Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$.
Problem
Source: Sharygin First Round 2013, Problem 9
Tags: geometry, geometric transformation, reflection, incenter, circumcircle, Asymptote, analytic geometry
aZpElr68Cb51U51qy9OM
09.04.2013 03:11
I swore to myself that I wouldn't bary this problem... oh well [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.000000000000000, xmax = 16.00000000000000, ymin = -9.000000000000000, ymax = 3.500000000000004; /* image dimensions */ /* draw figures */ draw((15.24000000000002,1.760000000000002)--(2.240000000000003,-5.020000000000005)); draw((2.240000000000003,-5.020000000000005)--(15.46000000000002,-4.560000000000005)); draw((15.46000000000002,-4.560000000000005)--(15.24000000000002,1.760000000000002)); draw(circle((10.00432791817439,-2.809330943754360), 5.729677129551685)); draw((4.548558575806221,-1.058475399742793)--(12.93144142419380,-2.201524600257211)); /* dots and labels */ dot((15.24000000000002,1.760000000000002),dotstyle); label("$A$", (15.32000000000002,1.880000000000002), NE * labelscalefactor); dot((2.240000000000003,-5.020000000000005),dotstyle); label("$B$", (1.700000000000002,-5.620000000000006), NE * labelscalefactor); dot((15.46000000000002,-4.560000000000005),dotstyle); label("$C$", (15.42000000000002,-5.300000000000006), NE * labelscalefactor); dot((12.93144142419380,-2.201524600257211),dotstyle); label("$I$", (13.02000000000001,-2.080000000000002), NE * labelscalefactor); dot((8.740000000000009,-1.630000000000002),dotstyle); label("$M$", (8.280000000000008,-1.360000000000002), NE * labelscalefactor); dot((4.548558575806221,-1.058475399742793),dotstyle); label("$I'$", (3.960000000000004,-0.9400000000000011), NE * labelscalefactor); dot((4.683535124008554,-4.934975328514078),dotstyle); label("$T_1$", (4.760000000000005,-4.820000000000006), NE * labelscalefactor); dot((15.32505972450844,-0.6835339040601216),dotstyle); label("$T_2$", (15.40000000000002,-0.5600000000000007), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The answer is $\boxed{90^\circ}$. We use barycentric coordinates. Let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$, and let $a = BC, b = CA, c = AB$, and let $s$ be the semiperimeter of $\triangle ABC$. Let $I$ and $M$ be the incenter of $\triangle ABC$ and midpoint of $AB$, respectively. Then $M = \left(\frac{a+b+c}{2} : \frac{a+b+c}{2} : 0\right)$ and $I = (a:b:c)$. Let $I'$ be the reflection of $I$ over $M$. Therefore $I' = (b+c: a+c:-c)$. Since $CT_1 = s-b$ and $CT_2 = s-a$, we have that $T_1 = (0:s-b:s-c)$ and $T_2 = (s-a:0:s-c)$. It is well-known that the general equation of a circle is \[a^2yz+b^2zx+c^2xy - (x+y+z)(ux+vy+wz) = 0.\] Substituting the coordinates of $C, T_1, T_2$ yields the equation of the circumcircle of $CT_1T_2$: \[a^2yz+b^2zx+c^2xy - (x+y+z)(b(s-c)x+a(s-c)y) = 0.\] Since $I'$ lies on this circumcircle, substituting the coordinates of $I'$ yields: \begin{align*}a^2(a+c)(-c)+b^2(-c)(b+c)+c^2(b+c)(a+c) - (a+b+c)(b(s-c)(b+c)+a(s-c)(a+c)) &= 0 \\ \Rightarrow \frac{1}{2} (a+b+c)(a+b+2c)(c^2-a^2-b^2) &= 0 \\ \Rightarrow c^2 - a^2-b^2 &= 0 \end{align*} Because $a^2+b^2=c^2$ if and only if $ABC$ is a right triangle with $\angle C = 90^\circ$, we are done.
pi37
09.04.2013 04:24
Note that $I'$ lies on the line through $I_C$ perpendicular to $AB$ (since the midpoint of $AB$ is also the midpoint of the intouch and extouch points). Call this line $\ell$; then it is concurrent with the line through $T_1$ perpendicular to $BC$ and through $T_2$ perpendicular to $AC$ at the Bevan point $V$. Then $CT_1VT_2$ are concyclic on the circle with diameter $CV$. Assume $I'\neq V$; then $I'$ also lies on this circle, so $\angle CI'V=\frac{\pi}{2}$. But the line $I'V$ is just $\ell$, which is perpendicular to $AB$, so $CI'\parallel AB$. But this is absurd, as $I'$ lies on the opposite side of $AB$ as $C$. Then $V=I'$, so $M$, the midpoint of $II'$, is the same point as $O$, the midpoint of $IV$. This implies $\angle ACB=\frac{\pi}{2}$.
Plops
02.08.2019 01:03
This might be silly, but can anyone explain how if pi37 wrote: $I'\neq V$; then $I'$ also lies on this circle.