Point $ I_1$ is the reflection of incentre $ I$ of triangle $ ABC$ across the side $ BC$. The circumcircle of $ BCI_1$ intersects the line $ II_1$ again at point $ P$. It is known that $ P$ lies outside the incircle of the triangle $ ABC$. Two tangents drawn from $ P$ to the latter circle touch it at points $ X$ and $ Y$. Prove that the line $ XY$ contains a medial line of the triangle $ ABC$. Author: L. Emelyanov
Problem
Source: Tuymaada 2008, Senior League, First Day, Problem 3.
Tags: geometry, geometric transformation, reflection, incenter, circumcircle, geometry proposed
pohoatza
17.07.2008 15:34
Let $ D$, $ E$, $ F$ be the tangency points of the incircle $ \rho(I)$ with the sides $ BC$, $ CA$ and $ AB$, respectively and let $ M$ be the midpoint of small arc $ BC$, not containing the vertex $ A$. Since $ MB = MI = MC$, the reflection $ M_{1}$ of $ M$ into the sideline $ BC$ is the circumcenter of the $ BI_{1}CP$. See that $ PD$ is the $ P$-altitude of triangle $ BPC$ and the reflection $ I_{1}$ of $ I$ into $ D$ lies on the circumcircle of $ BPC$. Thus, $ I$ is the orthocenter of $ BPC$ and since $ XY$ is the polar of $ P$ wrt. $ \rho$ and $ DE$ is the polar of $ B$, it follows that the lines $ XY$, $ DF$ and $ CI$ are concurrent. On the other hand, since $ XY$ is the polar of $ P$, we have that $ PI \perp XY$ and so $ XY \| BC$. In this case, the conclusion follows from http://www.mathlinks.ro/viewtopic.php?t=132338.
Erken
31.07.2008 19:34
It is pretty easy problem and in my opinion last year's third problem was much harder than this one. Here is a sketch of a solution: Suppose that $ AB=c,AC=b,BC=a$,$ p=\frac{a+b+c}{2}$ and $ r$ is incircle's radius. Since $ PBIC$ is cyclic and $ BD=p-b,DC=p-c$,it follows that \[ \boxed{PD=\frac{(p-b)(p-c)}{r}} \] Suppose that $ PD$ intersects $ XY$ at $ Z$. Then $ PZ\cdot PI=PX^2=PI^2-r^2$ and $ PI=PD-r=\frac{(p-b)(p-c)}{r}-r$. \[ \boxed{PZ=\frac{\frac{(p-b)(p-c)}{r}\left(\frac{(p-b)(p-c)}{r}-2r\right)}{\frac{(p-b)(p-c)}{r}-r}} \] So finally $ ZD=PD-PZ$ and it is enough to show that $ ZD=\frac{h_a}{2}$,but this is not really hard...
discredit
16.08.2008 17:46
pohoatza wrote: it follows that the lines $ XY$, $ DF$ and $ CI$ are concurrent. Why is this?
semy
28.09.2011 16:22
discredit wrote: pohoatza wrote: it follows that the lines $ XY$, $ DF$ and $ CI$ are concurrent. Why is this? By La Hire's theorem, the point $XY \cap DF = J$ is the pole of $PB$, so $IJ \perp PB$.
anantmudgal09
26.09.2015 02:22
Let $r$ be the inradius and $r_a$ be the $A$-exradius and $J$ be the $A$-excenter. Obviously, $BPCJ$ is a parallelogram, so $IP=DP-DI=r_a-r$. Let $P'$ be the inverse of $P$ in the incircle. We have $$IP'=\frac{r^2}{IP}=\frac{r^2}{r_a-r} \Longrightarrow DP'=\frac{rr_a}{r_a-r}=\frac{\frac{\Delta^2}{s(s-a)}}{\frac{a\cdot \Delta}{s(s-a)}}=\frac{\Delta}{a}=\frac{1}{2}\cdot h_a.$$Clearly, the polar of $P$ wrt the incircle is the line through $P'$ parallel to $BC$ which is the $A$-midline.