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Difference between revisions of "2020 CAMO Problems/Problem 3"

(Created page with "==Problem 3== Let <math>ABC</math> be a triangle with incircle <math>\omega</math>, and let <math>\omega</math> touch <math>\overline{BC}</math>, <math>\overline{CA}</math>, <...")
 
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==Solution==
 
==Solution==
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{{solution}}
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Let <math>R</math> be the intersection of <math>AT</math> with <math>\omega</math> other than <math>T</math>.
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Claim 1: <math>R</math> is the <math>D</math>-queue point of <math>DXY</math>. In particular, <math>R</math> lies on <math>(DXY)</math>.
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Proof: First, we claim that <math>X</math> and <math>Y</math> lie on <math>TE</math> and <math>TF</math> respectively. Let <math>X_0=TE\cap DF</math> and <math>Y_0=TF\cap DE</math>.
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By Brokard, <math>X_0Y_0</math> is the polar of <math>EF\cap DT</math> which is the line through <math>A</math> parallel to <math>BC</math>, so it coincides with the line <math>XY</math> which gives <math>X\equiv X_0</math> and <math>Y\equiv Y_0</math>.
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The fact that <math>\measuredangle TED = \measuredangle TFD = 90^{\circ}</math> is enough to imply that <math>T</math> is the orthocenter.
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Now,
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\begin{align*}
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(X, Y; A, \infty_{XY}) &\overset{T}{=} (E, F; R, T) \\
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&= -1
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\end{align*}
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so <math>A</math> is the midpoint of <math>XY</math>, which gives the claim.
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Denote <math>f_1</math> as the inversion centered at <math>T</math> swapping <math>(DXY)</math> with its nine-point circle <math>(AEF)</math>, and denote <math>f_2</math> as the inversion centered at <math>A</math> fixing <math>\omega</math>.
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Claim 2: <math>R</math> lies on <math>(APQ)</math>.
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Proof: Note that <math>f_1(Q)=P</math> which follows from the fact that <math>f_1(\omega)=XY</math>, so we are done with this claim by Power of a Point Theorem.
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Finally, we return to prove that <math>R</math> is the tangency point of <math>(DXY)</math> and <math>(APQ)</math>. Inverting at <math>f_1</math> and <math>f_2</math> in that order, it suffices to show that
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<cmath>
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f_2(f_1(DXY))=EF
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</cmath>
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and
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<cmath>
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f_2(f_1(DXY))=H'T
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</cmath>
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are parallel, where <math>H'</math> is defined to be the intersection of <math>AQ</math> and <math>\omega</math> other than <math>Q</math>.
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But note that <math>(Q, H'; E, F)=-1</math> and <math>Q</math> is the <math>D</math>-queue point of <math>DEF</math>, so it is well known that <math>H'</math> is the point on <math>\omega</math> such that <math>DH'\perp EF</math>. The parallelism then follows from here.
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~ alexanderchew
  
 
==See also==
 
==See also==

Latest revision as of 05:48, 29 September 2025

Problem 3

Let $ABC$ be a triangle with incircle $\omega$, and let $\omega$ touch $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Point $M$ is the midpoint of $\overline{EF}$, and $T$ is the point on $\omega$ such that $\overline{DT}$ is a diameter. Line $MT$ meets the line through $A$ parallel to $\overline{BC}$ at $P$ and $\omega$ again at $Q$. Lines $DF$ and $DE$ intersect line $AP$ at $X$ and $Y$ respectively. Prove that the circumcircles of $\triangle APQ$ and $\triangle DXY$ are tangent.

Solution

Let $R$ be the intersection of $AT$ with $\omega$ other than $T$.

Claim 1: $R$ is the $D$-queue point of $DXY$. In particular, $R$ lies on $(DXY)$. Proof: First, we claim that $X$ and $Y$ lie on $TE$ and $TF$ respectively. Let $X_0=TE\cap DF$ and $Y_0=TF\cap DE$. By Brokard, $X_0Y_0$ is the polar of $EF\cap DT$ which is the line through $A$ parallel to $BC$, so it coincides with the line $XY$ which gives $X\equiv X_0$ and $Y\equiv Y_0$. The fact that $\measuredangle TED = \measuredangle TFD = 90^{\circ}$ is enough to imply that $T$ is the orthocenter. Now, \begin{align*} (X, Y; A, \infty_{XY}) &\overset{T}{=} (E, F; R, T) \\ &= -1 \end{align*} so $A$ is the midpoint of $XY$, which gives the claim.

Denote $f_1$ as the inversion centered at $T$ swapping $(DXY)$ with its nine-point circle $(AEF)$, and denote $f_2$ as the inversion centered at $A$ fixing $\omega$.

Claim 2: $R$ lies on $(APQ)$. Proof: Note that $f_1(Q)=P$ which follows from the fact that $f_1(\omega)=XY$, so we are done with this claim by Power of a Point Theorem.

Finally, we return to prove that $R$ is the tangency point of $(DXY)$ and $(APQ)$. Inverting at $f_1$ and $f_2$ in that order, it suffices to show that \[f_2(f_1(DXY))=EF\] and \[f_2(f_1(DXY))=H'T\] are parallel, where $H'$ is defined to be the intersection of $AQ$ and $\omega$ other than $Q$.

But note that $(Q, H'; E, F)=-1$ and $Q$ is the $D$-queue point of $DEF$, so it is well known that $H'$ is the point on $\omega$ such that $DH'\perp EF$. The parallelism then follows from here. ~ alexanderchew

See also

2020 CAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All CAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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