU\ .$[/color]

[color=darkred][b]A equivalent enunciation.[/b] Let $ABC$ be a triangle for which denote the incircle $w=C(I,r)$, the $A\ -$ exincircle $w_a=C(I_a,r_a)$ and the intersection $D\in BC\cap AI\ .$ For a mobile line $d\ ,$ $D\in d$ define the intersections $E\in AC\cap d$ and $F\in AB\cap d\ .$ Ascertain the geometrical locus of the points $L_1\in IE\cap I_aF$ and $L_2\in IF\cap I_aE\ .$[/color]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=86344&start=40

[color=red][b][u]Three similar problems.[/u][/b]

$1.\blacktriangleright$ Let$ABC$ be a triangle and given are the points $D\in (BC)$, $E\in (CA)$, $F\in (AB)$. Prove that if two of the following sentencies are truly then and the other sentence is truly :
$1.\ \ \widehat {ADE}\equiv\widehat {ADF}\ .\ \ \ \ 2.\ \ AD\cap BE\cap CF\ne\emptyset\ .\ \ \ \ 3.\ \ AD\perp BC\ .$

$2.\blacktriangleright$ Let $ABCD$ be an isosceles trapezoid $(AB\parallel CD\ ,\ AD=BC)$. Denote $I\in AC\cap BD$ and the middlepoints $M$, $N$ of the sides $[AB]$, $[CD]$ respectively. For a mobile line $d$, $I\in d$ define the intersections $X\in AD\cap d$ and $Y\in BC\cap d$. Ascertain the geometrical locus of the points $L_1\in XN\cap YM$ and $L_2\in XM\cap YN$.
[u]Indication.[/u] For $S\in AD\cap BC$ the division $(M,I,N,S)$ is harmonically.

$3.\blacktriangleright$ Let $ABC$ be an isosceles triangle $(AB=AC)$ inscribed in the circle $w$. Denote the intersections $P\in BB\cap CC$, $M\in BC\in AP$ and $\{A,N\}=AP\cap w$. For a mobile line $d$, $N\in d$ define the intersections $X\in AB\cap d$ and $Y\in AC\cap d$. Ascertain the geometrical locus of the points $L_1\in MX\cap PY$ and $L_2\in MY\cap PX$.
[u]Indication.[/u] the division $(A,M,N,P)$ is harmonically.

[b][u]Generalization.[/u][/b] Given are the angle $\widehat {xAy}$, the ray $[Az\subset \left(\widehat {xAy}\right)$ and the points $\{B,C,D\}\subset [Az$ so that the division $(A,C,B,D)$ is harmonically, i.e. $C\in (AB)$, $B\in (CD)$ and $\frac{CA}{CB}=\frac{DA}{DB}$. A mobile line which passes through the point $B$ meets the rays $AB$ and $AC$ in the points $M$ and $N$ respectively. Ascertain the geometrical locus of the points $L_1\in MC\cap ND$ and $L_2\in MD\cap NC$.[/color][/hide]
[hide="Answer for the generalization."] [color=darkblue]The perpendicular line in the point $A$ on the ray [/color]$[Az.$[/hide]

In the initial problem, the points $B,C$ appear only to define the point $U$, for to define the harmonical division $(A,I,U,I_a)\ !$ Try to generalize this problem. It is a interesting problem.


geometry email MATHCOUNTS AMC AIME USA(J)MO USAMO

Could someone please give me a rundown on the classes offered, and how difficult each class is? i.e., the topics covered.

[quote="lifeisacircle"]http://www.awesomemath.org/AMSP_dailysch.pdf Check the 1st page for courses and pages 2-4 for the descriptions.[/quote]
I feel stupid for not checking there. <_<

Thank you.

[quote="hasan4444"][i][color=darkred]OK is there anyone take the introductory or medium level?[/color][/i][/quote]

i'd say the middle level classes&the highest level classes are the most crowded/packed/full ones.
but middle level classes are def the most crowded classes.

it doesnt mean that intro level classes are tiny either. there's about 20-25 people in intro level classes.
middle [well computational geometry had like 40 people last year] eeek so squished and..yeah
and im not sure about highest level classes, but there are plenty of people in each level.

Based on what I've seen from my friends who went, the olympiad geo class looks like it has a lot of pretty nice problems :) I'm looking forward to it.
Does anyone know how to sign up for classes? I still have yet to see some kind of sheet that I can fill out regarding them. Will it be fine if I just send them an email?

Please if anyone took Computational Geometry till me what is the previous background that is needed for this course. And if anyone took the introductory level of geometry please till me to what levels does he studied.

the introductory level classes are all mathcounts level. so if you can basically solve most mathcounts problems ur fine.

I would also recomend taking a few practice AIMEs and if ur score is 4-8

you should look at the types of problems you were not able to solve and pick an intermediate subject accordingly.

if it is greater than 10 , then just choose a level that focuses on proofs particularly number theory
because they show up quite frequently on USAMO or require a strong foundation in number theory.


geometry
incenter
angle bisector
geometry unsolved

Let $ABC$ be a triangle. The internal bisector of $\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$. Prove that if $AP+AB = CB$, then $API$ is an isosceles triangle.

Draw $PP'$ parallel to $IA$ so that $P'$ is on line $AB$. Then $\triangle PAP'$ is isosceles, which implies that $BC = AB+AP = AB+AP' = BP'$. this then implies that $\triangle P'BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\angle B$, $P'PC$ is also isosceles, with $PP' = PC$. it then follows, using similarity of triangles and the angle bisector theorem, that

$\frac{IA}{PP'}= \frac{BA}{BP'}= \frac{BA}{BC}= \frac{AP}{PC}= \frac{AP}{PP'}$,

from which $IA = AP$

I found the angles of triangle ABC.
I found that te angle <BAC is 90º and that <ABC=45.And after that  it is easy to show what we want.

That is wrong?

[color=darkblue][b]Proof (metric).[/b] Prove easily that $IA^{2}=\frac{bc(p-a)}{p}$ and $PA=\frac{bc}{a+c}\ .$ Thus,
$\boxed{\ PA+AB=BC\ }$ $\Longleftrightarrow$ $PA=a-c$ $\Longleftrightarrow$ $\frac{bc}{a+c}=a-c$ $\Longleftrightarrow$ $\underline{a^{2}=c(b+c)\ .}$
$\boxed{\ IA=PA\ }$ $\Longleftrightarrow$ $\frac{bc(p-a)}{p}=\frac{b^{2}c^{2}}{(a+c)^{2}}$ $\Longleftrightarrow$ $(a+c)^{2}(p-a)=bcp$ $\Longleftrightarrow$ $bc(a+b+c)-(a+c)^{2}(b+c-a)=0$ $\Longleftrightarrow$ $c\underline b^{2}-a(a+c)\underline b+(a-c)(a+c)^{2}=0\ .$ The discriminant is $\Delta =a^{2}(a+c)^{2}-4c(a-c)(a+c)^{2}=[(a+c)(a-2c)]^{2}\ .$
Thus, the roots of this equation in the variable $b$ are :  $\{\begin{array}{c}b_{1}=\frac{a^{2}-c^{2}}{c}\\\\ b_{2}=a+c\end{array}\ .$ Therefore, $AI=AP$ $\Longleftrightarrow$
 $(a+c-b)(c^{2}+bc-a^{2})=0$ $\Longleftrightarrow$ $\underline{a^{2}=c(b+c)\ .}$ Therefore, $\boxed{\ IA=PA\Longleftrightarrow PA+AB=BC\ }\ .$

[b]Remark.[/b] Denote the points $S\in AI\cap BC$ and $R\in AC$ for which $BR\parallel AI\ .$ Then exists the following chain of equivalencies :

$\boxed{\ PA+AB=BC\Longleftrightarrow AI=AP\Longleftrightarrow B=2C\Longleftrightarrow AS=\frac{bc}{a}\Longleftrightarrow a^{2}=c(b+c)\Longleftrightarrow BR=BC\ }\ .$[/color]

virgil :

  if AD    is th angle  bisector  of  A  then  :  $AD^{2}= bc(a+b+c)(b+c-a)/(b+c)$  no?  and also $AI/ID=b+c/a$ no?  using these formulas i always  get  $AI^{2}=bc(b+c-a)(b+c)/a+b+c$
  where am  i  wrong??

[quote="marko avila"]... $AD^{2}= bc(a+b+c)(b+c-a)/\underline{(b+c)}$  no? ... where am  i  wrong??[/quote] Must $(b+c)^{2}$

ohhh  of course!!   i was so  blind to see  it.  thank  you.

This is a very "standard" and beauty question.

From some standard notations, let
\[\angle A = 2*\alpha\]
\[\angle B = 2*\beta\]
\[\angle C = 2*\gamma\]
This is almost clear that $2(\alpha+\beta+\gamma)=\pi$

We need to show that $\angle AIP=\angle API$, or
$\angle IAB+\angle IBA=\angle IAP$ or even better,
$3\alpha+2\beta=\pi$
Or even $3\alpha+2\beta=2(\alpha+\beta+\gamma)$
Or even $\alpha=2\gamma$.

Let $Q$ be a point on line $BA$, such that $AP=AQ$ and $A$ is between $B$ and $Q$.
By the conditions of problem, $BC=BA+AP=BA+AQ=BQ$, and then $CBQ$ is isosceles. And by construction, $PAQ$ is also isosceles.
This fact has some interesting consequences:

$\angle{PAB}=\angle{PAB}=\alpha$ (external angle $\angle A$)
Let $T$ the common point of $CQ$and $BP$.
Then $BT$ is bissector of $\angle CBQ$, and so this is perpendicular bissector of $CQ$. Then $CP=PQ$.

Then $\angle ACB=\angle QCB-\angle ACQ=\angle CQB-\angle PQC=\angle PQB=\alpha$.
But this is the same as $\alpha=2\gamma$!

Good 'ol angle chasing.

Let $P'$ be a point on $AB$ such that $AP'=AP$. Define $X=\overline{P'C}\cap \overline{BI}$. Note that $$\frac{BA}{AP'}=\frac{BA}{AP}=\frac{BI}{IP},$$ by angle bisector theorem, so triangles $BAI$ and $BP'P$ are similar. It is given that $BP'=BA+AP'=BA+AP=BC$. Thus, $$\measuredangle AIP=\measuredangle P'PX=\measuredangle XPC=\measuredangle IPA.$$
Hence, triangle $API$ is isosceles as needed.

Dear Mathlinkers,

1. the circumcircle of AIC
2. R, Q the symmetric of C, B wrt BI (R, Q is on the last circle)
3. the triangle ARP is A-isoceles
4. AI is parallel RQ
5. we are done by considering two isoceles trapez...

Sincerely
Jean-Louis

My solution:
Let $T$ on the line $AB$ such that $AP=AT$ and $A \in BT$.
It follows $BC=BT$.
We have $\dfrac{AI}{TP}=\dfrac{BI}{BP}=\dfrac{AB}{AB+AP}=\dfrac{AB}{BC}=\dfrac{AP}{PC}=\dfrac{AP}{PT}$.
$\Rightarrow AP=AI$.
Q.E.D

[hide=Solution]
Let $BA$ be produced to $P'$ such, $AP=AP'$, and let, $BP \cap P'C=X \implies BX \perp P'C \implies PP'=PC \implies 90^{\circ}-\frac{B}{2}-\frac{A}{2}=90^{\circ}-\frac{B}{2}-C \implies \boxed{C=\frac{A}{2}}$
Therefore, $\angle AIP=\frac{A}{2}+\frac{B}{2}=\frac{B}{2}+C=\angle XPC=\angle API \implies \Delta AIP \text{ is isosceles}$[/hide]

Angle chasing yields $\angle AIP = A/2 + B/2$ and that $\angle API = 180 - A - B/2 = B/2 + C$. Now let $P'$ be the intersection of the rotation of segment $AP$ about point $P$ and the ray $BA$ so that points $B,A,P'$ are collinear in that order. Then $AP' P$ is isosceles so $\angle AP'P = \angle APP' = A/2$. By the length condition, it follows that $BP' = BC$, so $P' BC$ is isosceles. Line $BP$ is the angle bisector at vertex angle $B$, so $\angle BP'P = \angle BCP \implies A/2 = C$. Thus $\angle API = B/2+A/2$. The result follows.

Nice one!

Hint:
[hide]One can use trig, but if you want a constructive solution, take point $D$ inside $BC$ such that $BD=BA$.[/hide]

Opinion/style of solution:
[hide]My solution is quite short, and I did it constructively, but there was one detail that took me 24567 hours to see, and it is the final conclusion lol :)[/hide]

Take the point in the hint, and note that $ACDI$ is cyclic, but we want it to be an isosceles trapezoid, because if $DI || AC$, then $AI=AP=DC$, by the given condition. But since $\triangle AIB \equiv \triangle DIB$, we get that $IA=ID$, and $I$ belongs to the perpendicular bissector of $AD$, and $B$ also does, and because $P,I,B$ are collinear, $PA=PD=DC$, by the given condition. Now, note that, by congruent triangles $APDI$ is a parallelogram, because it's diagonals meet in their midpoint. And hence, because the diagonals are also perpendicular, $APDI$ is a losangle, completing the proof. $\blacksquare$

Sol:- Let (AIC) intersect BC at A' & AB at R. By symmetry BA'=BA. So A'C=AR=AP . Also R-P-A' are collinear (By ceva) . Angle A=2* Angle ARP= 2* Angle C.
So, Angle AIP= Angle (B+A)/2 = Angle ((B/2)+C)=Angle APB.


[quote="jb05"]We have $2y^2-y=2x^2-x$.[/quote]
 :?: Where did you get this? This is not equivalent to original equation... (pay attention to signs)

jb05, $-\frac{b}{2a}$ gives the value of $x$, not the value of $2x^2 - x$ (the value of $2x^2 - x$ for the $x$-value you calculated is $-\frac{1}{4}$ anyway) and moreover, because $2 > 0$, it gives the minimum, not the maximum.

[hide="Algebraic solution"]What we really have is $2x^2 - x = y - 2y^2 \implies x(2x-1) = y(1-2y)$.  If this value is zero, then $x = 0, \frac{1}{2}$ and $y = 0, \frac{1}{2}$, so we have $x - y = \frac{1}{2}$ our greatest value for this case.

If this value is positive, then $0 < y < \frac{1}{2}$.  The maximum value of $y(1-2y)$ is $\displaystyle\frac{1}{4}$, which produces $x < \frac{3}{4}$ (plug in to verify), so it is not the maximum.  The minimum value gives us $y = 0$ and the previous solution.

If this value is negative, then $0 < x < \frac{1}{2}$, and similar logic again gives us that this does not produce the maximum value, so $x - y = \boxed{\frac{1}{2}}$.[/hide]

[hide="Geometric solution"]We have $(2x^2 - x) + (2y^2 - y) = 0$.  Divide by two:  $(x^2 - \frac{x}{2}) + (y^2 - \frac{y}{2}) = 0$.  Complete the square:  $(x - \frac{1}{4})^2 + (y - \frac{1}{4})^2 = \frac{1}{8}$, so we have the graph of a circle whose center is $(\frac{1}{4}, \frac{1}{4})$ with a radius $\frac{1}{2\sqrt{2}}$.  The maximum value of $x - y$ on a unit circle (and by extension, any circle) occurs at an angle of $-\frac{\pi}{4}$ (this can be proven with a little trig), giving us a difference $x - y = \left( r \cos (-\frac{\pi}{4}) + \frac{1}{4} \right) - \left( r \sin (-\frac{\pi}{4}) + \frac{1}{4} \right) = \frac{1}{2\sqrt{2}} ( \sqrt{2} ) = \frac{1}{2}$. [/hide]

[quote="mollusk"]Find the maximum value of x-y if 2(x^2+y^2) = (x+y)[/quote]

[hide]let $x-y=k$, so we must maximize $k$

substituting, we get

$4y^2+(4k-2)y+2k^2-k=0$

since $y$ must be real (in dealing with maximizing, we cannot have imaginary parts)

then using the discriminant and simplifying a bit

$-4k^2+1 \ge 0$

$-\frac{1}{2} \le k \le \frac{1}{2}$

and clearly the max is $\boxed{\frac{1}{2}}$

i suppose that i have to show that maximum for $k$ is attained, but that part is trivial, all i need to do is substitute and find an ordered pair (x,y) that satisfies $k=\frac{1}{2}$[/hide]


trigonometry
function
inequalities
inequalities solved

That's a very easy problem  :lol:
Consider the function $f(x)=\frac{1-\sin(x)}{1+\sin(x)}$ for $0<x<\frac{\pi}{2}$
$f''(x) \geq 0$ so $f$ is concave on $[0;\frac{\pi}{2}]$
By applying Jenson inequality we get the result immediatly  :lol:

Let $f(x)=\frac{1-\sin x}{1+\sin x}$. Then $f''(x)=\frac{2(2\cos^2 x+\sin^2 x + \sin x)}{(\sin x + 1)^3}$, which is clearly positive for all $x \in [0, \frac{\pi}{2}]$. So we can use Jensen for $f(x)$. We get \[f(A)+f(B)+f(C) \geq 3 f(\frac{A+B+C}{3}) = 3 f(\frac{\pi}{6}) = 1\]

[quote="amir2"]Three angles $ A$, $ B$, $ C$ satisfy $ 0<A,B,C<\frac{\pi}{2}$ and $ A+B+C=\frac{\pi}{2}$.
Prove that:
$ \frac{1-\sin A}{1+\sin A}+\frac{1-\sin B}{1+\sin B}+\frac{1-\sin C}{1+\sin C}\geq 1$.[/quote]

Let $ x=\frac{\pi}{2}-A$, $ y=\frac{\pi}{2}-B$, $ z=\frac{\pi}{2}-C$. Then, cos x = sin A, cos y = sin B, cos z = sin C. Thus, the inequality we have to prove,

$ \frac{1-\sin A}{1+\sin A}+\frac{1-\sin B}{1+\sin B}+\frac{1-\sin C}{1+\sin C}\geq 1$,

becomes

$ \frac{1-\cos x}{1+\cos x}+\frac{1-\cos y}{1+\cos y}+\frac{1-\cos z}{1+\cos z}\geq 1$.

But $ 1-\cos x=2\sin^{2}\frac{x}{2}$ and $ 1+\cos x=2\cos^{2}\frac{x}{2}$; thus, $ \frac{1-\cos x}{1+\cos x}=\frac{2\sin^{2}\frac{x}{2}}{2\cos^{2}\frac{x}{2}}=\left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\right)^{2}=\tan^{2}\frac{x}{2}$. Similarly, $ \frac{1-\cos y}{1+\cos y}=\tan^{2}\frac{y}{2}$ and $ \frac{1-\cos z}{1+\cos z}=\tan^{2}\frac{z}{2}$. Hence, the inequality that we have to prove becomes

$ \tan^{2}\frac{x}{2}+\tan^{2}\frac{y}{2}+\tan^{2}\frac{z}{2}\geq 1$.

But this is a very easy consequence of Jensen: In fact, since the angles A, B, C lie in the interval $ \left]0;\;\frac{\pi}{2}\right[$, the angles $ x=\frac{\pi}{2}-A$, $ y=\frac{\pi}{2}-B$, $ z=\frac{\pi}{2}-C$ also lie in this interval; hence, the angles $ \frac{x}{2}$, $ \frac{y}{2}$, $ \frac{z}{2}$ lie in this interval, too. Since $ A+B+C=\frac{\pi}{2}$, we have

$ x+y+z=\left(\frac{\pi}{2}-A\right)+\left(\frac{\pi}{2}-B\right)+\left(\frac{\pi}{2}-C\right)=\frac{3\pi}{2}-\left(A+B+C\right)=\frac{3\pi}{2}-\frac{\pi}{2}=\pi$.

The function $ f\left(t\right)=\tan^{2}t$ is convex on the interval $ \left]0;\;\frac{\pi}{2}\right[$ (since the function tan t is convex on this interval, and the function $ t^{2}$ is convex and monotonically increasing). Hence, by the Jensen inequality,

$ \tan^{2}\frac{x}{2}+\tan^{2}\frac{y}{2}+\tan^{2}\frac{z}{2}\geq 3\tan^{2}\frac{\frac{x}{2}+\frac{y}{2}+\frac{z}{2}}{3}=3\tan^{2}\frac{x+y+z}{6}$
$ =3\tan^{2}\frac{\pi}{6}=3\cdot\left(\frac{\sqrt3}{3}\right)^{2}=1$.

And the problem is solved.

[b]EDIT:[/b] Haha, not the quickest one but the most elementary. ;)

  Darij


At the bottom of a glas bath lies a mirror covered by 20cm of water.In the air,30 cm above the water there hangs a lamp.Find the distance from the mirror at which an observer looking into the water will see the image of the lamp in the mirror

oops check out your rotating mirrors therad

hey u probably answered a completely different question of mine as such.
well u can refer old USSR physics olympiad problems for such questions
One piece of advice:why don't u post these either at india or Intro physics because they are really much more active than this forum :)


inequalities
function
real analysis
real analysis unsolved

Glaeser inequality refers to a nonnegative $ C^2$ function $ u$ defined on an
interval $ (\minus{}R,R)$. The inequality is in: Assume $ \left| {u''} \right| \leqslant M$. Then
\[ \left| {u'\left( 0 \right)} \right| \leqslant \begin{cases}
  \sqrt {2u( 0 )M} & {\text{if }} M \geqslant \dfrac{{2u\left( 0 \right)}}
{{R^2 }} \hfill \\
  \frac{{u( 0 )}}{R} \plus{} \frac{R}{2}M &{\text{if }} M < \dfrac{{2u\left( 0 \right)}}{{R^2 }}.
\end{cases}
\]
Following is a generalization for $ C^2$, nonnegative function $ u$ defined on a ball $ B_R$: Assume $ \max \left| {\Delta u} \right| \equal{} M$. Then
\[ \left| {\nabla u\left( x \right)} \right| \leqslant \begin{cases}
 C\sqrt {u\left( 0 \right)M} &{\text{if }} R \geqslant \sqrt {\frac{{u\left( 0 \right)}}
{M}}  \geqslant 2\left| x \right| \hfill \\
  \left| {\nabla u\left( x \right)} \right| \leqslant C\left( {\frac{{u\left( 0 \right)}}
{R} \plus{} \frac{R}
{2}M} \right) &{\text{if }} 2\left| x \right| \leqslant R < \sqrt {\frac{{u\left( 0 \right)}}
{M}}
\end{cases}
\]
for some constant $ C$ depending only on $ n$. So, what is the best constant $ C$ for $ x\equal{}0$ in the first inequality of the generalization ?

I'm all for finding best constants in important inequalities, but I'm not sure I want to know this one. Notice that you have a restriction $ |x|\le\frac12 R$, which is necessary because the estimate degenerates near the boundary. Here $ \frac12$ could just as well be any other number between $ 0$ and $ 1$. Since $ C$ depends on our arbitrary choice of a number between $ 0$ and $ 1$, I just don't see it as an important constant.

The asymptotic behavior of $ C$ as a function of $ n$ could be interesting, though.


[color=purple] On the plane let a triangle $ ABC$ and an arbitrary point $ P$ in space prove that :
  $ \frac {PA}{r_b \plus{} r_c} \plus{} \frac {PB}{r_c \plus{} r_a} \plus{} \frac {PC}{r_a \plus{} r_b}\ge1$
 With $ r_a,r_b,r_c$ are three exradii opposite vertex A,B,C respectively.
My 333rd posts  :D
[/color]

[color=purple]
  Let $ w_a,w_b,w_c$ are three internal bisectors from vertex A,B,C respectively.Then we have:
  $ \frac{MA}{w_a}\plus{}\frac{MB}{w_b}\plus{}\frac{MC}{w_c}\ge2$
[/color]


Find all $p$ primes so that $128p+1$ composite and $(128p+1)|2^{64p}-1$ .

Thank you Rust. It was a good job. :roll:


Is the level of difficulty of questions #26-38 on the 2007 exam similar to the difficulty of quarterfinals of other years?

I think they are a bit easier than quarterfinals questions. I could be wrong.

Conceptually, they are a little easier than the 2008 quarterfinals; mathematically, much easier than any quarterfinal


Pascal\u0027s Triangle
number theory proposed
number theory

Find all positive integers $n$ such that there are $\infty$ many lines of Pascal's triangle that have entries coprime to $n$ only. In other words: such that there are $\infty$ many $k$ with the property that the numbers $\binom{k}{0},\binom{k}{1},\binom{k}{2},...,\binom{k}{k}$ are all coprime to $n$.

by Lucas' Theorem, $p \not | \binom{k}{0},\binom{k}{1},\binom{k}{2},...,\binom{k}{k}$ iff $k=ap^l-1$ (where $p \geq a \geq 1$) for some $l$. As $ap^l=bq^m$ cannot hold (suppose $q>p$ but then $q|a$ so $p\geq a \geq q>p$ contradiction) $n$ must be a power of prime - clearly all power of primes work


number theory
greatest common divisor
least common multiple

prove that

gcd(a,b)=gcd((a+b),lcm(a,b))

let d be g.c.d. of $ (a,b)$
so, a and b can be written as $ a \equal{} dk$ and $ b \equal{} dl$ for certain natural no.s k and l
so, $ (a \plus{} b) \equal{} d(k \plus{} l)$
and l.c.m has factors of g.c.d of (a,b) so, l.c.m.can be written as d(m)
$ g.c.d.((a \plus{} b),(lc.m.(a,b))) \equal{} d$ as both have d as a factor

thus we need to prove that $ (k \plus{} l)$ and lcm dont have common factors.since ,l and k are independent digits with no common factors, so, it cannot have common factors with lcm unless a,b are same numbers.

[b]again[/b].lcm divides k but to completely have a factor of $ (k \plus{} l)$  'l' should also have a factor of k which is not true.


Prove that

\[\sqrt{2\sqrt{3\sqrt{4...\sqrt{(n-1)\sqrt{n}}}}}< 3\]

for every natural number $n$.

[hide="Hint"]Use:
$ab\leq (\frac{a+b}{2})^{2}$
and destroy the radicals :D [/hide]

This is equivalent to proving that $\sqrt{m\sqrt{(m+1)\sqrt{(m+2)\hdots\sqrt{(n-1)\sqrt{n}}}}}< m+1$, which is easy to prove using
[hide="the following method:"]Induction.[/hide]


probability
number theory
prime factorization

Question 1:

There are black marbles and white marbles in a bag. The probability of grabbing a black marble randomly from the bag is $%Error. "dispmode" is a bad command.
\frac{3}{5}$, and the probability of grabbing two black marbles in a row without replacement is $%Error. "dispmode" is a bad command.
\frac{1}{3}$. How many white marbles are there?(I couldn't quite get the exact solution for this problem).

Question 2:

How many positive integers can be multiplied into $10!=10*9*8*7*6*5*4*3*2*1$?
(There is a quick way to do this).

[quote="bedakilla"][quote="easyas3.14159..."]I'm confused on the first problem. $\frac{3}{5}*\frac{2}{4}$ doesn't equal $\frac{1}{3}$.[hide="or"]am I missing something?[/hide][/quote]
[hide]
It wouldn't be $%Error. "dispmode" is a bad command.
\frac{3}{5}*\frac{2}{4}$ because there would be one less marble from the [u]total amount of marbles[/u] too.[/hide][/quote]
Which is why it is 4 instead of 5.
[hide="Number 1"]$\frac{3}{5}$ and $\frac{1}{3}$ are both after being simplified. We could brute force and keep trying numbers that satisfy $\frac{3}{5} = \frac{x}{y}$ to do $\frac{x}{y} * \frac{x-1}{y-1} = \frac{1}{3}$. $\frac{6}{10} * \frac{5}{9} = \frac{30}{90}$ which is your answer. Since there are $6$ black marbles, there must be $\boxed {4}$ white marbles.[/hide]
[hide="Number 2"]$10!$ is also $5*2*3*3*2*2*2*7*3*2*5*2*2*3*2$ or $2^8*3^4*5^2*7$ so do $9*5*3*2$ which is $\boxed {270}$.[/hide]


For an integer $n$, denote $A =\sqrt{n^{2}+24}$ and $B =\sqrt{n^{2}-9}$. Find all values of $n$ for which $A-B$ is an integer.

We get $n\ge 3$, n=3,4 don't work and $n=5$ solution $A-B=3$-.
If $n>5$ $m=A-B<3$. $(m+\sqrt{n^{2}-9})^{2}=n^{2}+24$. Therefore $m^{2}+2m\sqrt{n^{2}-9}=33$ or $n=\sqrt{(\frac{33-m^{2}}{2m})^{2}+9}$. It is easy to chek, that m=1,2 give not solutions. Therefore $n=5,m=3$ unique solution.

Rust I think you are wrong since n=-5 is a solution. m does =3 though.

we get:$A^2-B^2=33$,and when $A-B=1$,a contradiction,so$A-B=3$

if $ A-B $ is integer$=>$ that $A+B$ is also integer, and $A^2-B^2=33=> A-B=3, A-B=1$
1 case) when $A-B=3$
2 case) when $A-B=1$, dear reader do it by yourself, and get all possible integer $n$;

We have A^2-B^2=33 then (A-B)(A+B)=33 which implies A-B=1 or 3.
When A-B=1,we get n^2=265 which contradicts since n is an integer. Thus,A-B=3.
Upon simplification,we obtain n=5 or -5. :D  :D


algebra
polynomial
parameterization
calculus
calculus computations

Hi all

I'm in the process of creating an online cricket game and I'm trying to calculate a decent formula for working out, for a particular delivery, whether a bowler has dismissed a batsman. I realise that many readers won't be familiar with the game of cricket. For the purposes of my problem, assume cricket is similar to baseball. A cricket bowler is similar to a pitcher. A cricket batsman is similar to a baseball batter.

I've come up with an extremely basic model for determining the result of a cricket delivery (same as a baseball pitch).

For each delivery, I calculate a bowler attack rating. lets call it BA. I then calculate a batsman defense rating. Lets call it BD. Based on some formula, we come up with a % chance that the bowler dismisses the batsman. Lets call it DC.

I've plotted a few numbers in a spreadsheet that tell me DC based on various BA and BD values. My problem is I have no idea how to make this into an equation that I can re-use.

eg. lets assume

when BA >= 4 x BD then DC = 10% (ie the bowler has a 10% chance of dismissing the batsmen. In baseball parlance, the pitcher has a 10% chance of dismissing the batter with a pitch)
when BA = 3 x BD then DC = 6%
when BA = 2 x BD then DC = 4%
when BA = 1.5 x BD then DC = 3%
when BA = BD then DC = 2%
when BD = 1.5 x BA then DC = 1.7%
when BD = 2 x BA then DC = 1.4%
when BD = 3 x BA then DC = 1.2%
when BD >= 4 x BA then DC = 1%

Given the above details, is there a way of creating an equation that i can use to determine DC, based on different values of BA and BD?

Thanks

I'm a baseball fan, and there are some very serious differences between baseball and cricket, including that nearly every batter (apart from a few historical freaks like Babe Ruth, Ted Williams, or Barry Bonds) has a significantly greater than 50% chance of making an out in any given plate appearance. But I have one question: why are you stratifying your players into a small number of discrete performance bands rather than using actual individualized statistical data?

The reason I just showed that simple table with a limited number of values was to illustrate my problem fairly simply.

I wanted to illustrate that the relationship between BA, BD and DC is not linear. Linear stuff even I can usually handle :)

Basically i want to end up with an equation where i can plug in any BD and BA values and come up with a resulting DC. Its been more than 10 years since I left university so apologies if my description of the problem is not clear.

The BA and BD variables are themselves calculated by separate formulae, based on batting/bowling skill, experience, fatigue, fitness, weather conditions, pitch conditions, morale and a number of other factors.

I guess an alternative to my question is..

Is there a tool out there where you can plot points on a graph, the tool calculates a line of best fit, and then the tool works out correct equation?

If I can find such a tool then i can play around with various data values until I get the graph curve that I want.

Isn't Microsoft Excel supposed to do such things?

[quote="ashario"]not that i am aware of.

it can plot a graph, but i dont think it can then automagically calculate the equation behind the plotted points.[/quote]As far as I know (and I have done this), MS Excel can find the line or curve of best fit for a given set points depending on whether you want exponential fit or polynomial fit and so on - there are several options actually. By this I mean, it finds out the relevant equation and also provides a host of parameters associated with that particular curve. I am not sure though if that's what you want.


I saw something very similar. I forgot to save my screenshot, but there was a guest viewing their profile.

'Viewing profile' can be a guest viewing an aopser's profile.

Whoa! So you can actually see what someone has been doing? HOW???

(Now don't get any ideas that I'm any sort of stalker... )

Under someone's username, click the little "online" or "offline".

[quote="jonathanchou711"]So really. How does some guest view a pm?[/quote]It's not quite that simple. That statistic is based on the last 5 minutes of interaction. So, if withing the last 5 minutes, someone who was viewing his PMs logged out then he will show up as guest.


geometry
number theory unsolved
number theory

The lengths of the altitudes of a triangle are positive integers, and the length of the radius of the incircle is a prime number.
Find the lengths of the sides of the triangle.

Please continue in http://www.mathlinks.ro/viewtopic.php?t=16637 (regardless of lack of solution).
You can have a look at the resource section for a lot of already posted TSTproblems ( http://www.mathlinks.ro/resources.php ).


Prove that the polynomial $ x^{4n}+x^{3n}+x^{2n}+x^n+1 $ is divisible by $ x^4+x^3+x^2+x+1 $ if and only if $ n $ is not divisible by 5.

Hint:[hide]It suffices to show that each of the zeros of x^4+x^3+x^2+x+1 are zeroes of the other function.  What is the special trait of the zeroes of x^4+x^3+x^2+x+1?[/hide]

Superhint:[hide]Multiply x^4+x^3+x^2+x+1 by x-1.[/hide]


[quote="bibobeo"]Prove that

$ \tan\frac {3\pi}{11} \plus{} 4\sin\frac {2\pi}{11} \equal{} \sqrt {11}$

[hide="My way ..."]
Square it up >.< (Lots of expanding and so on)
I wonder if there is any easier way?[/hide][/quote]

there's gotta be something with the fact that $ \frac{3\pi}{11} \plus{} 4\cdot\frac{2\pi}{11}\equal{}\pi$

i'm pretty tired right now (explains the lack of capitalization) so i'm not gonna pursue this lead but i'll probably return to this in the morning


What is the largest integer whose cube is less than 10,000?

How can you cube so fast without a calculator?

Well you can do a rough estimate, since $20^3$ is only a bit less than $10,000.$ Plus, a bunch of people memorize this.

Just do the cube root of 10,000 and round down to the integer.

Observe that $(10 \cdot 2)^3=20^3=8000.$ Furthermore, $(11 \cdot 2)^3=22^3=1331 \cdot 8 = 10648.$ Thus, it is clear that $21^3$ will be less than $10,000$ just thinking about the differences between consecutive cubes.


[quote="phongissimus"]What is the summation of (-1)^n*(2n+1)^(1/2)?[/quote]
A series can only converge if the corresponding series converges to 0. However, yours does not converge to 0, as (2n+1)^(1/2) increases. Therefore, your series diverges.
=Uncool-


prove identity:
 $ \sum_{k\equal{}0}^{m} \frac{ {m\plus{}k \choose k} }{2^{k}}\equal{}2^m$

I can prove it by algebraic manipulation, but I know that there is also a combinatorial proof.

rewrite it as $ \sum_{k \equal{} 0}^m 2^{m \minus{} (k \minus{} 1)}\binom{m \plus{} k}{k} \equal{} 2^{2m \plus{} 1}$.

the right side is the number of sequences of length $ 2m \plus{} 1$ containing only $ a$'s and $ b$'s. any such sequence necessarily contains more $ a$'s than $ b$'s or more $ b$'s than $ a$'s. the number of the latter kind is obviously equal to the number of the former kind. so we'll count the former kind. in such a sequence, there must be $ m \plus{} k$ $ a$'s, for $ k\geq 1$. so we'll count according to where the $ m \plus{} 1$-st $ a$ occurs. it can occur in any of the positions $ m \plus{} j$, where $ j \equal{} 1, 2, ..., m \plus{} 1$. the number of sequences where the $ m \plus{} 1$-st $ a$ occurs in position $ m \plus{} j$ is $ \binom{m \plus{} j \minus{} 1}{m}\cdot 2^{2m \plus{} 1 \minus{} (m \plus{} j)} \equal{} \binom{m \plus{} j \minus{} 1}{j \minus{} 1}2^{m \minus{} (j \minus{} 1)}$. the number of sequences with more $ b$'s and the $ m \plus{} 1$-st $ b$ occuring in the $ m \plus{} j$-th position is the same. so the number of sequences with the $ m \plus{} 1$-st occurrence of the more abundant letter in the $ m \plus{} j$-th position is $ \binom{m \plus{} j \minus{} 1}{j \minus{} 1}2^{m \minus{} (j \minus{} 2)}$. summing over $ j \equal{} 1, ..., m \plus{} 1$, we get exactly the above identity.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Find the indefinite integration
$\int sinlntanx$

I think that this integral cannot be expressed in terms of elementary functions.

[hide="Edit:  Well, if there were an x after that sin it would go a little something like this..."] Let $u = \ln \tan x, dv = \sin x \, dx$.  Then $du = \frac{\sec^{2}x}{\tan x}, v =-\cos x$ and we have

$\int u \, dv =-\cos x \ln \tan x-\int \csc x \, dx =-\cos x \ln \tan x+\ln (\csc x+\cot x)+C$ [/hide]

The integral reads $\int \sin(\ln (\tan x))dx$ and not $\int \sin x \ln (\tan(x))dx$.

Oh.  In that case, integrals.wolfram.com confirms your assessment.


:rotfl:  :  :mad:    :trampoline:
 hai, iam satya
please solve this inequlaity for me


analytic geometry
calculus
integration
geometry
perimeter
vector
trigonometry

Since nobody answered this question correctly from my contest, I'll decide to post it here as a challenge problem (I hope it isn't too hard):

Call a triangle graphed in the Cartesian coordinate system dk2ian if it has integral side lengths and integral coordinates. Prove that there are no dk2ian triangles with perimeter 43

I think that was a typo.
I actually got 12.5

if you use calculus, i think it will help. let $(a,b),(c,d),(e,f)$ be the verticies of the triangle, then define the three vectors that create this triangle:
$v_1=\sqrt{(a-c)^2+(b-d)^2}=m^2$
$v_2=\sqrt{(c-e)^2+(d-f)^2}=n^2$
$v_3=\sqrt{(a-e)^2+(b-f)^2}=p^2$
and since all three of the sides are ingtegers, it means that $v_1,v_2,v_3$ must be perfect squares. so now all you need to do is to prove that there is no $(m,n,p)\in\mathbb{N}$ such that:
$m^2+n^2+p^2=43 $and since $43$ is a prime that would be easy to solve...

wow, that is so much easier than my one... i had to case work to prove it my way.... there was 43 cases! lol

[quote="amirhtlusa"]that is the power of calculus! :blush:
by the way remember this, the best way to cheat on a problem is to use calculus on it. :D[/quote]

There's an even simpler way... :)

Well to start with instead of picking three arbitrary points we can say WLOG one of the vertices of the triangle is (0,0) and the other two are say (a,b) (c,d)

Then after that all you have to do is prove that there are no integer solutions to some big ugly equation. :P

Well that method isn't exactly calculus....  But where did the squares come from?

I noticed that too in his solution it doesn't make much sense to me.

dk2 what is the solution you have in mind?

[hide]If the dk2ian triangle can't exist with a vertex at (0,0) it can't exist at all.
So the vertecies are:
(0,0)
(a,b)
(c,d)
So:
P=sqrt(a2+b2)+sqrt(c2+d2)+sqrt(a2+b2+c2+d2-2ac-2bd)
To add up to an integer, the number in the sqrt must be a perfect square. [/hide]

[quote="dk2"][quote="mna851"]Yes, but the whole point of that problem is to set P=43 and show that there are no integer solutions to the resulting equation.[/quote]

Hint: Consider the area of any dk2ian triangle...[/quote]

[hide="Well..."]

$A = \frac{c \cdot h}{2}$.
$A = \sqrt{(\frac{43}{2})(\frac{43}{2}-a)(\frac{43}{2}-b)(\frac{43}{2}-c)}$.
$A = \frac{ab \sin C}{2}$.

What now? It's probably supposed to be heron's formula, but still, what do we do now?
[/hide]

[hide="hint"]43 is prime....[/hide]
edit--wait, does the area have to be an integer?
edit again--ok I reread the original problem statement and my hint is correct.

[quote="probability1.01"][hide="hint"]43 is prime....[/hide]
edit--wait, does the area have to be an integer?[/quote]

that's what I was wondering, but I don't think so...

[quote="dk2"][quote="mna851"]Ok I'll post my general solution if I'm too vague in certain areas ask me and I'll post more about it later.

[hide]So like everyone probably saw you can assign arbitrary coordinates to the vertices of the dk2ian triangle.  So let those vertices be (a,b) (c,d) (e,f).  So now we can say $\ s_1^2=(a-c)^2+(b-d)^2$ $\ s_2^2=(a-e)^2+(b-f)^2$ $\ s_3^2=(c-e)^2+(d-f)^2$.  Now we can identify pythagorean triples.  For the general pythagorean triple $\ a^2+b^2=c^2$ $\ c$ can be represented in general form as $\ c=r^2+s^2$ where $\ r$ and $\ s$ are integers.  So now we can assert the original statement and attempt to prove a contradiction.  $\ s_1 + s_2 + s_3=43$ then $\ u^2 + v^2 + w^2 + x^2 + y^2 + z^2=43$.  where u,v,w,x,y,z are all integers.  Now all you have to do is show that this equation has no solutions in integers and you've identified a contradiction.[/hide][/quote]

Think about probability's hint. If I say that the vertices of a triangle are (2,1), (3,5), and (4,10), how would you find the area?[/quote]
heron's formula,  :)

actually if you are given the vertices of a triangle in the plane, you can find two vectors, that are the two sides of the triangle, then half of their cross product would be the area of the triagle.

hmm... that hint didn't seem to work very well.

Try this:
If i told you that the vertices of a pentagon were A(1,2), B(0,7), C(5,8), D(10,6) E(8,1), then how would you find the area.

EDIT:
mna, your solution seems correct, but how would you go about proving the last statement of your problem. Thats a question in itself :P .

[quote="dangvh"]didnt xantos C. Guin get it right?? he got a 25 on group C[/quote]

What contest are you guys talking about?

The last step of my proof is very easy to prove just by common sense because the terms are all integer squares.  Formally, I guess you might be able to prove it modulo 4 or 8.  I didn't actually work it out so don't take my word for it.  Maybe I'll get around to it later.

[quote="dangvh"][quote="dk2"]hmm... that hint didn't seem to work very well.

Try this:
If i told you that the vertices of a pentagon were A(1,2), B(0,7), C(5,8), D(10,6) E(8,1), then how would you find the area.

[/quote]

shoelace method??[/quote]

Yes, so what would that tell you about a triangle with integral coordinates  :lol:


Even though I am probably not capable of understanding most of them I was wondering why the amc website still hasen't posted the usamo solutions?

Cause it takes them forever to do anything.  I think it was after USAMO, or shortly before, that I got AIME solutions.

THe usamo solutions are still not posted. Will they not post them this year?

They were posted at some point after the contest.  They may have taken them down already.

Looks like they were just posted...

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04USAMO_solutions.pdf


15 students join a summer course. Every day, 3 students are on duty after school to clean the classroom. After the course, it was found that every pair of students have been on duty together exactly once. How many days does the course last for?

there are $\binom{15}{2}$ pairs of people, and $\binom{3}{2}$ pairs a day, so the answer is $\frac{15*7}{3}=35$ days


I'm writing an essay for a History of Math class

I can choose any math related topic or person to write about as long as the subject matter is from the 18th century or afterwards. I can't choose Newton or Descartes...

Any ideas?
I really don't know that much about modern math ideas....

Anything is appreciated...

does it have to be someone really famous, who wrote theorems and formulas???? If not, then why dont you pick one of the modern mathematicians. E.G. Paul Erdos.

If you're going to do Paul Erdos, read the book 'The Man Who Loved Only Numbers" by Paul Hoffman.

I will read that someday, but my essay is due on Thursday (yes, I'm a procrastinator) so I don't have time to read it before the essay is due.

All other suggestions are welcome...


geometry
rectangle
combinatorics unsolved
combinatorics

Let $A_n$ be the number of ways to tile a $4 \times n$ rectangle using $2 \times 1$ tiles. Prove that $A_n$ is divisible by 2 if and only if $A_n$ is divisible by 3.

Who have any idea for this nice problem???

I'll give you a hint: recursion, and mod 6...

I know we have to do that,but i don't know how to do.Could you tell me more clearly?
  And a problem which have a same kind:
Insteed $4$ by $3$,determine its recursion!

Have a look at this: (sorry for the attachment, I didn't see any other way to present the pictures)

Somebody, please solve it.

Dear moderatons, why usually old problems are forgoten?. Isn't there way to make people solve them. ;)


What numeral is in the 100th decimal place in the decimal
representation of $ \frac{6}{7}$?


OK, so I don't get half the book, but I still want to figure out the problems, especially the "challenge" ones the author gave (you can't really do them unless you've read the book. (I'll waste a weekend typing an explanation of TNT if no one has read the book. (Yay! nested parentheses! (Sigh, haven't read it in such a long time, hardly remember what it's about...))))

Express the following statements in TNT:

b is a power of 2.
b is a power of 10.

[quote="leoxnlin"](you can't really do them unless you've read the book. [/quote]
[url=http://en.wikipedia.org/wiki/Typographical_Number_Theory]I wonder what this page is doing here, then...[/url]

By the way, it's my opinion that the problems aren't really the real meat of GEB.  But don't worry - I've read it repeatedly over a span of years and every time I keep finding new levels :)

Yes. The book is a masterpiece of structure, like a Brahms symphony. You really have to come at it with some musical background to appreciate that part.

Yeah, t0rajir0u, but, but.... the book has a better font  :lol: .

I know that that isn't the real meaning of the book (the topic title is misleading, sorry), but yeah, I didn't get the proof of the Incompleteness theorem... cause why does there have to be a proof-pair in order for the statement to be true?... I mean, it could be based on two previous statements in the proof,.... I mean,... AAH! I'm probably talking nonsense, I hardly remember anything.... BUT I do have a tiny musical background (i.e., occasionally deciding to try some Bach and failing)... I tried playing the endlessly rising scale thingy on the piano (with only two hands), but unfortunately I had some difficulty with that (couldn't get one hand really loud and the other hand really soft...  :( ) oh well.

The whole book makes me feel very sad about mathematics. It almost made me stop doing math (a year ago). I mean, after the Incompleteness theorem and those other thingies (Turch-Churing Thesis?), there would be no point to (pure) math... just proving theorem after theorem after theorem for no reason and knowing that you'll never get to any grand, unified, something-er-ma-bober. And if you get stuck on something, you'd never know if there was even a solution or not...

Sorry you had to waste your time reading my post.

[quote="leoxnlin"]I mean, after the Incompleteness theorem and those other thingies (Turch-Churing Thesis?), there would be no point to (pure) math... just proving theorem after theorem after theorem for no reason[/quote]

Well, I'd like to think that mathematics didn't stop after the proof of the Incompleteness Theorem.  There are certainly good reasons to continue proving theorems.  The Incompleteness Theorem just tells us that certain statements exist that can't be proven in a sufficiently powerful system; but why would be interested in proving [b]every[/b] true statement?  Instead, we pick and choose the most interesting and/or important ones, and those are (usually) quite provable.

I mean, the Laws of Thermodynamics don't stop you from decreasing entropy locally, do they?  :)

If you're reading for content, the book is quite dated by now. If you're reading for entertainment and art's sake, it's as good as it ever was.

The Incompleteness Theorem is not a practical issue for nearly all mathematicians. Statements that are verifiably unprovable are generally very artificial, except for a few set theory axioms- and the Axiom of Choice is accepted in mainstream mathematics despite being unprovable in a slightly smaller model of set theory. Incompleteness just says that no one set of axioms will answer all questions: a system cannot be consistent, interesting, and completely understandable all at once.

And then there's the Church-Turing thesis (was there more than 1?): there'll never be a surefire way to see if a statement is true or false... hmm... that's sort of bad, but then again, there'd be nothing more to math after that was discovered. So again leoxnlin's theory of stupidity (that stupidity is the key to happiness) prevails: the ideal case is if mathematicians were stupid (well, you know what I mean) so they [i]thought[/i] there was an ultimate algorithm to test theorems, but there really wasn't. Then everyone would be happy (sort of).

GEB is possibly my favorite book (actually not really, go [i]White Noise[/i]!, but close). I think that you may be missing the point slightly. Content-wise I think that the Incompleteness stuff is very small. Think about all the recursion things in there. In my opinion, it is really about how consciousness can exist at one level, when at lower levels everything is more machine-like. Noncontent-wise (structurally) the book is utter genius. He essentially has encoded the theme of the book into the structure of the book. I don't think a 100 page analysis of the structure and meta-devices used would do it justice.

If you really are interested in the Godel aspect of the book, there are lots of really interesting reads about the philosophical implications of the Incompleteness theorem. These may help alleviate some worries.


Lucas numbers $  L_0, L_1,..., L_n, ... $ are defined by $  L_0=2 \; ,\; L_1=1 $
and $ \displaystyle  L_{n+1}=L_n + L_{n-1} \; \; \; ,\; \; \; n\ge 1 \; . $  Let $\displaystyle S:=\sum\limits_{k=0}^{1001}  (-1)^{k+1}{2003 \choose k+1002}L_{(2k+1)2003}\; . $
 Prove or disprove  that there exists a positive integer $ K $  such that $\displaystyle S=K^{2003} $ .


trigonometry
ratio
geometry
angle bisector
trig identities
Law of Cosines

You may also recognize this problem from chapter three of AoPS vol. 2.  This is the form of the question given by kalva.

ABC is a triangle with:
 $BC=a, CA=b$. Show that the length of the angle bisector of C is
$(2ab cos C/2)/(a + b)$.

If no one else wants to do it...

[hide]Using angle bisector theorem, we can express the segments of side c as $bc/a+b$ and $ac/a+b$ on the sides of $b$ and $a$, respectively.  Let $x$ be the length of the bisector of angle C.

Now, we create three equations using law of cosines:

1)   $\frac{bc}{(a+b)^2} = b^2+x^2-2bxcos(C/2)$

2)   $\frac{ac}{(a+b)^2} = a^2+x^2-2axcos(C/2)$

3)   $c^2 = a^2+b^2-2abcosC$

After subtracting equation 2 from equation 1, we are left with:

      $\frac{c^2(b^2-a^2)}{(a+b)^2} = b^2-a^2-2bxcos(C/2)+2axcos(C/2)$

A little algebra leaves us with:

      $c^2 = (a+b)(a+b-2xcos(C/2))$

Equating this equation with equation 3, we show that:

      $x = \frac{ab(cosC+1)}{(a+b)(cos(C/2))}$

We now equate this with the desired value, and after simplifying, we show that the two values are equal.[/hide]

[hide="Triggy sol"]
Call the intersection of the angle bisector and AB $Q$, and $CQ=x$. Then by angle bisector theorem we have $(\frac{AQ}{BQ})^2=(\frac{b}{a})^2$
Then by the law of cosines we have:

$\frac{x^2 +b^2-2xb\cos(C/2)}{x^2+a^2-2ax\cos(C/2)}=\frac{b^2}{a^2}$

$a^2 x^2+a^2 b^2 -2a^2 x b \cos(C/2) = b^2 x^2 + a^2 b^2 -2 a x b^2 \cos(C/2)$

Getting rid of the  $a^2 b^2$ and manipulating we get:
$2a x b \cos(C/2) (b-a) = x^2 (b^2 - a^2) \implies x = \frac{2ab \cos(C/2)(b-a)}{b^2-a^2} = \boxed{ \frac{2ab \cos(C/2)}{a+b}}$[/hide]

jhcreinhardt, i think in your solution you could have avoided a little bit of tedious algebra if you used the ratio of the parts of $C$ instead of assigning them lengths... (oh and put parentheses on your lengths in the very beginning to avoid confusion) :)  but otherwise ours are about the same, yes?  :)

[quote="jhcreinhardt"]Nice, but I beat you by 1 minute.[/quote]

haha ouch :P

well i took like 5 minutes manually doing latex... so hah

[quote="jhcreinhardt"]Yes, quite similar, but yours did avoid some of the frustration that usually comes with geometry mixed with trig.  I've also seen a pretty elegant solution (solution manual of AoPS) involving the $(1/2)absinC$ area formula that simplifies out nicely.[/quote]

I see how that could work... yea, im sure that would be more elegant (like you said) than what either of us did  :D .


The 5 a.m. temperatures for seven consecutive days were $ \minus{}7^{\circ}$, $ \minus{}4^{\circ}$, $ \minus{}4^{\circ}$, $ \minus{}5^{\circ}$, $ 1^{\circ}$, $ 3^{\circ}$ and $ 2^{\circ}$ Celsius. What is the mean 5 a.m. temperature for the week in degrees Celsius?

Add them all and then divide by 7 and you get [b]-2 degrees Celsius[/b].

If you set $0^{\circ}$ as the mean temperature. Then the degrees in order are $-7 - 4 - 4 - 5 + 1 + 3 + 2 = -\frac{14}{7} = \boxed{-2}$

This method may not seem so great, but that is because the base temperature was $0^{\circ}$


I have one 5 dollar bill and two 10 dollar bills in bag A, and I have three 5 dollar bills and one 10 dollar bill in bag B. I randomly take one bill from bag A and place it in bag B. I then randomly select one bill from bag B. If this bill is a $10 bill, what is the probability that the bill taken from bag A was a 5 dollar bill?

(Please post answers in spoiler)

[hide]

let X be a "5 dollar bill"

let Y be a "10 dollar bill"

Prob. you choose X from A:  1/3

Prob you choose Y from A: 2/3

Prob you choose X from B given you chose X from A: 4/5

Prob you choose Y from B given you chose X from A: 1/5

Prob you choose X from B given you chose Y from A: 3/5

Prob you choose Y from B given you chose Y from A: 2/5

So (1/3)(1/5) of the time, you choose X from A and Y from B.

and (2/3)(2/5) of the time, you choose Y from A and Y from B.

Thus, the prob is (1/3)(1/5) / [(1/3)(1/5) + (2/3)(2/5)] = 1/5.

[/hide]


calculus
integration
trigonometry
limit
calculus computations

Evaluate the integral $ \int_{0}^{\infty} \frac {\sin ^{2} t}{t^{2}} \mathrm{d}t$ using laplace transforms

[hide]$ \int_0^{\infty} \frac{\sin^2t}{t^2}\,dt \equal{} \left[\minus{} \frac{\sin^2t}{t}\right]_0^{\infty} \plus{} \int_0^{\infty} \frac{\sin(2t)}{t}\,dt \equal{} 0 \plus{} \int_0^{\infty} \frac{\sin u}{u}\,du$

$ \equal{} \lim_{s \to 0^\plus{}}\int_0^{\infty} \frac{\sin u}{u} e^{\minus{}su}\,du \equal{} \lim_{s \to 0^\plus{}} \mathcal{L} \left(\frac{\sin u}{u}\right) \equal{} \lim_{s \to 0^\plus{}} \arctan \left(\frac1s\right) \equal{} \frac{\pi}{2}$.[/hide]

Like Carcul had shown, this integral boils down to evaulatng $ \displaystyle\int_0^\infty\frac{\sin t}{t}dt$.
You can solve this integral also by the following property of Laplace Transforms.
Property: $ \displaystyle\int_0^{\infty}e^{\minus{}st}\frac{f(t)}{t}dt\equal{}\int_s^{\infty}F(u)du$, where $ \displaystyle\mathcal{L}\{f(t)\}\equal{}F(s)$

Here, $ \displaystyle f(t)\equal{}\sin t$ which has the Laplace transform of $ \displaystyle F(s)\equal{}\frac{1}{s^2\plus{}1}$.
Then using the property above, and taking the limit as $ s\to 0^\plus{}$, we have $ \displaystyle\int_0^{\infty}\frac{\sin t}{t} dt \equal{}\int_0^{\infty}\frac{1}{u^2\plus{}1}du\equal{}\frac{\pi}{2}$.

But integrating by parts to get to $ \int_0^{\infty}\frac {\sin t}{t}\,dt$ makes for an integral that doesn't converge absolutely and for which justifying interchanges of order of integration may become delicate or doubtful. The original integral here does converge absolutely - why not stay with it?

Suggestion: keep it the way it is. Replace $ \frac {1}{t^2}$ by $ \int_0^{\infty}xe^{ \minus{} tx}\,dx.$ Use the identity $ \sin^2x \equal{} \frac12(1 \minus{} \cos 2x).$

Hi :) !!!

without Laplace transforms we can calcul that directly:

$ \int_0^{\plus{}\infty} \frac{sin^2(t)}{t}dt$ converge obviously than by integration by part we can find that:

$ \int_0^{\plus{}\infty} \frac{sin^2(t)}{t}dt \equal{} \int_0^{\plus{}\infty} \frac{sin(t)}{t}dt \equal{} \frac{\pi}{2}$ integral of Dirichlet.

for prove we can use the complexe integral $ \int_C \frac{e^z}{z}dz$ ....

and thinks


Determine for which positive integers $k$, the set \[X=\{1990, 1990+1, 1990+2, \cdots, 1990+k \}\] can be partitioned into two disjoint subsets $A$ and $B$ such that the sum of the elements of $A$ is equal to the sum of the elements of $B$.

Let $ S(W)\equal{}$ The sum of the elements of $ W$ and $ \# W\equal{}$ the number of elements of $ W$
$ S(X)\equal{}1990 (k\plus{}1) \plus{}k(k\plus{}1)/2$
It is necessary that $ 2|S(X)$ then $ 2| k(k\plus{}1)/2$ then $ 4| k(k\plus{}1)$ then $ k \equiv 0,\minus{}1 (4)$
$ 1) k\equal{}4m\plus{}1$
$ X \equal{} \{1990,1990\plus{}4m\plus{}1,$
$ 1990\plus{}1   , 1990\plus{}4m,$
$ 1990\plus{}2   , 1990\plus{}4m\minus{}1,$
$ \cdots,$
$ 1990\plus{}2m, 1990\plus{}2m\plus{}1\}$ and pick the first column for the elements for $ A$ and we are done.

$ 2) k\equal{}4m$
$ \#X\equal{}4m\plus{}1\equal{}\#A\plus{}\#B$
WLOG $ \#B\equal{}b \ge \#A\equal{}a$ then $ a\equal{}b\minus{}e$ with $ e \in \{0,1,2,\cdots \,b\minus{}1\}$
$ 4m\plus{}1\equal{}2b\minus{}e$ then $ e\equal{}2f\minus{}1$ with $ f \in \{1,2,\cdots \, 2m\}$
$ b\equal{}2m\plus{}f$, $ a\equal{}2m\plus{}1\minus{}f$

Let $ C\equal{}A\minus{}1990$ and $ D\equal{}B\minus{}1990$ then
$ Y\equal{}\{0,1,2,\cdots ,4m\}$ is splitted in two sets $ C$ and $ D$ such as
$ 1990 (2f\minus{}1) \equal{} S(D)\minus{}S(C)$
$ 4m(4m\plus{}1)/2\equal{} S(D)\plus{}S(C)$

$ 2m(4m\plus{}1)\minus{}1990 (2f\minus{}1) \equal{}2S(C)$
$ S(C) \ge  (2m\plus{}1\minus{}f)(2m\plus{}2\minus{}f)/2$

$ 2m(4m\plus{}1) \ge  (2m\plus{}1\minus{}f)(2m\plus{}2\minus{}f) \plus{}1990 (2f\minus{}1)$
$ 8m^2\plus{}2m \plus{}f (4m\plus{}3) \ge (2m\plus{}1)(2m\plus{}2)\plus{}f^2\plus{}1990 (2f\minus{}1)$
$ 8m^2\plus{}2m \plus{}f (4m\plus{}3) \ge 4m^2\plus{}6m\plus{}2\plus{}f^2\plus{}1990 (2f\minus{}1)$
$ 4m^2\plus{}f (4m\plus{}3) \ge 4m\plus{}2\plus{}f^2\plus{}1990 (2f\minus{}1)$
$ 4m^2\minus{}4m\minus{}2\plus{} f (4m\plus{}3) \ge f^2\plus{}1990 (2f\minus{}1)$

Notice that $ f(4m\plus{}3\minus{}f) \ge 0$ and
if $ 0 \le m \le 22$ then $ 4m^2\minus{}4m\minus{}2 \le 4 \times 22^2\minus{}4 \times 22 \minus{}2\equal{}1846 \le 1990$ contradiction then $ m \ge 23$
Let us analyze the case for $ k\equal{}23 \times 4\equal{}92$ for bigger $ k$ you can append any four numbers like: $ \{1990\plus{}92\plus{}1,1990\plus{}92\plus{}4\}$ and $ \{1990\plus{}92\plus{}2,1990\plus{}92\plus{}3\}$

$ S(X)\equal{}1990 \times 93\plus{}92 \times 93 /2\equal{}1990 \times 92\plus{}1990\plus{}4278\equal{}1990 \times 92\plus{}6268$

$ A\equal{}\{1990,1990\plus{}1,1990\plus{}2,\cdots,1990\plus{}44,1990\plus{}62,1990\plus{}92 \}$
$ \#A\equal{}1\plus{}44\plus{}1\plus{}1\equal{}47$
$ S(A)\equal{}1990 \times 47\plus{}44 \times 45/2\plus{}62\plus{}92\equal{}1990 \times 46\plus{}3134$
and $ 2 \times S(A)\equal{}S(X)$


Any suggestion how i can prove  a^2/(a^2+1)*b^2/(b^2+1)*c^2/(c^2+1) <= 1/8 ?

Thanks

Is the problem ${a^{2}\over a^{2}+1}\cdot{b^{2}\over b^{2}+1}\cdot{c^{2}\over c^{2}+1}\leqslant{1\over 8}$? Prove it by claiming it doesn't hold  :lol:

Put any $a^{2},b^{2},c^{2}$ which are all greater than one and you'll get a contradiction.

With the constraint that $|a|,|b|,|c|\leqslant 1$, it's easy:

$a^{2}\leqslant 1\iff{1\over a^{2}}\geqslant 1\iff 1+{1\over a^{2}}\geqslant 2\iff \frac{1}{1+{1\over a^{2}}}\leqslant{1\over 2}\iff{a^{2}\over a^{2}+1}\leqslant{1\over 2}$

Do that for $b$ and $c$ and multiply the obtained inequalities out.

Or there is some other constrainment you failed to include in the statement?

ohh. forgot about additional information:

a*b*c=1
a, b, c > 0

sorry but i dont understand what do you mean by "Now use AM-GM to get...."

The inequality between Arithmetic Mean and Geometric Mean. It's quite an indispensable tool for solving inequalities, so you should consult a textbook.

It states that for non-negative $a_{k}$, the following holds

$\frac{a_{1}+a_{2}+\dots+a_{n}}{n}\geqslant\sqrt[n]{a_{1}a_{2}\dots a_{n}}$

The equality holds iff $a_{1}=a_{2}=\dots=a_{n}$

Thank you very much Farenhajt.

after reading wikipedia's and yours explanation i came to conclusion that this problem not solved yet.
we have no information that a,b,c = 1    they can be 1/2, 1/2, 4 for example.

* sorry for my english

Yes, they can, but what's the problem? The only thing we used throughout is the given fact $abc=1$.

Excuse me, but what's your EXACT level in maths? Because I'm generally not in the mood to teach people their two-times-two's. And so far you do appear underqualified for this topic.

His solution is perfectly fine.  He used the fact that $abc=1$, not $a,b,c=1$. :wink:
Masoud Zargar

Thank you both. now i understand.

i thought that a1 have to be equal to a2 and a3 and so on.....


i need a review in logic because i 4got a lot of it. i learned it last year, but i need to know it for a test. Plz help me. (i mean logic as in inverses, converses, etc...)

An if-then statement is in the form, if $ p$, then $ q$, e.g. "If it is raining, then I will get a ride home from school." In that particular case, "It is raining" is statement $ p$ and "I will get a ride home from school" is statement $ q$.

The converse of an if-then statement is when you switch $ p$ and $ q$, so that it reads if $ q$, then $ p$. The converse of our example is "If I got a ride home from school, then it is raining." Converses aren't always true, though. For instance, I might get a ride home because it's snowing, or if I sprained my ankle.

Negating a statement is when take the opposite of it. If we negate statement $ p$, we would get "It is not raining." Something like "the fish is not red" would become "the fish is red". It is represented by $ \sim  p$. The inverse of an if-then statement is if $ \sim p$, then $ \sim q$. Our example would be "If it is not raining, then I will not get a ride home from school." This particularly example is also false, for reasons stated above.

The contrapositive of a statement is like a combination of converses and inverses, namely if $ \sim q$, then $ \sim p$. The contrapositive of our statement would be "If I do not get a ride home from school, then it is not raining." This statement would be true, in this case.

If an original statement is true, then the contrapositive will always be true. (Look at that, an if-then statement!  :P ) Similarly, if a converse is true, then so will the inverse.

thanks :) , but how do you do problem like PV(or)Q, P/\(and)Q, and those kind of problems?

You should check out [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting and Probability[/url]; there's a full chapter on Sets and Logic.  The rest of the book is also very useful :)

Also, it's not very easy to see what you're trying to ask in your newer problems; could you take the time to $ \text{\LaTeX}$ify and actually use correct grammar?

The 4 logic sections are:
If-Then
Converse
Inverse
Contrapositive

There's a lot more to logic than simple biconditional statements :wink:

Intermediate Counting and Probability will help you on this.  Also, [url=http://faculty.matcmadison.edu/alehnen/weblogic/logcont.htm]this site[/url] seems to be useful.

Is this even suitable for classroom math, given that AIME15 just said to use Intermediate Counting and Probability?


Solve the equation:
(2x+1)+(2x+4)+...+(2x+31)=110 (1)
where x is real

Answer

The $ n$th term in the sum is $ 2x+3n-2$, hence there are a total of $ 11$ terms (by setting $ 3n-2= 110$)

The formula for the sum of $ n$ terms of an arithmetic sequence is:

$ S_{n}= \frac{1}{2}n(a+l)$

$ \Rightarrow 110 = \frac{1}{2}\cdot 11(2x+1+2x+31)$

$ \Rightarrow 10 = \frac{1}{2}(4x+32)$

$ \Rightarrow 4x+32 = 20$

$ \Rightarrow x =-3$

Let the function be represented as[hide="..."]$ f(x)=(2x+b)-(2x+b)+k$
. . .$ \Rightarrow=(2x+b_{1})-(2x+b_{2})$
. . .$ \Rightarrow=(2x+1)-(2x+4)$
$ \to k=3$

$ k_{n}+b = 31$ ; $ b = 1...\Rightarrow 3n+1=31$ making the number of terms, $ n=10$

Obeying the logic of a constant difference in terms over a term difference....
$ k+2k+3k+...+10k = \frac{k(10\times11)}{2}= 55k$
There are 11 terms of $ 2x+b$, for the only thing changing is $ k$.

-->so, $ 11(2x+b)+55k = 110$
$ b=1$; $ k=3 \Rightarrow 22x+176=110$

$ \Rightarrow x=-3$
[/hide]

[hide]The left side is an arithmetic sequence, so
$ \frac{(4x+32)(11)}{2}=110$
$ 2x+16=10$
$ x=-3$[/hide]

mountain.dew...that is exactly the same way as I did it...

Sorry, I didn't see..

Anyways I like to try the questions too. :maybe:

[quote="wonki_noh"]$ (2x+1)+(2x+4)+...+(2x+31)=22x+3*\frac{10*11}{2}-11=110$
$ \Rightarrow 22x=-44$
$ \Rightarrow x=-2$[/quote]
I can't understend what did you do but your solution is wrong
The correct solution is
$ x=-3$.
We can find the sum in the first member of the equation using the formula for the sum of n terms of an arithmetic
sequence (as sludgethrower did).


There is a simple theorem:

If $ A, B \in M_n(\mathbb{R})$, and $ A^T\equal{}A, B^T\equal{}B$

then $ A$ is similar to $ B$ if and only if $ A$ is orthogonal similar to $ B$

Is this theorem having a Geometric View?

Let $ A,B$ be symmetric matrices.
$ A,B$ have orthonormal bases of eigenvectors $ (e_i),(f_i)$. There exists $ O\in{O_n}$ s.t. $ O(e_i)\equal{}f_ i$.
 If $ spectrum(A)\equal{}(\lambda_i),spectrum(B)\equal{}(\mu_i)$ then for all $ i$, $ OAO^{\minus{}1}f_i\equal{}\lambda_i{f_i},Bf_i\equal{}\mu_i{f_i}$.


Hola, pense que podiamos compartir links de los sitios de las olimpiadas en español, y de nuestras olimpiadas...

Aqui estan los que conozco:

Olimpiadas Colombianas De matematicas: (incluye una foto mia :D )
[url=http://olimpia.uanarino.edu.co/ocm/om98a.htm]Aqui[/url]

Mat latinos, [url=http://groups.yahoo.com/group/mat-latinos/interrupt?st=2&m=1&done=%2Fgroup%2Fmat-latinos%2Fmessage%2F600]Aqui[/url]

Olimpiadas Matematicas de Argentina [url=http://www.oma.org.ar/]Aqui![/url]

No se que mas, el caso es que cada cual puede dar links....

Hola,

Me ha parecido que la idea de Pascual es bastante buena, aun cuando parece que paso desapercibida para la mayoria. Decidi ponerla como un sticky por unos dias al menos para ver si hay mas propuestas. Lastimosamente la pagina de la Olimpiada de Costa Rica no contiene mas que informacion administrativa, y la pagina que existia antes, manejada por otro exolimpico parece que ha desaparecido, pero que tal el resto de la gente por aqui?

Buneo pondré la página oficial de la olimpiada nacional de méxico:

http://erdos.fciencias.unam.mx/omm/.

y ésta es la págin de la olimpiada en mi estado que esa yucatán:
http://www.uady.mx/sitios/matemati/omm/material.html.

espero que les sirva de algo. suerte a todos.

Aqui pongo lo que ahora es la pagina oficial de olimpiadas de matematicas venezolanas. En el link que dice entrenamiento pueden conseguir algunas guias interesantes hechas por profesores venezolanos.

http://www.acm.org.ve/

hace poco descubri este site q tiene libros gratis! http://www.rinconmatematico.com/libros.htm

Si leen portugues, aqui hay una buena coleccion de articulos relacionados, todos en portugues.

[url=http://www.obm.org.br/eureka/artigos/]Articulos Eureka[/url]

Si ustedes creen que no leen portugues, denle un chance, vieran que no es dificil, yo nunca he aprendido, sin embargo, como hispano hablante, leerlo es bastante natural.

com fórum e listas de treinamento/seleção da equipe brasileira para a IMO.

La página de la olimpíada de matemática uruguaya es
[url]http://www.reu.edu.uy/jpv/proyectos/cpm[/url]

Hola la pagina oficial de peru es:
http://www.pucp.edu.pe/~matemat/olimpiadas/

ah tambien hay otra en
http://usuarios.lycos.es/archimath/experiences.html

tambien hay una nueva pagina que ha salido la mia  :lol:  jejeje y es
http://www.cjeo-peru.com
Es reciente y aun se le esta haciendo mejoras pero agradeceria saber sus opinionessobre la pagina
gracias


Robert likes chocolate milk, so he decides to visit the milk bottling plant every day for a week to get the free samples. Unfortunately for him, the bottling plant sometimes bottles regular milk instead of chocolate milk, so each day the plant has a 2/3 chance of bottling chocolate milk. What is the probability that the bottling plant bottles chocolate milk exactly 4 of the 5 days he visits?