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Nice geometry problem
henderson   5
N Jun 15, 2015 by sunken rock
Source: JBMO Shortlist 2013 G4
Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $PE$ and $CQ$ are perpendicular.
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henderson
Jun 3, 2015
sunken rock
Jun 15, 2015
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Nice geometry problem
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Source: JBMO Shortlist 2013 G4
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henderson
312 posts
henderson
#1 h  Jun 3, 2015, 6:57 PM • 2 Y
Y by Adventure10, Mango247
Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $PE$ and $CQ$ are perpendicular.
This post has been edited 1 time. Last edited by henderson, Jun 3, 2015, 7:12 PM
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colinhy
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colinhy
#2 h  Jun 3, 2015, 7:35 PM • 2 Y
Y by Adventure10, Mango247
Solution:
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henderson
312 posts
henderson
#3 h  Jun 8, 2015, 6:51 PM • 2 Y
Y by Adventure10, Mango247
Can you explain why $BX = DC$?
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colinhy
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colinhy
#4 h  Jun 8, 2015, 7:20 PM • 2 Y
Y by Adventure10, Mango247
I think it is common knowledge that $BX = DC$ (you can check out Yufei's handouts for a proof, but it's just length chasing using the fact that tangents from a point are of equal length). Thus, if $M$ is the midpoint of $BC$, then $XM = MD$. Then, since $EC = 2 \cdot DC$ and $BC = 2 \cdot MC$, $BE = 2 \cdot MD = XD$.
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aleksam
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aleksam
#6 h  Jun 13, 2015, 4:15 PM • 2 Y
Y by Adventure10, Mango247
Let $X$ and $Y$ be the intersection of $CA$ and $CB$, respectivly, with the circle $k(I, CI)$ and $A'$, $B'$ and $C'$ are the projection of $I$ to sides $BC$, $CA$ and $AB$ respectivly. From the fact that $B'$ is the midpoint of $CX$, we have that $AX=XB'-B'A=(s-AB)-(s-BC)=BC-AB$, where s is the semiperimeter of the triangle $ABC$. Now, it is easy to see that triangles $IC'Q$ and $IA'C$ are congruent($IC'=IA', IQ=IC, <IC'Q=<IA'C$) and so $C'Q=A'C$, and we easily deduce $AQ=BC-AB=AX$. So, after short angle chasnig we find $<PCY=(<ABC-<ACB)/2$ and similarly $<XCQ=(<BAC-<ACB)/2$. Now, as in colinhy's solution we can get $<PEB=<ABC/2$, and as $<QCB=90-<ABC/2$ the result follows.
This post has been edited 1 time. Last edited by aleksam, Jun 13, 2015, 4:16 PM
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sunken rock
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sunken rock
#7 h  Jun 15, 2015, 1:12 PM • 2 Y
Y by Adventure10, Mango247
From second post, after having seen $BE=BP$, i.e. $\angle BPE=\frac{\hat B}2\ (\ 1\ )$ we can continue with $BQ=BC$, so $\angle BQC=\frac{\hat A+\hat C}2\ (\ 2\ )$, and adding $(1)$ and $(2)$ we get $PE\bot CQ$.

Best regards,
sunken rock
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