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Difference between revisions of "2024 AMC 8 Problems/Problem 15"

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==Problem==
 
==Problem==
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Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</math>. The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>, the point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent. (source: 2021 IMO)
 
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now go do this problem as a punishment for trying to cheat
 
  
 
==Solution==
 
==Solution==

Revision as of 13:02, 23 January 2024

Problem

Solution

2021 IMO 3j.png
2021 IMO 3i.png

We prove that circles $ACD, EXD$ and $\Omega_0$ centered at $P$ (the intersection point $BC$ and $EF)$ have a common chord. Let $P$ be the intersection point of the tangent to the circle $\omega_2 = BDC$ at the point $D$ and the line $BC, A'$ is inverse to $A$ with respect to the circle $\Omega_0$ centered at $P$ with radius $PD.$ Then the pairs of points $F$ and $E, B$ and $C$ are inverse with respect to $\Omega_0$, so the points $F, E,$ and $P$ are collinear. Quadrilaterals containing the pairs of inverse points $B$ and $C, E$ and $F, A$ and $A'$ are inscribed, $FE$ is antiparallel to $BC$ with respect to angle $A$ (see $\boldsymbol{Claim}$). Consider the circles $\omega = ACD$ centered at $O_1, \omega' = A'BD,$ $\omega_1 = ABC, \Omega = EXD$ centered at $O_2 , \Omega_1 = A'BX,$ and $\Omega_0.$ Denote $\angle ACB = \gamma$. Then $\angle BXC = \angle BXE = \pi – 2\gamma,$ $\angle AA'B = \gamma (AA'CB$ is cyclic), $\angle AA'E = \pi – \angle AFE = \pi – \gamma (AA'EF$ is cyclic, $FE$ is antiparallel), $\angle BA'E = \angle AA'E – \angle AA'B = \pi – 2\gamma = \angle BXE \implies$ $\hspace{13mm}E$ is the point of the circle $\Omega_1.$ Let the point $Y$ be the radical center of the circles $\omega, \omega', \omega_1.$ It has the same power $\nu$ with respect to these circles. The common chords of the pairs of circles $A'B, AC, DT,$ where $T = \omega \cap \omega',$ intersect at this point. $Y$ has power $\nu$ with respect to $\Omega_1$ since $A'B$ is the radical axis of $\omega', \omega_1, \Omega_1.$ $Y$ has power $\nu$ with respect to $\Omega$ since $XE$ containing $Y$ is the radical axis of $\Omega$ and $\Omega_1.$ Hence $Y$ has power $\nu$ with respect to $\omega, \omega', \Omega.$ Let $T'$ be the point of intersection $\omega \cap \Omega.$ Since the circles $\omega$ and $\omega'$ are inverse with respect to $\Omega_0,$ then $T$ lies on $\Omega_0,$ and $P$ lies on the perpendicular bisector of $DT.$ The power of a point $Y$ with respect to the circles $\omega, \omega',$ and $\Omega$ are the same, $DY \cdot YT = DY \cdot YT' \implies$ the points $T$ and $T'$ coincide. The centers of the circles $\omega$ and $\Omega$ ($O_1$ and $O_2$) are located on the perpendicular bisector $DT'$, the point $P$ is located on the perpendicular bisector $DT$ and, therefore, the points $P, O_1,$ and $O_2$ lie on a line, that is, the lines $BC, EF,$ and $O_1 O_2$ are concurrent.

2021 IMO 3.png
2021 IMO 3j.png

$\boldsymbol{Claim}$ Let $AK$ be bisector of the triangle $ABC$, point $D$ lies on $AK.$ The point $E$ on the segment $AC$ satisfies $\angle ADE= \angle BCD$. The point $F$ on the segment $AB$ satisfies $\angle ADF= \angle CBD.$ Let $P$ be the intersection point of the tangent to the circle $BDC$ at the point $D$ and the line $BC.$ Let the circle $\Omega_0$ be centered at $P$ and has the radius $PD.$ Then the pairs of points $F$ and $E, B$ and $C$ are inverse with respect to $\Omega_0$ and $EF$ and $BC$ are antiparallel with respect to the sides of an angle $A.$ $\boldsymbol{Proof}$ Let the point $E'$ is symmetric to $E$ with respect to bisector $AK, E'L || BC.$ Symmetry of points $E$ and $E'$ implies $\angle AEL = \angle AE'L.$ \[\angle DCK = \angle E'DL, \angle DKC = \angle E'LD \implies\] \[\triangle DCK \sim \triangle E'DL \implies \frac {E'L}{KD}= \frac {DL}{KC}.\] \[\triangle ALE' \sim \triangle AKB \implies \frac {E'L}{BK}= \frac {AL}{AK}\implies\] \[\frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.\] Similarly, we prove that $FL$ and $BC$ are antiparallel with respect to angle $A,$ and the points $L$ in triangles $\triangle EDL$ and $\triangle FDL$ coincide. Hence, $FE$ and $BC$ are antiparallel and $BCEF$ is cyclic. Note that $\angle DFE = \angle DLE – \angle FDL = \angle AKC – \angle CBD$ and $\angle PDE = 180^o – \angle CDK – \angle CDP – \angle LDE = 180^o – (180^o – \angle AKC – \angle BCD) – \angle CBD – \angle BCD$ $\angle PDE = \angle AKC – \angle CBD = \angle DFE,$ so $PD$ is tangent to the circle $DEF.$ $PD^2 = PC \cdot PB = PE \cdot PF,$ that is, the points $B$ and $C, E$ and $F$ are inverse with respect to the circle $\Omega_0.$ vladimir.shelomovskii@gmail.com, vvsss


(This was on the page of the 2021 IMO problem 3) Yes this person actually solved it. I just copy pasted for people that want to know. -Multpi12)

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