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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Problem 3
blug   2
N a few seconds ago by kokcio
Source: Polish Math Olympiad 2025 Finals P3
Positive integer $k$ and $k$ colors are given. We will say that a set of $2k$ points on a plane is $colorful$, if it contains exactly 2 points of each color and if lines connecting every two points of the same color are pairwise distinct. Find, in terms of $k$ the least integer $n\geq 2$ such that: in every set of $nk$ points of a plane, no three of which are collinear, consisting of $n$ points of every color there exists a $colorful$ subset.
2 replies
blug
Apr 4, 2025
kokcio
a few seconds ago
isogonal geometry
Tuguldur   3
N 43 minutes ago by whwlqkd
Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$. Let $\triangle P_1P_2P_3$ and $\triangle Q_1Q_2Q_3$ be their respective pedal triangles. Let\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]Prove that the points $X_1$, $X_2$ and $X_3$ lie on the line $PQ$.
3 replies
Tuguldur
Today at 4:27 AM
whwlqkd
43 minutes ago
Problem 1
blug   6
N 43 minutes ago by Tintarn
Source: Polish Math Olympiad 2025 Finals P1
Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
\[\begin{aligned}
\begin{cases}
    a+b+c+d=0,\\
    a^2+b^2+c^2+d^2=12,\\
    abcd=-3.\\
\end{cases}
\end{aligned}\]
6 replies
blug
Apr 4, 2025
Tintarn
43 minutes ago
4-variable inequality with square root
a_507_bc   11
N an hour ago by Apple_maths60
Source: 2023 Austrian Federal Competition For Advanced Students, Part 1 p1
Let $a, b, c, d$ be positive reals strictly smaller than $1$, such that $a+b+c+d=2$. Prove that $$\sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{ac+bd}{2}. $$
11 replies
a_507_bc
May 4, 2023
Apple_maths60
an hour ago
No more topics!
Show that the line AX is perpendicular to BC
Amir Hossein   10
N Sep 10, 2022 by HamstPan38825
Source: Middle European Mathematical Olympiad 2011 - Team Compt. T-6
Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$.
10 replies
Amir Hossein
Sep 6, 2011
HamstPan38825
Sep 10, 2022
Show that the line AX is perpendicular to BC
G H J
Source: Middle European Mathematical Olympiad 2011 - Team Compt. T-6
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Amir Hossein
5452 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$.
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sunken rock
4379 posts
#2 • 2 Y
Y by Adventure10, Mango247
Take $AD$ the third altitude, $D\in BC$. CX being tangent to that circle, $BX^2=BA\cdot BC_0\ (\ 1\ )$ ; similarly, $CX^2=CA\cdot CB_0\ (\ 2\ )$, but $BA\cdot BC_0=BC\cdot BD\ (\ 3\ )$ and $CA\cdot CB_0=BC\cdot CD\ (\ 4\ )$ hence $BX^2-CX^2=BC\cdot (BD-CD)= AB^2-AC^2$ or $AX\bot BC$.

Best regards,
sunken rock
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yunxiu
571 posts
#3 • 2 Y
Y by Adventure10, Mango247
$A{B^2} - X{B^2} = A{B^2} - AB \cdot B{C_0} = AB \cdot A{C_0}$$A{C^2} - X{C^2} = A{C^2} - AC \cdot C{B_0} = AC \cdot A{B_0}$Because $AB \cdot A{C_0} = AC \cdot A{B_0}$so$A{B^2} - X{B^2} = A{C^2} - X{C^2}$, hence $AX \bot BC$.
Attachments:
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Lyub4o
265 posts
#4 • 2 Y
Y by Adventure10, Mango247
Dear yunxiu,
I can't make the connection between $ A{B^{2}}-X{B^{2}}= A{C^{2}}-X{C^{2}} $ and $ AX\bot BC $.Is there a practicular formula for that?
Best regards
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dragon96
3212 posts
#5 • 2 Y
Y by Adventure10, Mango247
At above: Suppose you have AB _|_ CD. Let their intersection point be P. Then you can use the Pythagorean Theorem to show that $AB^2 + CD^2 = AD^2 + BC^2$.

If you start with the relation $AB^2 + CD^2 = AD^2 + BC^2$, then let the angle APB be $\theta$. You can use the Law of Cosines to backtrack and show that $\theta$ must be equal to $90^\circ$.
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Lyub4o
265 posts
#6 • 2 Y
Y by Adventure10, Mango247
dragon96 wrote:
$AB^2 + CD^2 = AD^2 + BC^2$.
I think the above is true if and only if triangles APD and CPB are similar,isn't it?
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dragon96
3212 posts
#7 • 4 Y
Y by Lyub4o, Adventure10, Mango247, and 1 other user
No, consider this:

If the two triangles are similar, here's a counter example:
[asy]defaultpen(linewidth(0.7));
pair A=origin, B=(12,0), C=(12,5), D=(0,5), P=(6,2.5);
draw(A--B--C--D--A--C^^B--D);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$",D, NW);
label("$P$", P, N);[/asy]

If the relation holds, here's a counter example:
[asy]defaultpen(linewidth(0.7));
pair P=origin, B=(8,0), A=(0,3), C=(0,-14), D=(-13,0);
draw(A--B--C--D--A--C^^B--D);
label("$A$", A, N);
label("$B$", B, E);
label("$C$", C,S);
label("$D$",D, W);
label("$P$", P, NE, fontsize(10));
markscalefactor=0.1;
draw(rightanglemark(A,P,D));[/asy]

From the second diagram, it should hopefully clear why if they're perpendicular, the relation holds.
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Lyub4o
265 posts
#8 • 2 Y
Y by Adventure10, Mango247
But look:
$ AB^{2}+CD^{2}= AD^{2}+BC^{2} $ <=> $ AP^{2}+BP^{2}+2AP.BP+DP^{2}+CP^{2}+2BP.CP=
AP^{2}+DP^{2}+BP^{2}+CP^{2} $ <=> $ AP.BP=CP.DP $ and with the right angle we get the two similar triangles.Where's the mistake?(if $ D $ and $ C $ are above $ P $ and $ D $ is above $ C $,which is the case in our problem)
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lazizbek42
548 posts
#9
Y by
Amir Hossein wrote:
Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$.
$$BX^2=BC_0 \cdot BA$$$$CX^2=CB_0 \cdot CA$$$$AB^2-BX^2=AC^2-CX^2$$Carnot theorem.
This post has been edited 1 time. Last edited by lazizbek42, Jan 18, 2022, 8:12 AM
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lazizbek42
548 posts
#10
Y by
$AA_0$ altitude.
$X$ is intersection $AA_0$ and $(BCB_0C_0)$.
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HamstPan38825
8857 posts
#11
Y by
By the Perpendicularity Lemma, it suffices for $$XB^2-XC^2 = AB^2-AC^2 \iff BC_0 \cdot BA - CB_0 \cdot CA = AB^2-AC^2.$$But this is clearly true as $AC_0 \cdot AB = AB_0 \cdot AC$.
This post has been edited 1 time. Last edited by HamstPan38825, Sep 10, 2022, 11:28 PM
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