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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
4-var inequality
RainbowNeos   4
N a few seconds ago by RainbowNeos
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
4 replies
RainbowNeos
Yesterday at 9:31 AM
RainbowNeos
a few seconds ago
An almost identity polynomial
nAalniaOMliO   6
N 2 minutes ago by Primeniyazidayi
Source: Belarusian National Olympiad 2025
Let $n$ be a positive integer and $P(x)$ be a polynomial with integer coefficients such that $P(1)=1,P(2)=2,\ldots,P(n)=n$.
Prove that $P(0)$ is divisible by $2 \cdot 3 \cdot \ldots \cdot n$.
6 replies
1 viewing
nAalniaOMliO
Mar 28, 2025
Primeniyazidayi
2 minutes ago
Hard diophant equation
MuradSafarli   1
N 4 minutes ago by FarrukhBurzu
Find all positive integers $x, y, z, t$ such that the equation

$$
2017^x + 6^y + 2^z = 2025^t
$$
is satisfied.
1 reply
MuradSafarli
33 minutes ago
FarrukhBurzu
4 minutes ago
Euler's function
luutrongphuc   2
N 24 minutes ago by KevinYang2.71
Find all real numbers \(\alpha\) such that for every positive real \(c\), there exists an integer \(n>1\) satisfying
\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]
2 replies
luutrongphuc
3 hours ago
KevinYang2.71
24 minutes ago
No more topics!
Concurrency
shobber   9
N May 6, 2024 by joshualiu315
Source: APMO 1992
In a circle $C$ with centre $O$ and radius $r$, let $C_1$, $C_2$ be two circles with centres $O_1$, $O_2$ and radii $r_1$, $r_2$ respectively, so that each circle $C_i$ is internally tangent to $C$ at $A_i$ and so that $C_1$, $C_2$ are externally tangent to each other at $A$.

Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.
9 replies
shobber
Mar 11, 2006
joshualiu315
May 6, 2024
Concurrency
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G H BBookmark kLocked kLocked NReply
Source: APMO 1992
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shobber
3498 posts
#1 • 2 Y
Y by Adventure10, Mango247
In a circle $C$ with centre $O$ and radius $r$, let $C_1$, $C_2$ be two circles with centres $O_1$, $O_2$ and radii $r_1$, $r_2$ respectively, so that each circle $C_i$ is internally tangent to $C$ at $A_i$ and so that $C_1$, $C_2$ are externally tangent to each other at $A$.

Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.
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yetti
2643 posts
#2 • 2 Y
Y by Adventure10, Mango247
Assign weights $m_{A1} = OO_1 = r - r_1, m_O = O_1A_1 = r_1$ to $A_1, O$, respectively, so that these 2 weights balance at $O_1$, which is then assigned weight $m_{O1} = m_{A1} + m_O = r - r_1 + r_1 = r$. Assign weight $m_{O2}$ to $O_2$, such that the weights $m_{O1}, m_{O2}$ balance at $A$:

$\frac{m_{O2}}{m_{O1}} = \frac{O_1A}{O_2A} = \frac{r_1}{r_2}$

$m_{O2} = m_{O1}\ \frac{r_1}{r_2} = \frac{r r_1}{r_2}$

Assign weight $m_{A2}$ to $A_2$, so that the weights $m_O, m_{A2}$ balance at $O_2$:

$\frac{m_{A2}}{m_O} = \frac{OO_2}{O_2A_2} = \frac{r - r_2}{r_2}$

$m_{A2} + m_O = m_{O2} = \frac{r r_1}{r_2}$

Solving for $m_{A2}$,

$m_{A2} = m_O\ \frac{r - r_2}{r_2} = \left(\frac{r r_1}{r_2} - m_{A2}\right) \frac{r - r_2}{r_2}$

$m_{A2} \left(1 + \frac{r - r_2}{r_2}\right) = \frac{r r_1}{r_2} \cdot \frac{r - r_2}{r_2}$

$m_{A2} = \frac{r_1(r - r_2)}{r_2}$

Let $OA$ meet $A_1A_2$ at $P$. The weights $m_{A1}, m_{A2}$ now balance at $P$, i.e.,

$\frac{A_1P}{A_2P} = \frac{m_{A2}}{m_{A1}} = \frac{r_1(r - r_2)}{r_2(r - r_1)}$

Since we also have $\frac{OO_1}{A_1O_1} = \frac{r - r_1}{r_1},\ \frac{OO_2}{A_2O_2} = \frac{r - r_2}{r_2}$, it follows that

$\frac{OO_1}{O_1A_1} \cdot \frac{A_1P}{PA_2} \cdot \frac{A_2O_2}{O_2O} = \frac{r - r_1}{r_1} \cdot \frac{r_1(r - r_2)}{r_2(r - r_1)} \cdot \frac{r_2}{r - r_2} = 1$

By Ceva's theorem, the cevians $A_2O_1,\ OA \equiv OP,\ A_1O_2$ in the triangle $\triangle OA_1A_2$ are concurrent.
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neverstop
9 posts
#3 • 3 Y
Y by Adventure10, Mango247, Stuffybear
in this problem, we just apply the ceva theorem with triangle $OO_1O_2$, with A in $O_1O_2$, $A_1$ in $OO_1$, $A_2$ in $OO_2$
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pontios
777 posts
#4 • 2 Y
Y by Adventure10, Mango247
shobber wrote:
Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.

This result is also true if the three circles are externally tangent.

That's what neverstop means and this is true because $\frac{s-a}{s-b} \cdot \frac{s-b}{s-c} \cdot \frac{s-c}{s-a} =1$
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Vo Duc Dien
341 posts
#5 • 2 Y
Y by Adventure10, Mango247
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1992Problem2

Vo Duc Dien
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erfan_Ashorion
102 posts
#6 • 1 Y
Y by Adventure10
oh nice problem!!
lemma 1:$O1,O2,A$ are on same line.
proof:we know that $\angle O2AO =90$ $  \angle O1AO =90$ so $O1,A,O2$ are on same line :)
lemma 2:$O,O2,A2$are on same line and $O,O1,A1$ are on same line.
proof:we know that $O2A2$ is perpendicular to line that tangent to circle$C$ and we know that $OA2$ is perpendicular to line that tangent to circle$C$ so $A2,O2,O$ are on same line :wink: !
proof of problem:
suppose $A2A1O2 = X$ and $O2A1O=Y$ and $OA2O1 = P$ AND $O1A2A1=Q$ we must proof:
$\frac{\sin X}{\sin Y} . \frac{\sin AOO1}{\sin AOO2} . \frac{\sin P}{\sin Q} =1$
we know that :
$\frac{\sin AOO1}{\sin AOO2} = \frac {r_1}{r_2} . \frac{OO2}{OO1}$
and we know that:
$\frac{\sin X}{\sin Y} . \frac{AA2}{A1O} = \frac {r_2}{OO2}$
and also it is right for $\frac{\sin P}{\sin Q}$and we know that $OA2=OA1$ so the problem is proof!
excuse me because im so lazy to right all of my solution but im sure that you know that my solution :wink:
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armpist
527 posts
#7 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Dear MLs

It is simple.

The concurrency point is this case is, obviously,
the Lemoine point (symmedian point) of triangle AA1A2.


r1 and r2 are two red herrings, while r isn't.


This post is dedicated to both Mr. Referendum - G. Papandreou and
Mr. M.I.T. - Darij, not necessarily in this order.

Friendly,
M.T.
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sunken rock
4390 posts
#8 • 2 Y
Y by Adventure10, Mango247
@armpist: very fine observation, although not quite 'obvious'; I think it can bring new good questions about this configuration.

Best regards,
sunken rock
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armpist
527 posts
#9 • 2 Y
Y by Adventure10, Mango247
@sunken rock

maybe the very first candidate should be the Sangaku of Gumma, even though there
are plenty of solutions by now.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=18266

It may lead to a new approach.

M.T.
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joshualiu315
2533 posts
#10
Y by
Apply Ceva's theorem on $\triangle OO_1O_2$. The following statement is equivalent to the desired lines being concurrent:

\[\frac{OA_1}{A_1O_1} \cdot \frac{O_1A}{AO_2} \cdot \frac{O_2A_2}{A_2O} = 1.\]
Then, notice that we can write everything in terms of $r$, $r_1$ and $r_2$. Since

\[\frac{r}{r_1} \cdot \frac{r_1}{r_2} \cdot \frac{r_2}{r} = 1,\]
we are done. $\square$
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