invariant
symmetry
quadratics
algebra
polynomial
Vieta
number theory solved

Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]

Do not mention that a very similar problem: $\frac{a^2+b^2}{ab-1}$ appeared not long ago in a USA TST. But it's very interesting what happens with this problem after you start solving it. It has a step curiously difficult. A very nice problem, like all the other problems given in this very difficult TST.

I think I have an idea about what a solution might look like. We consider ordered pairs $(a,b)$ s.t. $a\leq b$. We can go from the pair $(a,b)$ to a "smaller pair" $(a',b')=(\frac{a+b+a^3}{ab-1},a)$  until we eventually reach a "dead-end", i.e. a pair which can no longer become smaller by the above transformation.

A simple calculation shows that $f(a,b)$ is invariant wrt the transformation mentioned, so the value we're interested in remains the same. I also think that the "dead-end", as I've called it, must be one of the pairs $(1,2)$, $(1,4)$ and $(2,2)$. By analyzing the 3 cases we obtain $f(a,b)\in \{7,4\}$ (it's 7 in the first two cases and 4 in the third case).

I don't even know if this is th correct answer, let alone the correct idea for a solution, but it looks like this at 2 AM :).

The problem was proposed at the upmentioned romanian test, but, no matter the fact that the idea of the problem is well-known, at least for people which studied a little bit the questions from past IMOs (especially IMO '88 :D), only two contestants have managed to master the problem

you are right, Grobber ... the good idea, and also the good result. :D some more computations for clarification are required, but that's the main idea.

Well, I suppose I don't have to do the actual computations (I pity the contestants for that, as I often lose my interest in a problem after getting the main idea :D)..

Do you know who proposed the problem?

n As far as I know, it was proposed by Mircea Becheanu.

case1:$a=b$

$f(a,b)=\frac{3a^2}{a^2-1}=3+\frac{3}{a^2-1}\in\mathbb{N}$,we have $a^2-1=3$.Thus $a=2$.i.e. $(a.b)=(2,2)$,yielding $f(a,b)=4$.

case2:$a\neq b$
By symmetry of $a$ and $b$,WLOG $a>b\geqq1$,so we can set $a\geqq b+1$,thus we have

$f(a,b)=\frac{a^2+ab+b^2}{ab-1}=1+\frac{a^2+b^2+1}{ab-1}\cdots\ $   (☆)

$\leqq1+\frac{a^2+(a-1)^2+1}{ab-1}=1+\frac{2\left[(a-1)a+1\right]}{ab-1}\in\mathbb{N}$,since $(a-1)a+1$ is odd, we obtain $2|ab-1$,

i.e. $ab-1=1,2$,yielding $(a,b)=(1,2),(1,3)$

For $(a,b)=(1,2),f(a,b)=7$ and $(a,b)=(1,3)=\frac{3}{2}$, in addition to from （☆）

in using A.M$\geqq$ G.M and $a\geqq2,b\geqq1$,we obtain

$f(a,b)\geqq 1+\frac{2ab+1}{ab-1}\geqq 1+2+\frac{3}{ab-1}>3$.

Consequentry,The value of $f(a,b)$ is $4,7$.

kunny

kunny´s solution is wrong..

grobber, i don´t undestand your solution.
can you help me?  :)

Neither do I...I can't get how the procedure used by Grobber leads to the "dead-end"...and why this very procedure is justified in this case.
But still another question:
Could we in some way apply...or is there at least a connection to...Diophantine equations of the second order here:
[tex]ax^2+bxy+cy^2=k[/tex]? I looked it up at MathWorld and became curious about a possible connection.
Cheers!

[quote="Valentin Vornicu"]Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac{ a^2+ab+b^2} { ab- 1} .  \][/quote]
i've seen this problem approximately a year ago
 try to take a minimal and then show that there is x which is less than a   but it's pretty hard i think
  another way is to try to make it with the help of quadratic equation

[quote="Valentin Vornicu"]Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac{ a^2+ab+b^2} { ab- 1} .  \][/quote]
Let for some integer $n$ we have $n=\frac{a^2+b^2+ab}{ab-1}$.
Suppose $a>b>2$.
Then we have $a^2+b^2+ab=nab-n$, $a^2+a\cdot b(1-n)+b^2+n=0$. This equation has the second root (with respect to $a$), call it $a'$.
It's easy to prove that $0<a'<a$. We can repeat this while both $a,b>2$.
Hence all possible values of $n$ can be obtained while one of the numbers (i.e. b) is 0, 1 or 2, which is easy to handle.

[quote="armanf"][color=darkblue]Some Other questions with the same idea :

1: Find all natural numbers x,y,z such that x^2 + y^2 + z^2 = 3xyz

2: Suppose that m is a given natural number ,Show that there exist infinitly pairs (x,y) of natural numbers such that

xy | x^2 + y^2 + m

3: For which natural numbers n we can find integers a,b,c,d such that:

n = (a^2 + b^2 + c^2 + d^2)/(1 + abcd)

Good luck. :D [/color][/quote]

for the 2nd, saying WLOG that x is the greatest, take the couple $(x',y')=(\frac{y^2+m}x,x'$

I still can't find the complete solution :( I'm having troubles understanding why the new pair is "smaller" than the previous one. In IMO '88, that's clear, but here :blush: Does anyone dare to help me?

can anyone kindly give a complete solution?i am trying something like IMO 88 but it seems its not working here...

Of course Vieta-Jumping is working here  :)

See my proof in the attachment.

It's easy, not?
$a^2+b^2+ab$ is divisible by $ab-1$, so $a^2+b^2+2ab-1$ is also divisible by $ab-1$. $(a+b-1)(a+b+1)$ is divisible by $ab-1$. The rest is not so difficult.

With Alan Chen, Alexandru Girban, Anushka Aggarwal, Derek Liu, Dhrubajyoti Ghosh, Eric Shen (USA), Gene Yang, Jeffrey Kwan
Paul Hamrick, Sean Li, Zifan Wang:

Suppose that \[ k = \frac{a^2+ab+b^2}{ab-1} 	\iff a^2 - b(k-1) \cdot a + (b^2+k) = 0. \] We contend either $k=4$ or $k=7$. These are achieved by $f(2,2) = = 4$ and $f(2,1) = 7$.

Now fix $k$ and suppose $(a,b)$ is a minimal pair for that $k$. If $a=b$ we get $f(a,a) = \frac{3a^2}{a^2-1} = 3 + \frac{3}{a^2-1}$ which easily implies $a=2$ and hence $f(a,a) = 7$.

Else, assume $a > b$ and note that if $(a,b)$ is a solution then by Vieta jumping we get a neighboring pair \[ (a,b) \rightarrow \left( \frac{b^2+k}{a}, b \right) 	= \left( (k-1) \cdot b - a, b \right).  \] This forces $\frac{b^2+k}{a} \ge a$, so $k \ge a^2-b^2$. A calculation then gives \begin{align*} 	\frac{a^2+ab+b^2}{ab-1} &\ge a^2-b^2 \\ 	\iff a(2a+b) &\ge ab(a-b)(a+b) \\ 	\iff 2a+b &\ge b(a-b)(a+b) \\ 	\implies 2 &> b(a-b). \end{align*} This can only occur for $(a,b) = (2,1)$ hence $f(2,1) = 7$.

[b][color=red]Remark:[/color][/b]  	The Vieta chains for $k=4$ and $k=7$ are given respectively by 	\[ (2,2) \xleftrightarrow{a^2-6a+8} 		(2,4) \xleftrightarrow{a^2-12a+20} 		(10,4) \longleftrightarrow \dots \] 	and 	\[ \dots \longleftrightarrow 		(23,4) \xleftrightarrow{a^2-24a+23} 		(4,1) \xleftrightarrow{a^2-6a+8} 		(2,1) \xleftrightarrow{a^2-12a+11} 		(11,2) \xleftrightarrow{a^2-66a+128} 		(64,11) \longleftrightarrow \dots.  \] 	The proof above can be phrased conceptually 	as saying that $(2,1)$ and $(2,2)$ are the unique pairs 	(across all $k$) which do not have a neighbor of lesser sum.

I've no clue what exactly I'm doing here, new to Vietta jumping, but, well...
For $a \neq b$, say $a>b$.
Letting $f(a,b)=k \implies a^2+b^2+ab=kab-k \implies a^2+ab(1-k)+(b^2+k)=0$.
Say $(a,b)$ is the pair for the particular $k$ that is an output of $f$ such that $b$ is minimal.
Now say $a_0$ is another root of $x^2+xb(1-k)+(b^2+k)=0$, and clearly $a_0 \geq b$. Assume $a_0<a \implies a>a_0 \geq b$.
Notice $aa_0=b^2+k \implies b^2+k>ab \implies k \geq b(a-b)=ab-b^2$.
Also $a+a_0 =b(1-k) \implies 2a>b(k-1) \implies \frac{2a}b +1>k \geq ab-b^2 \implies 2a+b>ab^2-b^3 \implies \frac{b^3+b}{b^2-2} \geq a$.
As $a \geq b+1$, this implies $b^3+b \geq (b^2-2)(b+1)=b^3+b^2-2b-2 \implies 0 \geq b^2-3b-2 \implies b=1,2$ or $3$.

Thus we have 4 cases to deal with:
$b=1 \implies a-1 \mid a^2+a+1 \implies a-1 \mid a^2+a+1-a^2+1 \implies a-1 \mid 3 \implies a=4$ or $a=2$.
$b=2 \implies 2a-1 \mid a^2+2a+4=2a^2+4a+8-a(2a-1)=5a+8=10a+16-5(2a-1)=21 \implies a=1,2,4,11$.
$b=3 \implies 3a-1 \mid 91=13 \cdot 7 \implies$ Contradiction.
$a=b \implies a^2-1 \mid 3a^2 \implies a^2-1 \mid 3 \implies a=2=b$.
Checking all these cases give the only values of $k=f(a,b)$ as $4$ or $7$.

maybe i make mistake but if my solution be true it mean problem is too easy for TST.
$\[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \] $then should  $\frac {  (a+b)^2-1} { ab- 1}  $  be a integer. so
we can suppose that $a+b=t$ so:
$t^2-1=m(ta-a^2-1)$ then: $t^2-(ma)t+(m-1+ma^2)=0$ then we can check the possible roots and get solution.

We operate on the function $g(a,b) = f(a,b)-1=\frac{a^2+b^2+1}{ab-1}$. We first resolve some edge-cases.

[color=#f00][b]Claim:[/b][/color] If $b=1$, then $g(a,b)=6$.

[i]Solution[/i]: Note that
\[g(a,1)=\frac{a^2+2}{a-1}=a+1+\frac{3}{a-1},\]
so we get $a\in \{2,4\}$ and $g(a,1)=6$. $\fbox{}$

[b][color=#f00]Claim:[/color][/b] If $a=b$, then $g(a,b)=3$.

[i]Solution[/i]: Note that
\[g(a,a)=\frac{2a^2+1}{a^2-1} = 2+\frac{3}{a^2-1}.\]
This implies $a=2$, so $g(a,b)=3$. $\fbox{}$

We can now freely assume $a>b>1$ WLOG. Set $a/b=r$ so we get
\[g(a,b)=\frac{a^2+b^2+1}{ab-1}=\frac{r^2b^2+b^2+1}{rb^2-1} = r+\frac{b^2+r+1}{rb^2-1}=r+\frac{1}{r}+\frac{r+1+\frac{1}{r}}{rb^2-1} \le r+\frac{1}{r}+\frac{r+1+\frac{1}{r}}{4r-1}.\]
Then, note the inequality
\[4r-1>3r >r+1+1>r+1+\frac{1}{r},\]
so we must have
\[g(a,b)=\left\lceil r+\frac{1}{r}\right\rceil < r+\frac{1}{r}+1<r+2\]
because
\[\frac{r+1+\frac{1}{r}}{4r-1}<1\]
and $g(a,b)\in\mathbb{Z}$. Now, we use bruteforce to make Vieta Jumping work; we take some choice of $(a,b)$ and either reduce $|a|+|b|$ or get stuck at some particular points. WLOG assume $a>b$ as we already dealt with the other cases. Let $g(a,b)=k$. We can assume $k\not\in\{3,6\}$ as we have already found constructions for those $k$.

We write
\[a^2+b^2+1=kab-k.\]
Rewrite this as
\[a^2-(kb)a+b^2+k+1=0.\]
Thus, the other root of this quadratic in $a$ is
\[\frac{b^2+k+1}{a} < \frac{b^2+k+1}{b(k-2)} = \frac{b}{k-2}+\frac{k+1}{b(k-2)}.\]
We take the time to note that as $k\ne 3$ and $k\ge r+\frac{1}{r}>2$, we have
\[\frac{b^2+k+1}{a}<\frac{b}{2}+\frac{k+1}{2(k-2)} = \frac{b}{2}+\frac{1}{2}+\frac{3}{2(k-2)}\le \frac{b}{2}+\frac{1}{2}+\frac{3}{4} = \frac{b}{2}+\frac{b}{2}+\frac{1}{4}=b+\frac{1}{4}.\]
As the latter expression is an integer and cannot be $b$ (as that would give $k=3$, which is disallowed), we get that $|a|+|b|$ is monotonically decreasing when we Vieta Jump, absurd. Thus, $k\in \{3,6\}$ are the only possibilities and so $f(a,b)\in \{4,7\}$.

Let's look at $g(a, b)=f(a, b)-1=\frac{a^2+b^2+1}{ab-1}.$ Note that by $\text{AM-GM}$ we have $g(a, b)>2.$ Consider the solution $(a_0, b_0)$ with minimal sum, for which $g(a_0, b_0)=k$ is an integer. Assume WLOG $a_0\geq b_0.$ Consider the function $h(a)=a^2-kb_0a+b_0^2+k+1.$ We know that there is a number $a_1\in \mathbb{R},$ such that $$h(a_1)=0, a_0+a_1=kb_0, a_0a_1=b_0^2+k+1.$$ From here we easily deduce $a_1\in \mathbb{Z}_{> 0}$ and therefore it must be the case $a_1\geq a_0.$

[b]Case 1:[/b] Assume that $b_0\geq 2$ and recall that $k\geq 3.$ We obtain $$h(b_0)=b_0^2(2-k)+k+1\leq 4(2-k)+k+1=9-3k\leq 0,$$ so we must have $a_1\geq b_0\geq a_0\geq b_0,$ so $a_0=b_0$ and now by simple polynomial division we obtain $(a_0, b_0)=(2, 2),$ and so $f(a, b)=g(a, b)+1=4.$

[b]Case 2:[/b] Let $b_0=1.$ We then have $a_0-1\mid a_0^2+2,$ and so $a_0\in \{2, 4\}.$ In both cases we deduce $f(a, b)=g(a, b)+1=7.$ $\blacksquare$

We claim the only possible integer values are $4$ and $7$. First, note that when $a=b$, the fraction becomes $\frac{3a^2}{a^2-1} = 3 + \frac{3}{a^2-1}$ which is only an integer for $a=2$. This yields $f(a,b)=4$. Now, assume that $a>b>0$. We will use Vieta jumping. We have
\[ \frac{a^2+ab+b^2}{ab-1} = k \implies a^2 - (k-1)b \cdot a + b^2 + k .\]
Therefore, we can jump from the pair $(a,b)$ to $\left(\frac{b^2+k}{a},b)\right)$. This process always produces a smaller pair when $\frac{b^2+k}{a} < a$ or when $k < a^2 - b^2$. In this case we have,
\[ \frac{a^2+ab+b^2}{ab-1} < a^2 - b^2 \iff \frac{2a+b}{a+b} < b(a-b) \]
Note that $b(a-b) \ge 2 \ge \frac{2a+b}{a+b}$ when $b \ge 2$. Therefore, the only case when this breaks is when $a=1$ and $b = 2$. This yields $f(a,b) = 7$, so we are done.

The only possible values of $f(a, b)$ are $4,7$, which can be obtained with $(a,b) = (2,1), (2,2)$ respectively.

Let $k = \frac{a^{2} + ab + b^{2}}{ab-1}$. Then, we have $a^{2} - a(kb-b) + (k+b^{2}) = 0$. Suppose $a>b>0$ is a solution. Then,$(\frac{k+b^{2}}{a}, b)$ is also a solution. Observe that for $b\geq 2$, or $a-b\geq 2$ then we have
\[2a+b < b(a-b)(a+b) = a^{2}b - b^{3} \Rightarrow 2a^{2} + ab < a^{3}b - ab^{3}\]
\[\Rightarrow a^{2} + ab + b^{2} < a^{3}b - ab^{3} - a^{2} + b^{2} \Rightarrow a^{2} > \frac{a^{2} + ab + b^{2}}{ab-1} + b^{2} = k+b^{2}\]
Therefore, for all $b\geq 2$ or $a-b\geq 2$, when we flip the larger of $(a,b)$, it always decreases it. Thus, we need to check when $a = b$, or when $b = 1$ and $a-b < 2$. When $a = b$, we need $\frac{3a^{2}}{a^{2} -1}$ is an integer, or $3 + \frac{3}{a^{2} - 1}$ is an integer. The only case where this is possible is $a = b = 2$. When $b = 1$ and $a-b < 2$, this is when $a = 2$. For both these cases, flipping the larger one does not decrease, so all solutions $(a,b)$ can be continuously flipped to reach one of these two. Since $k$ is constant when flipped, the possible values of $k$ are $f(1,2), f(2,2) = 4, 7$.

Rewrite the problem as $\tfrac{a^2+b^2+1}{ab-1}=k \in \mathbb{Z}$ which is the same as $a^2-(bk)a+b^2+k+1=0$; given any solution $(a,b)$ with $a \ge b$ we Vieta jump to $(\tfrac{b^2+k+1}{a},b)$. It suffices to find all minimal solutions, which have $\tfrac{b^2+k+1}{a} \ge a$, so $k \ge a^2-b^2-1$. Letting $a=b+t$ for some $t \ge 0$, we have $$\frac{a^2+b^2+1}{ab-1}=\frac{2b^2+2bt+t^2+1}{b^2+bt-1} \ge a^2-b^2-1=2bt+t^2-1$$ which is equivalent to $$(2t)b^3+(3t^2-3)b^2+(t^3-4t)b-2t^2 \le 0.$$ When $t=0$ we see that all $b$ work, while when $t=1$ we can easily see that we must have $b=1$. There's no solutions when $t \ge 2$ for size reasons (for instance, $(2t)b^3+(3t^2-3)b^2+(t^3-4t)b-2t^2 \ge 3t^2-3-2t^2 >0$). When $t=0$ we have $a=b$, and so $\tfrac{a^2+b^2+1}{ab-1} =\tfrac{2a^2+1}{a^2-1} \in \mathbb{Z} \implies \tfrac{3}{a^2-1} \in \mathbb{Z} \implies a=b=2$. When $t=1,b=1$ we get $a=2$ which indeed works. So the only minimal pairs with $a \ge b$ are $(a,b)=(2,2),(2,1)$, which in the context of the original problem give $f(a,b)=4,7$ respectively.

WLOG $a\ge b$. Suppose $k=f(a,b)-1=\tfrac{a^2+b^2+1}{ab-1}$. Then
\[ a^2-(kb)a+(b^2+1+k) = 0. \]
So if $(a,b)$ works, then $(kb-a,b)$ works. Also, if $(a,b)$ works, then $(a,ka-b)$ works by symmetry. Consider the pair $(a,b)$ with minimal $a+b$ that works. Then we must have $kb-a \ge a$, so
\begin{align*}
    kb-a \ge a &\implies k=\frac{a^2+b^2+1}{ab-1} \ge \frac{2a}{b} \\
    &\implies a^2b+b^3+b \ge 2a^2b - 2a \\
    &\implies 2a+b\ge a^2b-b^3=b(a-b)(a+b) \\
    &\implies 2 > b(a-b).
\end{align*}
Therefore, $(a,b)=(2,1)$ or $a=b$. In the latter case, $f(a,b)=7$, and in the latter case, $f(a,a)=3 + \frac{3}{a^2-1}$, which is an integer if and only if $a=2$, in which case $f(2,2)=7$. The answer is 3 and 7.

First, setting $b=1$ gives $a^2+a+1-(a+2)(a-1)=3$ must be a multiple of $a-1$, forcing $a=2,4$, both of which cases gives $f(a,1)=7$.
Setting $a=b$ similarly gives that $3a^2-3(a^2-1)=3$ is a multiple of $a^2-1$, similarly forcing $a=2$, and $f(2,2)=4$. Now, we claim that these are the only possible values.

WLOG assume $a>b>1$ from here onwards, and let $(x,y)$ be a solution to $f(x,y)=k$ such that $x+y$ is minimized. We have $x^2+xy+y^2=k(xy-1)\rightarrow x^2-(k-1)xy+y^2+k=0$; treat this as a quadratic in $x$, and with Vieta's formulas, we know that the other solution is $ky-y-x$, which must be equal to $\frac{y^2+k}{x}$. The former gives that it is an integer, and the latter gives that it is positive, and therefore another valid solution; for a contradiction to not arise, we need $\frac{y^2+k}{x}\ge x$, or $k\ge x^2-y^2$.

$x^2+xy+y^2\ge (x^2-y^2)(xy-1)$ can be rearranged to $2x+y+y^3\ge x^2y$. Now, let $\frac{x}{y}=r$; the inequality can be again rewritten as $2r+1\ge (r^2-1)y^2$, from which we can solve $r\le \sqrt{\frac{y^4+y^2+1}{y^4}}+\frac{1}{y^2}<\sqrt{\frac{y^4+2y^2+1}{y^4}}+\frac{1}{y^2}=\frac{y^2+2}{y^2}\le \frac{y+1}{y}$, which makes it impossible for $x$ to be an integer.

We claim that $\boxed{f(a,b) = \frac { a^2+ab+b^2} { ab- 1} \in [4,7]}$
WLOG,assume that $a \ge b$,and let $k=f(a,b)-1=\tfrac{a^2+b^2+1}{ab-1}$.
Expanding it,\[ a^2-(kb)a+(b^2+1+k) = 0. \]
So if $(a,b)$ works, then $(kb-a,b)$ works. Also, if $(a,b)$ works, then $(a,ka-b)$ works by symmetry.
By Vieta Jumping,consider the pair $(a,b)$ with minimal sum:-
\begin{align*}
kb-a \ge a &\implies k=\frac{a^2+b^2+1}{ab-1} \ge \frac{2a}{b} \\
&\implies a^2b+b^3+b \ge 2a^2b - 2a \\
&\implies 2a+b\ge a^2b-b^3=b(a-b)(a+b) \\
&\implies 2 > b(a-b).
\end{align*}
Therefore, $(a,b)=(2,1)$ or $a=b$. In the former case, $f(a,b)=7$, and in the latter case, $f(a,a)=3 + \frac{3}{a^2-1}$, which is an integer if and only if $a=2$, in which case $f(2,2)=4$.$\blacksquare$

The answer is $f(a,b) \in \{4,7\}$. We can check that $f(2,1)=7$ and $f(2,2)=4$. Further, if $a=b$, then
$$f(a,b)=\frac{3a^2}{a^2-1}=3+\frac{3}{a^2-1},$$
so $a=2$ and $f(a,b)=4$. We can also check that if $\min\{a,b\}\leq 2$ we only have $f(a,b) \in \{4,7\}$. Now suppose that $k \neq 4,7$ is in the range of $f$, and pick $(a,b)$ such that $f(a,b)=k$, $a>b$, and $a+b$ is minimal, so $b>2$.\\
Treating $b$ as a constant, $f(a,b)=k$ rewrites as the following quadratic in $a$:
$$a^2-(k-1)ba+(b^2+k)=0.$$
By Vieta jumping, $(\tfrac{b^2+k}{a},b)$ is also a solution, and by minimality we must have
$$\frac{b^2+k}{a}\geq a \implies k \geq a^2-b^2 \geq a^2-(a-1)^2=2a-1.$$
This forces
\begin{align*}
    \frac{a^2+ab+b^2}{ab-1} &\geq 2a-1\\
    3a^2 &\geq 2a^2b-ab-2a+1\\
    3a^2 &> 2a^2b-3a^2\\
    3a^2 &> a^2b,
\end{align*}
which evidently only holds for $b \leq 2$, contradiction. Thus such a $k$ does not exist, so $f(a,b) \in \{4,7\}$ as desired. $\blacksquare$

[quote=Sprites]We claim that $\boxed{f(a,b) = \frac { a^2+ab+b^2} { ab- 1} \in [4,7]}$[/quote]

btw this is saying that $4 \leq f(a,b) \leq 7$ :P

The answer is $f(a, b) \in \{4, 7\}$.

Fix $k = \frac{a^2+ab+b^2}{ab - 1}$. Then the equation simplifies to $$a^2+(b-kb)a + b^2+k = 0.$$ This means that if $(a, b)$ yields a particular value of $k$, then $\left(\frac{b^2+k}a+b\right)$ yields this value of $k$ too. WLOG $a > b$.

[b][color=#f00]Claim.[/b][/color] If $a, b \geq 2$, $b^2+k < a^2$ unless $a=b$.

[i]Proof.[/i] The inequality simplifies to
\begin{align*}
a^2b-a &> b^3+a+b \\
b(a+b)(a-b)& > 2a+b \\
ab+b^2 &> 2a+b
\end{align*}
as $a-b > 0$. On the other hand the final display is clearly true. $\blacksquare$

This means that all values in the range of $f(a, b)$ can be generated by $f(a, a)$ or $f(\leq 2, \leq 2)$. The only such values are $4, 7$, as needed.

Answer: $4, 7$.

If $\min(a, b) \le 3$, then one can check that all pairs $(a, b)$ such that $f(a, b)$ is an integer are $(4, 1), (2, 1), (11, 2), (4, 2), (2, 2)$ and we can check the case $a = b$.  Let $(a, b)$ be a pair such that $f(a, b)$ is an integer and $a > b > 3$ and let $c = f(a, b)$. Then we have $a^2 + ab + b^2 = cab - c$, or equivalently $a^2 + a(b - cb) + (b^2 + c) = 0$. Let $a_1$ be a root of $x^2 + x(b - cb) + (b^2 + c) = 0$ other that $a$. Then $a_1 + a = cb - b$ and $aa_1 = b^2 + c$. Thus $a_1$ is an integer and $a_1 > 0$. We'll prove that $b  > a_1$. Assume not. Then we get $b \le a_1 = \frac{b^2 + c}{a}$, or equivalently $c \ge ab - b^2$. Therefore $a^2 + ab + b^2 \ge (ab - 1)(ab - b^2)$. By expanding, we get $a(2b + b^3) \ge a^2(b^2 - 1)$, so $2b + b^3 \ge a(b^2 - 1) \ge (b + 1)(b^2 - 1)$. Thus we get $0 \ge b^2 - 3b - 1$. Hence $b \le 3$, a contradiction. Thus we have $a_1 < b$.

Thus if we replace $(a, b) \to (a_1, b)$, then $\min(a, b)$ decreases. Therefore repeating this process until $\min(a, b) \le 3$, we obtain $c = 4$ or $7$. This completes proof. $\blacksquare$


ARML
geometry
3D geometry
MATHCOUNTS
AMC 8

i am.

its in santa teresa high school, i think

When is the Tokutomi Contest happening this year? Our school hasn't received any news about it. I did it last year and was $1$ point from receiving a prize.  :furious:

I never heard of it so I guess my schools not doing it

Its a small local one I believe.  Only schools from San Jose/Santa Clara can participate I think.

my school is in the santa clara county (gunn high)

[quote="Azaz"]It's the 20th of january.  Your school should have gotten a reply of confirmation from them already.  The deadline of registration was 1 week ago.[/quote]

Our coach said that the deadline was the 12th. So we registered.

How hard is the tokutomi contest is it like MC level or more like AMC 8/10 level?

i don't know.  itll be my first time there.  if i go.  im not sure yet.

I'm going! If they accept our almost-late registration.

Hmmmm... I remember someone saying that it was AMC 8 level. I don't think I am going though.

Um does anyone know which other schools are coming???

Lol I got the same one wrong for the same reason  :D . What were the highest scores for 9th graders? I didn't stay for the awards ceremony.

18, but they didn’t tell us the scores. I only know because I am a friend of the person who got 1st place for 9th graders and he told me that’s what he got.

Wait, sorry I got mixed up. The grand winners was a 9th grader and a 7th graders who both got 20, but 1st place for 9th graders, not including anyone who got in the top 5 for the junior division, got 18. Confused? Me too :)

This is how they did the scoring. First, they found the top 5 people in each division. Then, they found the top people of each grade that are not part of the top 5.

Oh yeah, how did you guys do the tiebreaker. I know the answer was 13, but I thought it was kind of hard to prove it. I just counted :)

wait chenhsi, you got a 19 and still got only 3rd in the 8th grade division? Hmm I think I got a 19 yet nobody told me I got an award or anything but that guy said that he got an 18 and got 1st place in 9th....weird.....

[quote="chenhsi"]i took it today :) I just got home...

Im from Sunnyvale Middle School and I got 3rd place for 8th graders :)

Not as good as my mom wanted, but who cares. The test was really easy, I thought it would be really hard so I was studying like AMC 12 level math, but then when I took the test I realized that it was actually easier than AMC 8, like AMC 6/7 or something :(  I got stupid number 17 wrong, because I found the percentage of the area of the nonshaded region instead of the percentage of the area of the shaded region :(  :mad: lol

Edit: Oh yeah, I know Homestead went.

Hey vishal, I saw you there. I was the kid that was bothering you and Alvin and following you two around. Remember, I tried to make you lose the race between you and the other kid in your rubix cube competition :) and i sat next to you last year at MathCounts State[/quote]

Yeah, that "other kid" would be me... :lol:

I left early. Someone tell me what happened?

I am almost ready to inflict violence on something (maybe myself. No. That would be emo)

I got 17 wrong for the exact same reason, and I got number 4 wrong. I counted red balls, not yellow. I "should" have gotten a perfect score. Yea, right.

[quote="chenhsi"]did you get in the top 5 of the junior division? im not really sure, i dont really get it myself. I [i]think[/i] alvin got 18, at least thats what he said. But knowing Alvin, who knows. Hey vishal, do you know?[/quote]

Alvin got 18 and I got 19. And yet he got 1st in 9th grade and I got 2nd in 9th grade. :furious:

I only got a 19 because I thought that white was not a color. :fool:

Senior division should be better next year. I bet the scores will be more further apart, since they don't even have a tiebreaker. The problems from this year senior seem interesting.

...and this doesn't help me at all, btw. What actually happened?

[quote="chenhsi"]i remember how he wanted to copy off of me on the Alg 2/Trig FInals last year. I just gave him the wrong answers :) As a result, I got a much higher grade than him. Laugh at him for me.

Edit: Hey vishal, I though you said you got 20 on the Tokutomi Poll.[/quote]

Umm... That was last year, and that topic was one year old. Don't you look at the dates?

[quote="sonny"]Vishal, can you post this year's senior division test? I just wanna look over how difficult it is and whether I should consider coming back next year.[/quote]

I took junior this year, so I tried to get a senior test from other people. They let me copy the questions, and I have attached it.

Anyways, the senior division questions are actually interesting and unambiguous, unlike the junior division questions. the toughest are like ARML Indiv 5-7 questions, so they shouldn't be too hard. So I recommend that you come back next year.

I have the junior paper, so I can put it up here:

EDIT: LaTeXed the senior paper!

I was a 6th grader (snuck in as a 7th grader)and I got a 19 yet I still did not receive a prize(not in top 5).

Did you make a typo on #12 in the Senior Test, vishalarul?Is it possible to give an answer as no solution? By the way, how long are we allowed to take this test?

wtf? why the sudden interest in tokutomi?

did it just happen again this year?  :huh:


probability
ARML
function
algebra
binomial theorem

You flip a coin 15 times.  What is the probability that there will be atleast a string of 9 or more heads?
I manage to do this (list out about all the cases) and used a caculator and did it in like 10 mins or so...but this was an arml practice problem and I cant use a caculator.  Is there an easier way?  Thanks?

You can set up a recursion.
From there, you use a generating function method to get the coefficient on a particular term, and that is the amount of seq. that satisfy the properties.

If you want a bigger hint, I think I can provide one in 5 minutes or so.

[quote="Bobtheball"]I'd think you could use binomial theorem to find the probabilities of 9 and up heads, then within each of those, you can use combinations to find how many of the 9, 10, 11, etc. heads results would contain a string of 9 or more heads.[/quote]

ya thats what I did, but I cant caculate that fast....especially counting all the sets.

[hide="Hint"]
For each case, we have secured and unsecured spaces.

We can break down into cases by having each case represent a position of the first H of the string in the sequence.
[/hide]

[hide="Answer"]Case 1: The first H is in position 1
The 9 H's are fixed into the first 9 slots and the remaining 6 slots are variable, giving $2^{6}=2(2^{5})$ possibilities.

Case 2: The first H is in one of positions 2 through 7 (6 cases)
The 9 H's, as well as the slot before the first H, are fixed (the latter is a T); the remaining 5 slots are variable, for $6(2^{5})$ possibilites.

Thus, there are a total of $(2+6)(2^{5})=2^{8}$ possible ways of accomplishing this.

Clearly, there are a total of $2^{15}$ outcomes, so $\boxed{P=\frac{2^{8}}{2^{15}}=\frac{1}{128}}$

[/hide]

Zuton, what you did, i think thats for just 9 in a row, I have atleast 9 in a row, so you may not have THHHHHHHHHT......you can have all H's if you want

[hide]
Since there are only 15 flips (which is less than $2\cdot 9$), there is at most one superstring of at least 9 consecutive heads. So we can put the first 9 heads in the superstring in a giant pile.

(For example, if we denote the giant pile by [b]G[/b], then the string

HTHHHHHHHHHHHTT (that's 11 H's in a row)

becomes HT[b]G[/b]HHTT.)

The number of ways to get 15 flips with a superstring is simply the number of ways of arranging a giant pile along with 6 other flips, which may be heads or tails. We also require that the [b]G[/b] not be immediately preceded by an H (otherwise, the giant pile wouldn't be the [i]first[/i] 9 consecutive heads.) Here there are 7 places to put the giant pile and $2^{6}$ ways to choose the outcome of the other flips. For the bad case of an H coming before the [i]G[/i], we can add the preceding H to the pile and get a pile of 10 heads. (It doesn't matter whether an H comes before this pile, since it's already bad.) The number of bad cases equals $6\cdot 2^{5}=3\cdot 2^{6}$.

Since the number of ways of getting 15 flips with at least 9 consecutive heads is equal to $4\cdot 2^{6}=2^{8}$, and there are $2^{15}$ total outcomes, the probability is $\frac{1}{2^{7}}=\boxed{\frac{1}{128}}$.

Hmm... that's exactly what zuton got...
[/hide]

No, if you look, the other flips remain variable... the flips after the forced string of 9 [i]could[/i] be heads. However, by counting the strings by their starting head, we make sure that we get every string without ever overcounting.


geometry
rectangle
quadratics
algebra
algebra proposed

Let $ m,n\in \mathbb{N}^*$. Find the least $ n$ for which exists $ m$, such that rectangle $ (3m \plus{} 2)\times(4m \plus{} 3)$ can be covered with $ \dfrac{n(n \plus{} 1)}{2}$ squares, among which exist $ n$ squares of length $ 1$, $ n \minus{} 1$ of length $ 2$, $ ...$, $ 1$ square of length $ n$. For the found value of $ n$ give the example of covering.

ok the sum squares area is:
$ n*1^2\plus{}(n\minus{}1)*2^2\plus{}...\plus{}1*n^2\equal{}(n\plus{}1)*(1^2\plus{}2^2\plus{}...\plus{}n^2)\minus{}(1^3\plus{}2^3\plus{}...\plus{}n^3)\equal{}\frac{(n\plus{}1)^2n(2n\plus{}1)}{6}\minus{}\frac{(n(n\plus{}1))^2}{4}\equal{}\frac{n(n\plus{}1)^2(n\plus{}2)}{12}\equal{}(3m\plus{}2)(4m\plus{}3)$
or
$ 12m^2\plus{}17m\plus{}6\minus{}\frac{n(n\plus{}1)^2(n\plus{}2)}{12}\equal{}0$
solving quadratic equation we get:
$ m\equal{}\frac{\minus{}17\plus{}\sqrt{289\minus{}48*(6\minus{}\frac{n(n\plus{}1)^2(n\plus{}2)}{12})}}{24}$
$ m\equal{}\frac{\minus{}17\plus{}\sqrt{1\plus{}4n(n\plus{}1)^2(n\plus{}2)}}{24}$
$ m\equal{}\frac{n^2\plus{}2n\minus{}8}{12}$
from that we deduce that $ n\equiv 8 \bmod{12}$
so the smallest such n is 8 and for that n m=6 for that the example is very easy.
So we are done.


Find all real-valued functions on the reals which satisfy $ f( xf(x) \plus{} f(y) ) \equal{} f(x)^2 \plus{} y$ for all $ x, y$.

1.$ f(x)$ is biject function .
2. for $ x\equal{}0$ then $ f(f(x))\equal{}x$
For $ y\equal{}0$ then $ f(xf(x))\equal{}f(x)^2$
For $ x\to\ f(x)$
$ f(xf(x))\equal{}x^2$
So $ f(x)^2\equal{}x^2$
Easy to check $ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}x$


At What % do we pass any topic? 70%? of 75%?

I have a Venn diagram with score 75%. My other two unpassed topics are counting in independent, and counting in lists. I got a problem with a binomial theorm, and expected value, and pascals triangle. Except, i dont see those topics on my report.
Also, if we pass a topic, do we get a new topic to replace it?
and, how many topics do we juggle at a time?
i see probability, counting, and an occasianal binomial theorm, pascals triangle, expected value............(This is for me, at least,)

It's different passing score for each topic; it depends on how other users perform in it.


geometry
Pythagorean Theorem
similar triangles
power of a point

In the figure $ DE \equal{} 6$, $ EF \equal{} 2$. Find the area, in terms of $ \pi$.

The area of what?

And if I may ask, what is this the "problem of the week" for?

The area of the circle. This problem of the week came from our schools math club.

[quote="ben_b"][quote="123456789"]So this isn't drawn to scale at all...[/quote]

Not even close. I am a little changed when it comes to creating computer graphics.[/quote]

It's hard to visualize this when the perpendicular signs are on angles that aren't even close to 90 degrees. I can't tell if the segments are connected or not.

Well, since CDF is a right triangle, and it's inscribed in a circle, segment FD (which measures 8) must be the diameter of the circle.

So the area is $ \pi (8/2)^2\equal{}\pi 4^2\equal{}\boxed{16 \pi}$

Yea, The Zuton Force's solution is correct. I was confused by the diagram.

[quote="The Zuton Force"]Not quite. Recall that it is the [i]hypotenuse[/i] that is the diameter of the circle. Hence, CD is the diameter, not FD.

The original poster likely wants us to assume that AB is also a diameter, since it passes through point "O".

If that is the case, then [hide="this is a possible solution"]Let the length of the radius be r. Then, OD = r, and by Pythagorean Theorem, OE = $ \sqrt {6^2 \minus{} r^2}$.

By Power of a Point about point E, $ (6)(2) \equal{} (AE)(EB) \equal{} (r \plus{} OE)(r \minus{} OE) \equal{} r^2 \minus{} OE^2$

Thus, $ 12 \equal{} r^2 \minus{} (36 \minus{} r^2) \equal{} 2r^2 \minus{} 36 \implies 2r^2 \equal{} 48 \implies r^2 \equal{} 24 \implies \boxed{A \equal{} \pi r^2 \equal{} 24 \pi }$[/hide]
If that is not the case, there is not enough information given.

[b]Sidenote:[/b] [i]"POTW" suggests that is the problem of the week for some place. If this is for school, then you probably don't want to just casually mention "power of a point" -- you should prove that the formula works. To do so, consider similar triangles in quadrilateral AFDB[/i]  :)[/quote]

Well, Power of a Point is a theorem that was taught in my math class, and it was completely valid to cite this theorem in a problem.


A teacher with a math class of 20 students splits the class into two equal groups of 10. If the groups are chosen randomly, with the only restriction that there are exactly 10 people in each group, what is the probability that Camilla and Cameron, two students in the class, are in the same group?

The original problem is #486 from The Art of Problem Solving, Volume 1. I thought up of this variation on it but wasn't sure how to solve it, can anyone help?

There is $ \binom{20}{10}$ possible groups....

the number of groups where there is one of the girls but not the other is $ \binom{18}{9}*2$

so the probability is $ 1 \minus{} \frac {\binom{18}{9}*2}{\binom{20}{10}} \equal{} \frac{9}{19}$...

am I right? ...

I'm confused as to why the number of groups where there is one girl but not the other is \[ \left(\begin{array}{cc}18\\9\end{array}\right) * 2\] instead of just \[ \left(\begin{array}{cc}18\\9\end{array}\right)\]

if I get it corectly that is because  let' s take those two girls away from the rest of the students. now we have $ \binom{18}{9}$ ways how we can arrange the other students.
we get our 10 students group by puting one girl in the first group and the other girl in the second .

but we could also put the first girl in the second group and the second girl in the first group....

that's why i  $ *2$

[hide]
For whatever group one girl is in, the other girl has a $ 9/19$ chance of being in the same group, since there are 9 spots left on one group and 10 in the other.[/hide]

:lol:  much more elegant solution alkjash

For next time, you can type $ \binom{n}{k}$ with[code]\binom{n}{k}[/code]

Never mind I figured it out you just do \binom{18 }{8} + \binom{18}{10} , all over \binom{20}{10} because if you put two girls in one group that leaves 8 spots open in one group and 10 in the other
sorry don't know how to write with the math terms.

Just put $ \$$ symbols around everything you want math-ized.

[quote]just wondering, how would you do it without using complements. i understand your way but i just want to see it from another perspective.[/quote]

just a question was that 'your' refering to me or to alkjash?

Call the two groups A and B. The probability that Camilla is in group A is 1/2, and the probability that Cameron is in group A is 1/2. So the probability that they are both in group A is 1/4. Likewise the probability that they are both in group B is 1/4. Therefore the probability that they are in the same group, A or B, is 1/4 + 1/4 = 1/2.

[hide]The above argument is incorrect. The answer is not 1/2, it is in fact 9/19. What is wrong with the argument?[/hide]

i think alkjash's solution explains that pretty well ... because if we would have a bag with 10 balls that have 'A' on them and 10 balls that would have 'B' on them then when Camilla would get 'A' there would be 10 balls with 'B' left but only 9 balls with 'A'

[hide]I agree that alkjash's solution is correct, but that doesn't [i]exactly[/i] explain what's wrong with the solution that I gave. Could you point out precisely which sentence in my solution is incorrect and why?[/hide]

[quote="gauss202"]Call the two groups A and B. The probability that Camilla is in group A is 1/2, and the probability that Cameron is in group A is 1/2. So the probability that they are both in group A is 1/4. [/quote]

$ \text{Camilla is in group A}$ and $ \text{Cameron is in group A}$ are [b]not independent events.[/b]  This is because there are only a fixed number of spots in each group of $ 10$.  This is the distinction between drawing balls out of a bag with and without replacement.


If r can make r + 1/r be an integer , prove that r^n + 1/(r^n) is an integer.

Thanks!

hello, i think this is wrong, take $ r\equal{}\frac{1}{2}$ and $ n\equal{}2$:
$ \left(\frac{1}{2}\right)^2\plus{}\frac{1}{\left(\frac{1}{2}\right)^2}\equal{}\frac{17}{4}$
Sonnhard.

Excuse me but I may be missing the point.

The expression is only an integer when $ r^n \mid 1 \Leftrightarrow r \equal{} \pm 1$ (if r is an integer)

Oh now I saw its "any number". Still, this isn't right.

[hide="Hint"]Proceed by strong induction on n.

[hide="Solution"]

The base case if for n=1, which showing $ r^1\plus{}\frac{1}{r^1}$ is an integer, and this is given.

So now assume the inductive hypothesis: $ r^{n\minus{}1}\plus{}\frac{1}{r^{n\minus{}1}}$ is an integer. Therefore, $ (r^{n\minus{}1}\plus{}\frac{1}{r^{n\minus{}1}})(r\plus{}\frac{1}{r})\equal{}r^n\plus{}\frac{1}{r^n}\plus{}r^{n\minus{}2}\plus{}\frac{1}{r^{n\minus{}2}}$ is an integer, and by the inductive hypothesis, $ r^{n\minus{}2}\plus{}\frac{1}{r^{n\minus{}2}}$ is an integer, so $ r^n\plus{}\frac{1}{r^n}$ is an integer.[/hide][/hide]

[quote="Dr Sonnhard Graubner"]hello, i think this is wrong, take $ r \equal{} \frac {1}{2}$ and $ n \equal{} 2$:
$ \left(\frac {1}{2}\right)^2 \plus{} \frac {1}{\left(\frac {1}{2}\right)^2} \equal{} \frac {17}{4}$
Sonnhard.[/quote]

But $ \frac{1}{2} \plus{} \frac{1}{\frac{1}{2}}$ is not an integer.

I think some of the posters made the mistake of confusing the notation $ r \plus{} 1/r$ with $ (r \plus{} 1)/r$ (I've seen that in some baaaaaad books).

Clearly $ r \plus{} 1/r \equal{} r \plus{} \frac {1} {r}$, while $ (r\plus{}1)/r \equal{} \frac {r\plus{}1} {r}$. Notations are not to be interpreted, but obeyed.

[quote="aadil"]that still doesnt explain why $ \frac {1}{2} \plus{} \frac {1}{\frac {1}{2}}$ is not an integer[/quote]

But $ \frac{(\frac{1}{2} \plus{} 1)}{\frac{1}{2}}$ [b]is [/b]an integer. This is what was meant in the first place by Sonnhard.

This is getting pretty off topic...

But if you interpret it as $ \frac{r\plus{}1}{r}$, then why interpret it also as $ r^n\plus{}\frac{1}{r^n}$ instead of $ \frac{r^n\plus{}1}{r^n}$? Either way, I was just pointing out a mistake with Sonnhard's correction.

i think the problem is correct in its given form. take for example r=2+(3^1/2)


calculus
integration
floor function
logarithms
limit
calculus computations

evaluat: $ \int^1_0 \frac {1}{x} \minus{} \lfloor\frac {1}{x}\rfloor$.
 :)

Cute.

[hide="Edited for clarity"]$ \int_{\frac {1}{m \plus{} 1}}^{\frac {1}{m}} \frac {1}{x} \minus{} \left\lfloor\frac {1}{x}\right\rfloor \,dx \equal{} \minus{} \frac {1}{m \plus{} 1} \plus{} \ln\frac {m \plus{} 1}{m}$, so your integral is $ \sum_{m \equal{} 1}^\infty \minus{} \frac {1}{m \plus{} 1} \plus{} \ln\frac {m \plus{} 1}{m} \equal{} \lim_{m \to \infty} 1 \plus{} \ln m \minus{} H_m \equal{} 1 \minus{} \gamma$.  (We use $ \ln a \plus{} \ln b \equal{} \ln ab$ to telescope that portion of the summation, and $ H_m : \equal{} \sum_{i \equal{} 1}^m \frac{1}{i}$.)[/hide]

JBL, are you sure about those boundaries of yours? The upper one doesn't tend to $ 1$ as $ m \rightarrow \infty$.


geometry
circumcircle
function
cyclic quadrilateral
angle bisector
geometry unsolved

1996 United Kingdom #3.

Let ABC be an acute triangle and O its circumcenter.  Let S deonote the circle through A, B, O.  The lines CA and CB meet again at P and Q, respectively.  Prove that the lines CO and PQ are perpendicular.

[hide="Solution"]The angles $ <PAB$ and $ <BQP$ are supplementary, so $ <BQP=<CAB$ (as directed angles modulo $ \pi$).  In other words, the line PQ makes the same angle with the line CQ as the tangent to the circumcircle of ABC through C.   Hence PQ is parallel to the tangent, so perpendicular to OC.[/hide]

There are two things I don't understand why it is true.

[hide="Why???"]
1) $ <BQP=<CAB$ in directed angles modulo $ \pi$.  If I am not mistaken shouldn't it be $ <BQP=<BAC$.

But this is what I don't get.

2) How do we derive the second sentence just from the fact that directed angles modulo $ \pi$ $ <BQP=<CAB$, or from what I think $ <BQP=<BAC$.[/hide]

Please Help

Thanks

Well, you are right that it should be $ BAC$, not $ CAB$.  As for the other question, it appears the solution writer skipped a step about arc/angle chasing.  Consider all of the other pains taken, the writer shouldn't have skipped this.

[hide]The angle the tangent makes with $ CQ$ is just $ m\angle{BAC}$, because both angles are half of $ m\overarc{BC}$.  Then we have by the cyclic quadrilateral that $ m\angle{BAC}=m\angle{PQC}$.

[/hide]

However, I'm kinda wondering why they made $ S$ be the circle through $ A, B, O$...the result holds for any circle through $ A, B$.  We have $ \triangle{ABC}\sim \triangle{QPC}$, and so therefore, flipping $ CO$ around the angle bisector gives the line whose "function" (if you like to think of it this way) in triangle $ ABC$ is the same as the function of $ CO$ in triangle $ QPC$.  But flipping this line gives the altitude, and so we're done.

Isn't this a bit easy for an olympiad problem?


1) The sequence $ 1, 2, 3, 4, 6, 7, 8, 9, 11, ...$ is formed by omitting every fifth natural number.  Find the sum of the first 100 terms of this sequence.

Answer: [hide]$ 6250$[/hide]

2) It is impossible to find two positive integers such that the square of one is exactly twice the square of the other.  However, there are integers such that the square of one is one less than twice the square of the other.  Find the product of the smallest two positive integers where this is true.

Answer: [hide]$ 35$[/hide]

[hide="1"]
It is easy to see that the numbers omitted are 5*the natural numbers.
The sum becomes the sum of the first 125 natural numbers - the sum of the first 25 multiples of 5.
The first part of that sum is $ \frac{(125)(126)}{2}$
The second part is $ 5\frac{(25)(26)}{2}$
Simplify the sum to $ (125)(63)\minus{}(125)(13)\equal{}(125)(50)\equal{}6250$

[/hide]

[hide="2"]
$ a^2\equal{}2b^2\minus{}1$
$ a^2\minus{}b^2\equal{}b^2\minus{}1$
$ (a\minus{}b)(a\plus{}b)\equal{}(b\minus{}1)(b\plus{}1)$
Now...how to solve from there....  hmm doesn't work.

Okay, let's try a table
$ b$~~~~~4~~9~~16~~25~~36~~49
$ 2b^2\plus{}1$~7~~17~31~~49~~71~~97
Ding ding ding for [b]different[/b] positive integers we get $ 5\cdot 7\equal{}35$

For the same integers obviously $ 0\cdot 0\equal{}0$ works
[/hide]

[quote="Tiger3"]
2) It is impossible to find two positive integers such that the square of one is exactly twice the square of the other.  However, there are integers such that the square of one is one less than twice the square of the other.  Find the product of the smallest two positive integers where this is true.
[/quote]

[hide="2"]
We have $ a^{2}\equal{}2b^{2}\minus{}1$ or $ a^{2}\minus{}2b^{2}\equal{}\minus{}1$.  Taking mod 2 of both sides we get $ a^{2}\equiv 1 (\mbox{mod }2)$.  Then, if we let a=1, then b=1, which is technically the smallest possible pair, which would make the answer 1.  If the numbers must be different, we move on.  3 and 5 give us irrationals for some reason, probably cause this problem involves squares, but finally we get $ a\equal{}7$ and $ b\equal{}5$ as a solution.  The product is $ \boxed{35}$.
[/hide]

Hmm, lists are probably easier

Really? I don't think lists would be much better...although most people won't think of mods.


[quote="salubadshah"]hey guys

could someone please help me with these questions

1- if $n\geq3$ then prove that $n!+3$ isn't a prime
2- Prove that for every k, a positive integer, k consecutive positive integers that are not prime can be found.

Any hints would be really helpful, thanks in advance![/quote]
[hide="1"]If $n\geq3,$ then $n!$ will be a multiple of $3.$ Adding another three to it will keep it a multiple of three, so if $n\geq3,$ then $n!+3\equiv0\pmod{3},$ and $n$ is not prime.[/hide]

this is what i can conculde from what you gave me

Since $n\geq3$ then $n!$ is multiple of 3, so $3|n!$ and $3|3$ but $n! = n(n-1)(n-2)(n-3)!$ which means that $n!$ has more than 2 factors hence $n!+3$ has more than 2 factors or $3|n!+3$ has more factors other than 3. Therefore, $n!+3$ is not prime number. Does this make sense, can someone help me to refine what i'm trying to say, tons of thanks!

[hide]
For the first question, just consider what happens to the expression when $n\geq 3$. You can always factor out a $3$, so the expression is never prime.
for the second question, it's kind of the same idea, suppose that $k=10$. You can have $11!+2, 11!+3,...,11!+11$, which are all composite.
[/hide]


pls ans this
 if my flight takes off at 2:00 am from 18N70E and it takes 10 hrs to reach 36N70W then wht is the loacl time at the destination.

It would be 12:00 pm plus the time change... not sure what it is because I don't know where 18N7OE and 36N7OW are...

Um...notice how it says latititude and longitude in the title.  Does that help?

Oh! :blush:  :D  Well, too lazy to go and figure the rest out... :D

wtf is that? transfer from 2 US fleets? with the speed of sound? 12690km in 10h from indian ocean to bermuda triangle?

nice :)

btw OHO's correct

[quote="OHO"]$2: 00am+10-\frac{70-(-70)}{15}=2: 40am$[/quote]

Notice the fact that you can't be 2/3 of the way through a time zone. I'm guessing the answer would be more like 3:00


function
limit
real analysis
real analysis unsolved

Let $\{f_{n}\}$ be a sequence of continuous functions which conveges uniformly to a function $f$ on a set $E$. Prove that \[\lim_{n\to\infty}f_{n}(x_{n})=f(x)\]
for every sequence of points $x_{n}\in E$ such that $x_{n}\to x$, and $x\in E$. Is the converse of this true?

Assume $x_{n}\to x\in E$ and $x_{n}\in E$. Fix $\epsilon >0$, from hypothesis there exist $M$ such that $|f_{n}(x)-f(x)|<\epsilon\forall x\in E, \forall n>M$ and $|f(x_{n})-f(x)|<\epsilon\forall n>M$. Then, $|f_{n}(x_{n})-f(x)|\leq |f_{n}(x_{n})-f(x_{n})|+|f(x_{n})-f(x)|< 2\epsilon\forall n>M$. Done for first part. But for second part? Can who?

It is not quite clear to me what the converse statement is

Me too  :oops:  , but I think it must is

Let $\{f_{n}\}$ be a sequence of continuous functions on $E$ , and a function $f$ defined on $E$. Prove that if
\[\lim_{n\to\infty}f_{n}(x_{n})=f(x) \]
for every sequence of points $x_{n}\in E$ such that $x_{n}\to x$, and $x\in E$ then $\{f_{n}\}$ conveges uniformly to the function $f$ on the set $E$.

The converse is true if E is compact.

http://www.math.rutgers.edu/courses/412/412-s09/412solns1.pdf


[b]If Microsoft Built Cars[/b]

1.Every time they repainted the lines on the road, you'd have to buy a new car.

2.Occasionally your car would just die on the motorway for no reason, accept this, restart and drive on.

3.Occasionally, executing a manoeuvre would cause your car to stop and fail to restart and you'd have to reinstall the engine. For some strange reason, you'd just accept this too.

4.You could only have one person in the car at a time, unless you bought a "Car 95" or a "Car NT". But then you'd have to buy more seats.

5.Other car manufacturers would make a car that was powered by the sun, was twice as reliable, five times as fast, twice as easy to drive - but it would only run on five percent of the roads.

6.The Macintosh car owners would get expensive Microsoft upgrades to their cars which would make their cars go much slower.

7.The oil, engine, gas and alternator warning lights would be replaced with a single "General Car Fault" warning light.

8.People would get excited about the "new" features in Microsoft cars, forgetting completely that they had been available in other cars for many years.

9.We'd all have to switch to Microsoft gas and all auto fluids but the packaging would be superb.

10.The airbag system would say "Are you sure?" before going off.

11.If you were involved in a crash, you would have no idea what happened.

12.They wouldn't build their own engines, but form a cartel with their engine suppliers. The latest engine would have 16 cylinders, multi-point fuel injection and 4 turbos, but it would be a side-valve design so you could use Model-T Ford parts on it.

13.There would be an "Engium Pro" with bigger turbos, but it would be slower on most existing roads.

14.Microsoft cars would have a special radio/cassette player which would only be able to listen to Microsoft FM, and play Microsoft Cassettes. Unless of course, you buy the upgrade to use existing stuff.

15.Microsoft would do so well, because even though they don't own any roads, all of the road manufacturers would give away Microsoft cars free, including IBM!

16.If you still ran old versions of car (ie. CarDOS 6.22/CarWIN 3.11), then you would be called old fashioned, but you would be able to drive much faster, and on more roads!

17.If you couldn't afford to buy a new car, then you could just borrow your friends, and then copy it.

18.Whenever you bought a car, you would have to reorganise the ignition for a few days before it worked.

19.You would need to by an upgrade to run cars on a motorway next to each other.

[b]If Linux made cars[/b]
1. The engine would be completely inaccesible: it's in a different partition
2.  It would take 10 minutes to start
3.  If you made any changes you would have to make public the design
4.  It would take 5 hours of tweaking after buying just to get it to run [/b]

[quote="tetrahedr0n"]Sorry, but this post is typical Microsoft-bashing. By no means is MS a perfect company, but this post doesn't make much sense. The comparison is in itself flawed of course, but here are my comments:

1. No, you'd have to update your car. Note that every major software comany does the same ... thats why we have "Adobe Acrobat 6" and "Photoshop 9." Its in no way a Microsoft thing.

2. I have yet to find a piece of software that won't crash occassionally, and yet to find a person that doesn't just restart and move on with life.

3. Sure, if by "some manouver" you crash your car into a tree, repairs might be needed. Of course, reinstalling Windows is a half hour process anyone can do, while reinstalling an engine is insanely difficult.

4. ? XP has access for any number of users, and theres no charge for that.

5. Which company provides a piece of software/hardware not powered by electricity?

6. This just contradicts 5. Why wouldn't they want their cars to run on more roads? It would be useless without the upgrade.

7. ? Debug mode will give you details.

8. Name these features.

9. You can always use 3rd party products, and of course Microsoft will not reccomend it. Similarly, Ford might introduce their brand of oil and reccomend it, but you're free to use 3rd party oil.

10. ? What is the airbag supposed to be in Windows?

11. How many people in car accidents know what happened?

12. Car companies don't build their own engines, and they do form deals with other companies to use them. I don't see how this relates.

13. Pentium Pro? When was that in use?

14. One word: RealPlayer

15. This just sounds like... "hey, Microsoft is succesful."

16. Don't get it.

17. Not recently, and when would you need to buy Windows now anyway?

18. Windows works out of the box, immediately.

19. What people drive two cars at once.[/quote]

I dont see you proving me wrong anywhere, mainly that other companies are doing the samething. The point of the post is, to see what would happen if Microsoft builds a car.

[quote="GoBraves"]I dont see you proving me wrong anywhere, mainly that other companies are doing the samething. The point of the post is, to see what would happen if Microsoft builds a car.[/quote]Actually the point of your post was to prove how bad of a company MS is :) But eventually it ended up being a "company like all the other ones" :)

The first time that I saw this kind of message was may be seven or eight years ago, in a newspaper that published (probably by an hoax, and a stupid jurnalist that didn't confirm source) that says something like that Bill Gates said in a conference that he was very proud about the new products of the company, and that he thought that if Ford would develop that fast we would have cars that would give 1000 miles per galon and cost $200, and therefore, a list like thatone was the answer of the Ford company.

Best,

Aha, so that's the story behind the Ford using Linux in their on-board computers ... :D ...

then your car would automatically tell you, when you opened the door, how many ways you are your four passengers could sit in the six seats, provided that someone 16 or over drives, and Aunt Marge and Uncle Max don't sit next to each other.

[quote="filletwho"]What if AoPS made cars?[/quote]
Then your car would be smarter than you are.

Maybe we need a top ten list of what would happen if AoPS made cars...

[quote="djimenez"][quote="themonster"][b]If Linux made cars[/b]
2.  It would take 10 minutes to start
[/quote]

Well, actually Linux (at least in my machine, with Fedora 3) is just a little slower to start than windows XP (and is something like 2:50 vs 2:30 if we include the logging to a user account) and also, linux spend less than 30 seconds turning down, but Windows spend around 3 minutes!!!

Best,[/quote]Mine spends exactly 18 seconds to completely shutdown. It's a matter of tweaks. Linux without tweaks is very weak. Windows XP without tweaks is acceptable. With tweaks, both are ok :P

If AOPS made cars:

1: There would be a new car model every few months.
2: Cars would crash occasionally, like in November.
3: Half of the car drivers wouldn't have even tried the cars.
4: I wouldn't drive one. :)

There isn't much to write about AoPs making cars- too few anologies that can be made.


geometry circumcircle invariant symmetry trigonometry ratio conics

What if I say that there exists a projective transformation leaving the circumcircle of triangle ABC invariant and mapping triangle ABC to an equilateral triangle, that the symmedian point K of triangle ABC becomes the center of this equilateral triangle under this transformation, and that your problem then follows trivially from symmetry?  :D

Well, I don't see a more nice solution right now. But the method above is really a good way if one cannot find anything better ("7 marks are 7 marks").

  Darij

I can solve it applying direct calculations using sine law and trigonometric version of Ceva.
Just calculate $ AM/BM $ and $AN/BN$ then
$ sinAQN/sinPQN $ and $ sinAPM/sinQPM $
then trigonometric Ceva that's all.

This is true with any point H (not necessarily on the symedian AK)
Kind regards. Jean-Pierre

Dear prowler,
    Can you tell me more specificly how you computed them ? Thanks  :)
 Dear Darij,
    Can you tell me more about your proof? Thanks :)
 Dear jpe,
    I think you 're right. But do you have a proof ? Maybe a version of Ceva ? ;) Thanks.
   I 'll give a pure synthetic proof to the first problem in my next post (of course without any computation)  :D )

Hope for no misprints.
Applying sine law in  $AVM$ and $BVM$ we get:
$ \frac {AM}{BM} = \frac {sin(ABH)*sin(ACH)}{sin(BCH)*sin(BVU)}$ (1)

Applying sine law in  $APM$ and $BPM$ we get:
$ \frac {sin(APM)}{sin(BPM)} = \frac {AM*sin(BAU)}{BM*sin (b)}$  (2)

 Putting (1) in (2) we get:
$ \frac {sin(APM)}{sin(BPM)} = \frac {sin(ABH)*sin(ACH)}{sin(BCH)*sin(b)}$ (3)

Analogously
$ \frac {sin(AQN)}{sin (BQN)} = \frac {sin(ABH)*sin(ACH)}{sin(CBH)*sin(j)}$ (4)

It is enough to prove Ceva trigonometric theorem:
$\frac {sin(APM)}{sin(BPM)}* \frac {sin(BQN)}{sin (AQN)}* \frac {sin(QAH)}{sin (PAH)} = 1$

Form (3) and (4) it is enough to prove:
$\frac {sin(CBH)}{sin(BCH)}* \frac {sin(j)}{sin (b)}* \frac {sin(QAH)}{sin (PAH)} = 1$ (5)

Applying sine law in  $AVH$ and $AUH$ we get:
$\frac {sin(QAH)}{sin (PAH)} =  \frac {VH*AU*sin(AHC) }{HU*AV*sin(AHB)}=
\frac {sin(BCH)*sin(AHB)*sin(AHC) }{sin(CBH)*sin(ACH)*sin(AHB)} $ (6)

Putting (6) in (5) we have to prove:
$\frac {sin(j)*sin(AHB)*sin(AHC) }{sin(b)*sin(ACH)*sin(AHB)}=1 $ (7)

 But  this directly goes from sine law in  $AHB$ and $AHC$. q.e.d.

As  JPE said we dont need condition about Lemoine point.
So it is a good problem, maybe you will show your solution treegoner.

Unfortunately, I didn't find a synthetic proof. I've just made an easy barycentric computation : if H =(p,q,r), the line MP is
mp = -aaqrx+p(aar+ccp)y+bbppz = 0
Similarly, for NQ
nq = -aaqrx + ccppy + p(aaq+bbp)z = 0
So mp-nq = 0 is ry-qz=0, which is the line AH.
Note that it follows immediately from the equations above that the line joining the common point of BH and NQ to the common point of CH and MP is the tangent at A to (O).
Kind regards. Jean-Pierre

Maybe someone can find solution with harmonic ratios ?

Probably, because it is a purely projective property.
I don't know if the following can help to a solution, but if we consider the triangles AUC and BAV, it follows from Pascal's theorem that the common point of the lines UB and NQ lies on the tangent at A to (O).
Using the triangles AVB and CAU, we get the same conclusion for the common point of the lines VC and PM.
Kind regards. Jean-Pierre

Wow, every emoticon has a very nice hat  :)
  Here is my solution to the original problem :
     The key idea is the following property :
 [b] Property[/b] : The tangent at $A$, $BC$ and $UV$ are concurrent at $X$
    Since H lies on the line that passes through $A$ and the intersection of the tangents at $B$ and $C$, let 's construct the circle $(O')$ that intangents to $(O)$ and intersects $AB$ and $AC$ at $D$ and $E$ such that $HD, HE$ are the tangents of $(O')$.
     Then by some angle chase, we get $ADHU$ and $AEHV$ are cyclic.
     Thanks to the property, we can construct a circle $(o)$ that passes through $B$ and $C$, and has $UV$ the radical axis of it and $(O')$.
     We have $QB.QC = QA . QV$. It follows that $Q$ lies on the radical axis of $(o)$ and $(AEVH)$. Now we will show that $N$ lies on that too. Consider the three circles $(o), (O'), (AEHV)$. The radical axis of $(o)$ and $(O')$ is $UV$, the radical axis of $(O')$ and $(AEVH)$ is $AE$. This implies what we want.
     Similarly, we have $MP$ is the radical axis of $(o)$ and $(ADHU)$. Since $AH$ is the radical axis of $(ADHU)$ and $(AEHV)$, the conclusion follows.

This topic is old, but I have some remarks.

[color=blue][b]1.[/b][/color]

[b]Initial problem:[/b]

[quote="treegoner"]Let $ABC$ be a triangle that is inscribed in the circle $(O)$. Let $K$ be the Lemoine point (i.e the isogonal conjugate point of the centroid $G$). Let H be a point lying on $AK$. Denote by $U,V$ the intersections of $BH$ and $CH$ with $(O)$. Let $M, N$ be respectively the intersections of $UV$ with $AB$ and $AC$. Let $P, Q$ be respectively the intersections of $BC$ with $AU$ and $AV$. Prove that $MP, NQ, AH$ are concurrent at $X$[/quote]

[quote="treegoner"] Dear Darij,
    Can you tell me more about your proof?[/quote]

In fact, my proof in post #2 uses the following fact:

[b]Lemma 1.[/b] [b](a)[/b] For any two (non-degenerate) triangles ABC and A'B'C' inscribed into one and the same circle k, there exists a projective transformation $\varphi$ which leaves the circle k fixed, and transforms the triangle ABC into the triangle A'B'C'.

[b](b)[/b] This projective transformation $\varphi$ maps the symmedian point of triangle ABC to the symmedian point of triangle A'B'C'.

[i]Proof of Lemma 1.[/i] I won't show Lemma 1 [b](a)[/b] here - my proof is quite non-elementary, but I am almost sure that there is a better one, since the fact is quite famous.

As for Lemma 1 [b](b)[/b], it is trivial if one remembers that the symmedian point of a triangle is the perspector of this triangle and its tangential triangle (i. e., the triangle formed by the tangents to the circumcircle of the triangle at its vertices). The projective transformation $\varphi$, leaving the circle k invariant and transforming the triangle ABC into the triangle A'B'C', will map the tangents to the circle k at the points A, B, C to the tangents to the circle k at the points A', B', C', and thus it will map the tangential triangle of the triangle ABC to the tangential triangle of the triangle A'B'C', so that the symmedian point of triangle ABC, being the perspector of the triangle ABC and its tangential triangle, will be carried to the symmedian point of triangle A'B'C', which is the perspector of the triangle A'B'C' and its tangential triangle. Hence, Lemma 1 [b](b)[/b] is proven.

Now, how to solve the initial problem using Lemma 1? Well, let k be the circumcircle of triangle ABC, and let A'B'C' be an arbitrary [b]equilateral[/b] triangle inscribed into this circle k. After Lemma 1 [b](a)[/b], there exists a projective transformation $\varphi$ which leaves the circle k fixed, and transforms the triangle ABC into the triangle A'B'C'. Let's see what happens to the configuration of the initial problem under this transformation $\varphi$. After Lemma 1 [b](b)[/b], the projective transformation $\varphi$ maps the symmedian point K of triangle ABC to the symmedian point K' of triangle A'B'C'. Clearly, the point H lying on the line AK is mapped to a point H' lying on the line A'K'. The points of intersection U and V of the lines BH and CH with the circumcircle k of triangle ABC are mapped to the points of intersection U' and V' of the lines B'H' and C'H' with the circumcircle k of triangle A'B'C' (since we know that the circle k is fixed under our projective transformation $\varphi$). And so on. Finally, we see that the assertion of the problem, namely that the lines MP, NQ and AH are concurrent, is equivalent to the concurrence of the lines M'P', N'Q' and A'H' (in fact, three lines are concurrent if and only if their images under a projective transformation are concurrent). Thus, altogether, we see that, after the application of our projective transformation $\varphi$, we are left with the same initial problem, but formulated for the triangle A'B'C' and not for the triangle ABC. But the triangle A'B'C' is equilateral, and for an equilateral triangle, the initial problem is trivial (everything is symmetric with respect to the line AH). So the problem is solved.

By the way, this is not the only problem which can be reduced to the case of an equilateral triangle using an appropriate projective transformation. There are lots of other examples for this strategy. I, personally, don't like this method of solving problems, but I would always use it on an olympiad unless I would have a better solution.

[color=blue][b]2.[/b][/color]

Jean-Pierre Ehrmann (jpe) generalized the initial problem as follows:

[quote="jpe"]This is true with any point H (not necessarily on the symedian AK)[/quote]

Now, this can be generalized even further. In fact, if the point H is just an arbitrary point, then the initial problem becomes a purely projective problem (i. e. all conditions and assertions of the problem can be formulated in terms of intersecting lines and joining points)... except of one thing: The five points A, B, C, U, V are supposed to lie on one circle. But for a projective problem, there is no difference between [i]five[/i] points lying on one circle and just any [i]five[/i] points in general position: any five points lie on one conic, and using an appropriate projective transformation, this conic can be transformed into a circle (unless it is a degenerate conic). Hence, the condition that the five points A, B, C, U, V have to lie on one circle must be superfluous to the problem. So we get the following generalized version of the problem:

[color=blue][b]General problem:[/b] Let ABC be a triangle, and H an arbitrary point in its plane. Denote by U and V two arbitrary points on the lines BH and CH. Let M and N be the points where the line UV meets the lines AB and AC, respectively. Let P and Q be the points where the line BC meets the lines AU and AV, respectively. Prove that the lines MP, NQ and AH are concurrent.[/color]

And this problem admits a very easy solution using projective transformations:

Since the general problem is purely projective, we can apply a projective transformation which maps the line AH to the line at infinity. In other words, in order to solve the general problem, it is enough to solve it for the case when the line AH is the line at infinity. In this case, of course, the points A and H are infinite points. The lines BM, CN, PU and QV, all passing through the point A, must therefore be parallel to each other; also, the lines BU and CV, both passing through the point H, must be parallel to each other.

Now, consider the two triangles BPU and CQV. Since BP || CQ (this is trivial, since the lines BP and CQ coincide), PU || QV (this was proven above) and UB || VC (this is just a rephrase of BU || CV), these two triangles are homothetic. In other words, there exists a homothety h which maps the triangle BPU to the triangle CQV. This homothety must therefore map the points B, P, U to the points C, Q, V, respectively. Let this homothety h map the point M to a point M'. Then, CM' || BM, QM' || PM and VM' || UM (since a homothety maps every line to a parallel line).

Now, from CM' || BM, it follows that the point M' lies on the parallel to the line BM through the point C. But since CN || BM, the parallel to the line BM through the point C is the line CN, and thus, we see that the point M' lies on the line CN. Also, since VM' || UM, but the point V lies on the line UM, it follows that the point M' lies on the line UM, too. Hence, the point M' is the common point of the lines CN and UM. But the common point of the lines CN and UM is the point N. Thus, we get M' = N. And our result above, QM' || PM, becomes QN || PM. In other words, MP || NQ. This yields that the lines MP and NQ intersect on the line at infinity. But the line at infinity is the line AH; thus, the lines MP and NQ intersect on the line AH. In other words, the lines MP, NQ and AH are concurrent.

So the general problem is solved for the case when the line AH is the line at infinity. Since the problem is a projectively invariant one, and every line can be mapped to the line at infinity by an appropriate projective transformation, it follows that the general problem is solved for all cases. $\blacksquare$

  Darij


trigonometry
geometry
circumcircle
geometry proposed

If   $ A\leq\frac{\pi}{2}$,  then

$ R(1\plus{}\cos A)\geq m_a$

First let's prove that in any triangle $ ABC$ there is the relation : $ \boxed{\ m_a\equal{}R\sqrt{1\plus{}2\cos A\cos(B\minus{}C)\plus{}\cos^2 A}\ }$ .

Indeed, $ m_a^2\equal{}\frac{2(b^2\plus{}c^2)\minus{}a^2}{4}\equal{}\frac{b^2\plus{}c^2\plus{}2bc\cos A}{4}\equal{}$ $ \frac{4R^2\sin^2 B\plus{}4R^2\sin^2 C\minus{}2\cdot 4R^2\sin B\sin C\cos A}{4}$

$ \Longleftrightarrow\ m_a^2\equal{}R^2(\sin^2 B\plus{}\sin^2 C\minus{}2\sin B\sin C\cos A)$

$ \Longleftrightarrow\ m_a^2\equal{}R^2\left(1\minus{}\cos^2 B\plus{}1\minus{}\cos^2 C\plus{}\underline{2\cos A\cos B\cos C\minus{}2\sin B\sin C\cos A}\minus{}2\prod\cos A\right)$

$ \Longleftrightarrow\ m_a^2\equal{}R^2\left(2\minus{}\cos^2 B\minus{}\cos^2 C\plus{}\underline{2\cos A\cos(B\minus{}C)}\minus{}2\prod\cos A\right)$ . Now we take into account that in

$ \triangle ABC$ is true the identity $ \boxed{\ \cos^2 A\plus{}\cos^2 B\plus{}\cos^2 C\plus{}2\prod\cos A\equal{}1\ }$ and the last relation becomes :

$ m_a^2\equal{}R^2(1\plus{}2\cos A\cos(B\minus{}C)\plus{}\cos^2 A)$ $ \Longleftrightarrow\ \boxed{\ m_a\equal{}R\sqrt{1\plus{}2\cos A\cos(B\minus{}C)\plus{}\cos^2 A}\ }$ .

Since $ \measuredangle A\le\frac{\pi}{2}$ and $ \cos(B\minus{}C)\le 1\ \implies\ m_a\le R\sqrt{1\plus{}2\cos A\plus{}\cos^2 A}\equal{}R\sqrt{(1\plus{}\cos A)^2}$ $ \Longleftrightarrow\ \boxed{\ m_a\le R(1\plus{}\cos A)\ }$ .

[quote="Mateescu Constantin"]First let's prove that in any triangle $ ABC$ there is the relation : $ \boxed{\ m_a \equal{} R\sqrt {1 \plus{} 2\cos A\cos(B \minus{} C) \plus{} \cos^2 A}\ }$ .

Indeed, $ m_a^2 \equal{} \frac {2(b^2 \plus{} c^2) \minus{} a^2}{4} \equal{} \frac {b^2 \plus{} c^2 \plus{} 2bc\cos A}{4} \equal{}$ $ \frac {4R^2\sin^2 B \plus{} 4R^2\sin^2 C \minus{} 2\cdot 4R^2\sin B\sin C\cos A}{4}$

$ \Longleftrightarrow\ m_a^2 \equal{} R^2(\sin^2 B \plus{} \sin^2 C \minus{} 2\sin B\sin C\cos A)$

$ \Longleftrightarrow\ m_a^2 \equal{} R^2\left(1 \minus{} \cos^2 B \plus{} 1 \minus{} \cos^2 C \plus{} \underline{2\cos A\cos B\cos C \minus{} 2\sin B\sin C\cos A} \minus{} 2\prod\cos A\right)$

$ \Longleftrightarrow\ m_a^2 \equal{} R^2\left(2 \minus{} \cos^2 B \minus{} \cos^2 C \plus{} \underline{2\cos A\cos(B \minus{} C)} \minus{} 2\prod\cos A\right)$ . Now we take into account that in

$ \triangle ABC$ is true the identity $ \boxed{\ \cos^2 A \plus{} \cos^2 B \plus{} \cos^2 C \plus{} 2\prod\cos A \equal{} 1\ }$ and the last relation becomes :

$ m_a^2 \equal{} R^2(1 \plus{} 2\cos A\cos(B \minus{} C) \plus{} \cos^2 A)$ $ \Longleftrightarrow\ \boxed{\ m_a \equal{} R\sqrt {1 \plus{} 2\cos A\cos(B \minus{} C) \plus{} \cos^2 A}\ }$ .

Since $ \measuredangle A\le\frac {\pi}{2}$ and $ \cos(B \minus{} C)\le 1\ \implies\ m_a\le R\sqrt {1 \plus{} 2\cos A \plus{} \cos^2 A} \equal{} R\sqrt {(1 \plus{} \cos A)^2}$ $ \Longleftrightarrow\ \boxed{\ m_a\le R(1 \plus{} \cos A)\ }$ .[/quote]

Very nice Dear [color=red][b]Constantin[/b][/color]!!!!

[quote]$ m_a\begin{array}{c} \le \\
\ > \end{array}R(1 + \cos A)\Longleftrightarrow$ $ A\begin{array}{c} \le \\
\ > \end{array}90^{\circ}$ .[/quote]
[b]Method 1.[/b] Let $ w = C(O,R)$ be the circumcircle of $ \triangle ABC$ and let $ M$ be the midpoint of  $ [BC]$ .

 Prove easily that $ OM = R\cdot \{\begin{array}{ccc} \cos A & \mathrm{if} & A\le 90^{\circ} \\
 \\
- \cos A & \mathrm{if} & A > 90^{\circ}\end{array}$ . Therefore,

 IF $ A\le 90^{\circ}$ , then $ AM\le AO + OM$ , i.e. $ m_a\le R + R\cos A$ .

IF $ A > 90^{\circ}$ , then $ AM > AO - OM$ , i.e. $ m_a > R + R\cos A$ .

[b]Method 2.[/b] Denote $ X\ .s.s.\ Y\ \Longleftrightarrow\ XY > 0\ \mathrm{or}\ X = Y = 0$ (same sign) . Therefore,

$ [m_a - R(1 + \cos A)]\ .s.s.\ 4m_a^2 - 4R^2(1 +$ $ \cos A)^2\ .s.s.\ 2(b^2 + c^2) - a^2 - 4R^2(1 + 2\cos A + \cos^2A)\ .s.s.$

$ 2(b^2 + c^2) - 8R^2(1 + \cos A)\ .s.s.\ \sin^2B + \sin^2C -$ $ 1 - \cos A\ .s.s.\ 1 - \cos 2B + 1 - \cos 2C - 2 - 2\cos A\ .s.s.$

$ - \cos 2B - \cos 2C - 2\cos A = 2\cos (B - C)\cos A - 2\cos A =$ $ - 2\cos A[1 - \cos (B - C)]\ .s.s.\ \cos 90^{\circ} - \cos A\ .s.s.\ (A - 90^{\circ})$ .

In conclusion, $ [m_a - R(1 + \cos A)]\ .s.s.\ (A - 90^{\circ})$ , i.e. $ \boxed {\ m_a\begin{array}{c} \le \\
\ > \end{array}R(1 + \cos A)\Longleftrightarrow\ A\begin{array}{c} \le \\
\ > \end{array}90^{\circ}\ }$ .


How many square units are in the shaded triangle shown on the 3 by 2 grid of unit squares? Express your answer as a common fraction.

[asy]size(50);
draw((0,0)--(3,0)--(3,2)--(0,2)--cycle);
draw((0,1)--(3,1));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
filldraw((2,0)--(3,0)--(2,2/3)--cycle,black);
draw((0,2)--(3,0));[/asy]

The height is $ \frac{2}{3}$ and the base is $ 1$, so the area is $ \boxed{\frac{1}{3}}$.


AMC USA(J)MO USAMO geometry rectangle Gauss Putnam

Alright I really need some clarification.

For question 3, we know that one of the following is right:

0 < k < 1/2   ,     2 < k

0 < k < 1      ,     1 < k

The first possible answer is true if fractals (a sequence of lines forming an infinite polygon) cannot be used.  The second is true if fractals can be used.

So here's the question:
ARE FRACTALS ALLOWED????

I'm thinking no, because fractals aren't covered in high school math, but maybe I'm wrong.
Also, if anyone can show me a way to achieve the second answer without fractals (some people said they've seen this solution but I don't believe them) please show me.

The second is also possible if fractals can't be used. I'm sure you'll see a solution if you go to the Math Jam tomorrow. k=1 can't be done unless you take an infinite number of sides. Polygons have only a finite number of sides by definition.

this picture isn't mine, it was posted on Mathlinks, but it shows an example for 1 < k < 2.

in fact the whole solution can be read here http://www.mathlinks.ro/viewtopic.php?t=5396

We'll have a full solution in the Math Jam tonight, with an explanation as to how an investigative student might have found it.

We'll give a full explanation that will explain that tonight.

[quote="GreenTwilight"]Oh!  Well, that explains everything.  I have no idea how I was supposed to get that, but cool.  At least someone has finally showed me an answer, and I kind of get it now.

Does anyone know if there has to be all right angles?  It would seem so from what I now understand about the problem.[/quote] hmm, indeed Iura is one of the best math students in the world (scored 28p in IMO last year, and he was only 8th grade!! still 4 more IMOs to go for him) so I guess that is the only explanation one can obtain.

Other people who solved the problem: Le Viet Bao (42p in IMO 2003), Andreis (25p in IMO 2003) :) All have the same explanation: "it just came to me".

[quote="GreenTwilight"]Oh!  Well, that explains everything.  I have no idea how I was supposed to get that, but cool.  At least someone has finally showed me an answer, and I kind of get it now.

Does anyone know if there has to be all right angles?  It would seem so from what I now understand about the problem.[/quote]

I don't think that it does explain everything.  It's not true that the picture posted (two right angles in the cut through the rectangle) works for all rectangles.  For some values of k there needs to be more "steps" in the cutting process.  It is however possible to prove that we can take a sufficient number of steps in our cut and make it work.

[quote="rrusczyk"][quote="Valentin Vornicu"]All have the same explanation: "it just came to me".[/quote]

For those of you who want to see how the solution could 'just come to you' -> come to the Math Jam tonight and we'll explain how you might have seen the solution.[/quote]
I personally don't think that this kind of problems (olympiad question) can be explained. The ideas just flow in one's mind, and only the gifted are capable of solving these in the given time (or even without any given time).

The Olympiad questions are usually chosen such that they select the most gifted students, not necesarily the ones with the most training and best memory.

No one can make any class, or any course, on how to takle (most of the) problems no. 3,6 of the IMO, USAMO, Romanian Olympiad, Russian Olympiad, because they are usually problems chosen BECAUSE they have original ideas.

[quote="Valentin Vornicu"]
I personally don't think that this kind of problems (olympiad question) can be explained. The ideas just flow in one's mind, and only the gifted are capable of solving these in the given time (or even without any given time). [/quote]

I think we'll have to agree to disagree.  There are ways to arrive at solutions to these problems without just 'seeing them', and without being particularly gifted.  This problem #3 is a great example.  Like many others, I figured that the solution was just the similar rectangles.  Only while rigorously going through proving it did the step ladder appear - and it wasn't the result of a magical insight, but the result of rigorous investigation.

Sometimes it's very hard to see where the insight comes from - this is the first step to learning to teach, which is its own skill (and not necessarily one that's always highly correlated with the ability to do).  In the problem at hand, there is a very clear way to arrive at the solution in a step-by-step manner.  Some may jump to the solution with great insight, but the less gifted among us can still get there.

I also disagree that only the truly gifted can do these types of problems, unless we are to agree that the universe of gifted students is *extremely* large.

Olympiad problem solving skills can be taught.  Isn't that why you're writing a book?  (I hope so, and look forward to another resource at that level.)

What we can agree on is that olympiad problems are chosen to require insight. However, training is a very important part of developing that insight.  (Also, the problems aren't particularly 'original' - they're usually twists on established ideas.  That said, they are generally excellent, and the writers of such problems should be greatly admired for that skill.)

I'm gonna have to side with Valentin on this one.  I don't believe that you can teach anyone all the math in the world, and then that person would be able to solve any modern problem.  I think that the people who got the question all had to "just see it."  Even during math jams tonight, when you do show us how to go about solving the problem, there isn't one set way to solve the problem.  Math can't be taught so that you'll see the problem and immediately know how to solve it.  The same holds true for any of the questions on USAMO, not just Question #3.  The whole point of USAMO is that you can't just know how to do the questions, or know how to approach it.  You have to be VERY insightful and creative, a quality that not many people have.  If we could be assured of the answer by knowing methods beforehand, then we would see somewhere around 200 perfect scores.  We don't, because there is no definite way of approaching any USAMO question, nothing you can KNOW to give you exactly what you need to know how to solve a USAMO question.  Knowing a lot of math certainly helps, but you have to be somewhat of a brilliant mind to figure out questions like #3 on USAMO.

I think there are some problems to which it's very hard to explain how to come up with solutions, but I don't think #3 on the USAMO is one of them. The solution involves something of a natural extension of the case k>2, and so I think someone who works hard on this problem can figure out what is going on.

[quote="ComplexZeta"]I think there are some problems to which it's very hard to explain how to come up with solutions, [/quote]

Very true - this is one of the great joys/challenges of teaching - those problems that make me think 'how am I ever going to teach how to do that'.  Figuring out how to teach those problems is a problem solving process very similar to doing problems.  This is why we advocate students trying to teach other students - by trying to figure out how to teach somebody, you will learn better how to use the tools yourself.  Obviously, this works best on topics you understand mostly but haven't entirely mastered.

I personally don't believe in "genius" (as in some mystical ability to have ideas "come to you" -- that sounds like magic to me).  I think the top students are those who learn how to explore in a lot of good ways with a lot of good tools.  The IMO is a much harder competition these days.  Is this because there are genetically more geniuses (by a serious leap) or because there are more and better materials for learning to understand those methods of exploration and how all the tools can be used together?  I believe that latter is the vast majority of the explanation.

Very good point Matthew. And also, how does one explain how certain smaller countries, like Romania for instance, consistently perform so well on the IMO? Is it that they have more "geniuses" per capita than the other countries? (I know Valentin will say yes! lol j/k)

But no, it's because they have more training at Olympiad problems from a younger age - hence they [i]learn[/i] how to approach such problems. This is the whole reason for having Olympiad training camps. It may take many years, but one can learn to do such problems.

A good discussion of this "genius" vs. "learning" debate can be found in Steve Olson's book, [u]Countdown[/u], about the 2001 IMO.

Dear Gauss,

Yes I will say that (unexplicably) Romania "produces" many geniuses. Take a look at Caraiani Ana, or Ciprian Manolescu, or Hornet Stefan or Manea Mihai. Look at the Priceton's Putnam team which has been Romanianized in the past couple of years (and it appears that it will be for some time on from now :D). Look that the AMC/USAMO director was Titu Andreescu (a Romanian) for almost a decade, Dan Voiculescu which just received the USA Academy Award for High Merits in mathematics, etc. etc. etc.

It's not the training that makes these students best, because most of the training sesions in Romania are hold only for (at most) 3 weeks time (compared with US, China, Russian IMO training). Yes, good training can help you solve problems 1,2,4,5 on IMO, and with a little luck you might struck 1p on no. 3 or 6, and get a gold medal, but usually it takes a genius mind to solve any of these problems (3 or 6).

You probably don't know this, but problem 3 in 2000 was one of the hardest problems in IMO history. Only some of the students solved it, about 15, but most of them (!!!) solved at most another problem in the contest. Those guys had no training, but had extraordinary minds. Also lots of students performed very well on all other tehnique based problems, but did not solve this one.

There are geniuses in every country, not only in Romania, China/Taiwan, USA, Russia, South Korea, Iran, Bulgaria, Vietnam...

Of course there are! :) but my point was that there are a lot of Romanian geniuses ;)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group of cardinal $p^{n}m$ where $m<2p$. Prove that $G$ has a normal subgroup of order $p^{n}$ or $p^{n-1}$.

Hi.

If there is only one p-sylowgroup we have a normal subgroup of order $p^{n}$.

So we assume that there is more than one p-sylowgroup. Since $m<2p$ we can deduce that there are m=p+1 of them.

Let $S_{1}$ and $S_{2}$ be two different p-sylowgroups of G. Then $|S_{1}S_{2}|=\frac{|S_{1}\parallel S_{2}|}{|S_{1}\cap S_{2}|}\leq |G|=p^{n}m<p^{n}\cdot 2p\leq p^{n}\cdot\ pp=p^{n+2}$
Let $|S_{1}\cap S_{2}|=p^{i}$ then
$\frac{|S_{1}\parallel S_{2}|}{|S_{1}\cap S_{2}|}=\frac{p^{n}\cdot p^{n}}{p^{i}}$
$\Rightarrow  p^{2n-i}<p^{n+2}$
$\Rightarrow  p^{n-i}<p^{2}$
$\Rightarrow  i\in\{n,n-1\}$

i=n means that $S_{1}=S_{2}$ which is a contradiction to our assumption $S_{1}\neq S_{2}$
So $|S_{1}\cap S_{2}|=p^{n-1}$ which means that $U: =S_{1}\cap S_{2}$ is a normal subgroup of $S_{1}$ and $S_{2}$. Therefore the normalisator $N(U)$ contains at least two p-sylowgroups and so it contains all p+1 p-sylowgroups. Therefore $p^{n}$ and m and hence $|G|$ are divisors of $|N(U)|$.
$\Rightarrow  N(U)=G$

q.e.d.


Predictions, anyone? I'm talking about Blue Group here, which is NOT the frosh group and it's also NOT the winners... the 18 or so in between.

I'm hoping for $n$ where $n$ is my score, even though it'll probably be higher...

If anyone wants to make a winners poll... I've limited myself to one forum poll per week, [size=42] because otherwise I go on a rampage[/size]

20 seems reasonable.
but i chose 17 bcz my score is a lil bit higher than that..


\[ \underline {\underline {\mathrm {Problema\ propusa\ (clasa\ a\ VII\ -\ a)\ .}}}  \]\[ \mathrm {Autor\ : \ Claudiu\ Andone\ ,\ Cugir\ .}  \][color=darkred] Pe laturile $[AB]$ si $[AC]$ ale unui tringhi [u]isoscel[/u] $ABC\ (AB=AC)$ se construiesc in exteriorul acestuia [u]triunghiurile dreptunghice si isoscele[/u] $BAD\ (AB\perp AD)$ respectiv $ACE\ (CA\perp CE)\ .$
$1.\blacktriangleright$ Sa se arate ca $DB\perp DE$ $\Longleftrightarrow$ $m(\widehat {BAC})=120^{\circ}\ .$
$2.\blacktriangleright$ [b][u]Incercati o generalizare[/u][/b] la un triunghi oarecare ![/color]

[hide="Va uitati aici dupa ce ati incercat si nu ati reusit."]
[color=darkblue][b]O usoara generalizare.[/b] Pe laturile $[AB]$ si $[AC]$ ale unui triunghi [u]oarecare[/u] $ABC$ se construiesc in exteriorul acestuia [u]triunghiurile dreptunghice si isoscele[/u]  $BAD\ (AB\perp AD)$ respectiv $ACE\ (CA\perp CE)\ .$ Notam proiectia $P$ a punctului $C$ pe dreapta $AB$. Sa se arate ca urmatoarele afirmatii sunt echivalente :
$1.\blacktriangleright DB\perp DE\ .$
$2.\blacktriangleright a^2=b^2+2c^2\ \ \left(\cos A=-\frac{c}{2b}\right)\ .$
$3.\blacktriangleright A\in (BP)$ si $AB=2\cdot AP\ .$[/color][/hide]

Intr-adevar, aceasta problema a fost data la concursul La scoala cu ceas.Pentru implicatia directa puteam presupune ca m(<BDE)>90 sau m(<BDE)<90.Luam in considerare doar primul caz si construim punctul D' astfel incat m(<BD'E)=90 si se ajungea usor la o contradictie( nu mai tin minte, parca obtineam ca masura unui unghi e nula;nu am figura in fata).Pentru reciproca nu stiu daca mi-au dat ceva, am incercat sa consider figura din hartie.Apoi am indoit figura astfel incat BD sa fie perpendiculara pe planul triunghiului si observam ca indoitura se suprapunea pe dreapta AB.Asa ca m(<BAC)=120.Nu stiu daca e corect ce spun.Va rog frumos domnule profesor Nicula sa ma corectati unde e gresit.Va multumesc! :)

RobertuX comes back, this time with another hat. ;)


Is given a triangle . If all angles are greater than 59 degree pove that they all are smaller than 62 degree.

[hide="Proof"]Let $ x_1,x_2,x_3$ be angles of a triangle such that:

$ x_1,x_2,x_3>59$

$ x_1\plus{}x_2\plus{}x_3>177$

$ x_1\plus{}x_2\plus{}x_3\equal{}180$

Then, $ x_1\plus{}x_2>118$

If, $ x_3>62$

Then $ x_1\plus{}x_2\plus{}x_3>180$, a contradiction.

[/hide]

[hide="Answer"]
We have three angles $ a_i$, and we have $ 59<a_i$. We sustain to prove $ a_i<62$.

We can see that $ 177 > a_1\plus{}a_2\plus{}a_3 \equal{} 180$.

We can use an argument of contradiction. We can assume that $ a_3>62$.

If we take the minimal value for $ a_1$ and $ a_2$, we can see that $ 59(2)\equal{}118<a_1\plus{}a_2$.

Now, we can combine these two equations. We can now see that $ a_1\plus{}a_2\plus{}a_3>180$, which contradicts the fact that all angles of a triangle must be exactly 180. $ \Box$
[/hide]

My own solution  :

Let be $ \alpha,\beta,\gamma$ the angles of this triangle.
We have $ \alpha > 59$
             $ \beta > 59$
             $ \gamma > 59$

Let be $ \alpha = 59$ and $ \beta = 59$

Than:

$ \gamma < 180 - (59 + 59) = 62$

Let be $ \alpha = 59$ and $ \gamma = 59$

Than:

$ \beta < 180 - (59 + 59) = 62$

Let be $ \beta = 59$ and $ \gamma = 59$

Than:

$ \alpha < 180 - (59 + 59) = 62$

SO: $ \alpha < 62$ $ \beta < 62$ $ \gamma < 62$    :)

Elio, your solution is wrong. Here are the mistakes:

The angles cannot be equal to $ 59$, you contradited yourself by saying $ \alpha,\beta,\gamma > 59$. Then you said $ \alpha,\beta \equal{} 59$.

$ \gamma > 180 \minus{} (59 \plus{} 59) > 62$ does not show that $ \gamma > 62$, in fact, $ \gamma \equal{} 62$, which also contradicts the proof.

Note that a triangle can have two angles equal to $ 59$ and one angle equal to $ 62$. Its the case in which two angle are greater than $ 59$ that the third angle must be less than $ 62$.

Try redoing it and posting.

Also, #H34N1, your proof is correct but how can you justify:

$ 177>a_1\plus{}a_2\plus{}a_3\equal{}180$, this is essentially saying that $ 177>180$.

I am soory   :blush:

That was not what I mean .

Now I fixed it.  :)

[quote="JZambrano201"]$ x_1 \plus{} x_2 \plus{} x_3 > 177$[/quote]This step is unnecessary.

10000th User, your right, that step was unnecessary. Elio (n), your proof is still incorrect, the variables cannot equal $ 59$ as you stated, they must be greater than $ 59$.

Everyone has the [i]idea[/i] of the proof -- however, the [i]presentation[/i] of this idea has been very muddled.  Here're two ways to write it:

[hide="As a contradiction"]Let the measures of the three angles of our triangle be $ a, b$ and $ c$.  Assume for the sake of contradiction that one of the angles has measure greater than 62.  Without loss of generality, $ a > 62$.  We are given that $ b > 59$ and $ c > 59$.  Thus $ 180 \equal{} a \plus{} b \plus{} c > 62 \plus{} 59 \plus{} 59 \equal{} 180$, a contradiction, so our assumption was incorrect and every angle of the triangle has measure less than 62.[/hide]

[hide="Directly"]We know $ b > 59$ and $ c > 59$, so $ \minus{}b < \minus{}59$ and $ \minus{}c < \minus{}59$.  Therefore $ a \equal{} 180 \minus{} b \minus{} c < 180 \minus{} 59 \minus{} 59 \equal{} 62$.  Since $ a, b$ and $ c$ are interchangeable we have also $ b < 62$ and $ c < 62$, as desired.[/hide]

I don't understand what is JZambrano201 saying. I don't think my proof is wrong.

Explain:

$ \alpha \plus{} \beta \plus{} \gamma \equal{} 180$ is constant

Whether $ \beta$ and $ \alpha$ are equal to 59 , $ \gamma$ should be smoller then 62

IF $ \beta$ and $ \alpha$ are greater than 59 ,than  surely $ \gamma$ must be greater than 59 and smoller than 62.

And so this predicate serve for the 2 angles ($ \beta$ and $ \alpha$) too.

Conclusion: $ 62>\alpha,\beta,\gamma>59$

Am I clear?   :wink:

Sory if I'm not being clear. I don't mean to start an argument, but here are the problems:

You said: Whether $ \alpha$ and $ \beta$ are equal to 59 ,  should be smoller then 62.

This makes no sense because, if $ \alpha,\beta\equal{}59$, then $ \gamma\equal{}62$. It is not smaller than $ 62$, thus contradicting your proof.

Also, if an angle is equal to $ 59$ then it immediately contradicts the proof because the angles must be [b]greater[/b] than $ 59$.

I hope you see my point. Also, check out JBL's proofs.

\[ \alpha_1 \plus{} \alpha_2 \plus{} \alpha_3 \equal{} 180^{\circ}
\]
[hide="My solution"]Setting $ \alpha_i \equal{} 59^{\circ} \plus{} \delta_i$ such that $ \delta_i$ is a positive real number, we find that
\[ \delta_1 \plus{} \delta_2 \plus{} \delta_3 \equal{} 3^{\circ}\Leftrightarrow 0 < \delta_i < 3^{\circ}
\]
so,
\[ \alpha_i \equal{} 59^{\circ} \plus{} \delta_i\Longrightarrow 59^{\circ} < \alpha_i < 62^{\circ}
\]
[/hide]  :D

edit: [quote="JZambrano201"]Also, if an angle is equal to $ 59$ then it immediately contradicts the proof because the angles must be [b]greater[/b] than $ 59$.[/quote]

This can be justified by limit definition or even some simple lingo I think. Such an approach works because (If you extend my format) we officially allow

$ \lim_{\delta_{1,2}\to 0} \sum \delta_i \equal{}3^{\circ}\Rightarrow \delta_3\equal{}3^{\circ}$ Thus, $ \alpha_3\equal{}59^{\circ}\plus{}\delta_3\equal{}62^{\circ}$ which is maximized.

[quote="igiul"]edit: [quote="JZambrano201"]Also, if an angle is equal to $ 59$ then it immediately contradicts the proof because the angles must be [b]greater[/b] than $ 59$.[/quote]

This can be justified by limit definition or even some simple lingo I think. Such an approach works because (If you extend my format) we officially allow

$ \lim_{\delta_{1,2}\to 0} \sum \delta_i \equal{} 3^{\circ}\Rightarrow \delta_3 \equal{} 3^{\circ}$ Thus, $ \alpha_3 \equal{} 59^{\circ} \plus{} \delta_3 \equal{} 62^{\circ}$ which is maximized.[/quote]This is totally absurd.

Consider
\[ f(x)\equal{}\left\{\begin{array}{cc}1&\text{if }x\ne0\\0&\text{if }x\equal{}0\end{array}\right.\]

$ \lim_{x\to0}f(x)\equal{}1$ but $ f(0)\equal{}0$...
Elio(n)'s proof is not correct. Are you guys going to keep arguing?

:huh: Of course it is. He used circular reasoning. Besides your cited function has absolutely no application here and I fail to see how it is at all relevant; first of all, this is a case in which
\[ \lim_{x\to b} f(x) \equal{} f(b).
\]
Secondly, perhaps you should have looked more closely at what I'm saying; I don't think you got it. If you let $ \delta_1,\delta_2$ tend towards zero, $ \alpha_1,\alpha_2$ both approach fifty-nine. If you acknowledge that $ \delta_i$ is still greater than zero though very small, there is no reason why this isn't fine. You are effectively showing the 'limit' of the maximum, as in the value towards which the angle [b]approaches[/b], but [b]is never equal to[/b] because we are on working on the interval $ \delta_i \in(0,3)\Leftrightarrow \alpha_i \in(59,62)$ [b]not[/b] $ \delta_i \in[0,3]\Leftrightarrow \alpha_i \in[59,62]$. When doing a min/max problem [i]you always check the endpoints[/i]. If the interval is open then you check the limit of the endpoints (which happens to be the case) to account for the possibility of a function being discontinuous there (which isn't the case).

edit1: typos
edit2: I think the phrase I'm looking for is the 'upper bound' of the range. You proabably get what I'm trying to say though.

[quote="igiul"]:huh: Of course it is. He used circular reasoning.[/quote] :huuh: If you look closely at his proof, that's not circular reasoning. Circular reasoning happens when you prove what you assumed, for example. His solution is by cases in which he took wrong assumptions, as pointed out by JZambrano201.
[quote="igiul"]Besides your cited function has absolutely no application here and I fail to see how it is at all relevant;[/quote]The function I've provided maybe too hypothetical to illustrate my point that there is no equality and by no means you use limit concept to replace inequalities... maximizing $ \alpha_3$ is not even part of the problem.... Please see JBL's solution.
[quote="igiul"]Secondly, perhaps you should have looked more closely at what I'm saying; I don't think you got it.[/quote]Oh I do know. If you don't think so, PM me.
[quote="igiul"]If you let $ \delta_1,\delta_2$ tend towards zero, $ \alpha_1,\alpha_2$ both approach fifty-nine. If you acknowledge that $ \delta_i$ is still greater than zero though very small, there is no reason why this isn't fine. You are effectively showing the 'limit' of the maximum, as in the value towards which the angle [b]approaches[/b], but [b]is never equal to[/b] because we are on working on the interval $ \delta_i \in(0,3)\Leftrightarrow \alpha_i \in(59,62)$ [b]not[/b] $ \delta_i \in[0,3]\Leftrightarrow \alpha_i \in[59,62]$. When doing a min/max problem [i]you always check the endpoints[/i]. If the interval is open then you check the limit of the endpoints (which happens to be the case) to account for the possibility of a function being discontinuous there (which isn't the case).

edit1: typos
edit2: I think the phrase I'm looking for is the 'upper bound' of the range. You proabably get what I'm trying to say though.[/quote]Thank you for agreeing JZambrano201's. That is what he is saying about Elio (n)'s solution, that the angles cannot be equal to 59. And why make things complicated where there is a simpler proof? Even your solution is a much simpler version than having an 'upper bound' argument. Besides... the limit you took in the edit part after your solution... is a two-sided limit and it's still wrong. ;)

[quote="mathemonster"]lets the angles = a , b, c.
if we try to maximize an angle of the triangle, lets say c, then 59 + 59 + c = 180
so c = 62. However, a and b are greater than 59 so c < 62. QED[/quote]This is the kind of argument that we would normally 'brainstorm', if that's the closest meaning I can think of. However, as JBL said, it lacks rigor, the argument is muddled.

ok is this more rigorous:
call the angles of the triangle a, b, c,
assume that b> 62
we are given that a > 59 and b>59
summing the three inequalities we get a+b+c > 180
this is false because the sum of the angles in a triangle equal 180.
contradiction


geometry
trigonometry
parallelogram
trig identities
Law of Sines

Let H be the orthocenter of ABC. Prove that the circum radii of ABC, ABH, ACH, and BCH are equal.

How do you do this without law of sines or trig or analytic geo?

I think I got it, not sure tho....
[hide]

Construct Parallelograms AHCX, AHBY, and BHCZ.

Drawing the intersection of BH and AC P, AH and BC Q, we can show:
$ \angle AHP \cong \angle BHQ$ from vert angles
And due to right triangles we have $ \angle HBC \cong HAC$
Similar logic shows:
$ \angle HBC \cong \angle HAC$
$ \angle HAB \cong \angle HCA$
$ \angle HBA \cong \angle HCB$

and then we extend to the Parallelograms and we find (cutting a little algebra out):
$ \angle AYB \equal{} 180\minus{}\angle HCA \minus{} \angle BCH$
And so now we know (from sum of interior angles, and what we know about the angles) $ \angle YBC \equal{} \angle YAC \equal{} 90$
Because these two are right triangles, they have the same circumcircle. Therefore the circumradii of $ \triangle ABY$ ,$ \triangle ABH$,$ \triangle ABC$ are all equal. Apply the same logic for the remaining sides and you get all of those radii are equal.
QED

[/hide]

[hide="hint"]Consider the nine-point circles of all four triangles.[/hide]


induction
combinatorics proposed
combinatorics

there are some identical squares with sides parallel, in a plane. Among any $k+1$ of them, there are two with a point in common. Prove they can be divided into $2k-1$ sets, such that all the squares in one set aint pairwise disjoint.

I remember Iura's idea on this problem.
Consider the leftmost square $X$. Let $A$ and $B$ are two right its vertices. Then all squares can be divided into three parts: squares containing $A$, squares containing $B$ and square nonintersecting with $X$. For squares from thord part we know that from any $k$ of them there are two intersecting squres. Therefore, by induction, third part can be partitioned into $2k-3$ subsets with required property.

That's all.