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BdMO Online Forum • View topic - collinearity from russia

collinearity from russia

For discussing Olympiad level Geometry Problems

collinearity from russia

Let triangle $ABC$ has intersection of $A$-tangent with side $BC$ to be V. Suppose X,W are points in this plane so that both of $BX,CW$ are perpendicular to $BC$ and $BX = AX, CW = AW$. Show $X,W,V$ collide.

Power of point and homothety.

Fm Jakaria

Posts: 77
Joined: Thu Feb 28, 2013 11:49 pm

Re: collinearity from russia

In my solution, I have replaced $W$ with $Y$, and $V$ with $Z$. So we need to prove that $Y,X,Z$ are collinear.

Let $BX\cap ZY=P$. Now it is clear that $\dfrac{BZ}{ZC}=\dfrac{BA\cdot \sin\angle BAZ}{CA\cdot \sin\angle CAZ}=\dfrac{BA\cdot \sin\angle C}{CA\cdot \sin\angle B}$. Thus by using sine law in $\triangle ABC$, $\dfrac{BZ}{ZC}=\dfrac{BA^{2}}{CA^{2}}=\dfrac{c^{2}}{b^{2}}$. Again since $\triangle ZBP\sim \triangle ZCY$, we have $\dfrac{BP}{CY}=\dfrac{c^{2}}{b^{2}}$. Let $O$ be the circumcenter of $\triangle ABC$. Then, a simple angle chasing shows that $\triangle ABC\sim \triangle COY$. So, $\dfrac{CO}{CY}=\dfrac{AB}{AC}=\dfrac{c}{b}$. Now, $\dfrac{BP}{CY}=\dfrac{BP}{CO}\cdot \dfrac{CO}{CY}=\dfrac{BP}{BO}\cdot \dfrac{c}{b}$. But $\dfrac{BP}{CY}=\dfrac{c^{2}}{b^{2}}$. So, $\dfrac{BP}{BO}=\dfrac{c}{b}=\dfrac{AB}{AC}$. And, a simple angle chasing shows that $\angle PBO=\angle BAC$. So $\triangle PBO\sim \triangle BAC$. Thus $\angle BOP=\angle C$. But $\angle BOA=2\angle C$. So $OP$ is the perpendicular bisector of $AB$. Thus $PA=PB$ and thus $P\equiv X$, which proves our claim.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

SANZEED

Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

Re: collinearity from russia

My solution:

Let $S,T$ be circles through $(A,B),(A,C)$, respectively; with centers $X,W$. Then $BC$ is their common external tangent. Also $I$, the intersection of $BC$ and $XW$; is their external center of similitude. So letting the second point of intersections of $IA$ with $S,T$ be $R,U$; $RB$ is parallel to $AC$. Then
$\angle URB = \angle UAC = \angle UCK$, for $K$ any point in the ray through $C$ opposite to ray $CB$. So
$BCUR$ is cyclic. Then $IU.IR =IC.IB$....(1)

Now consider power of point $I$ wrt $S,T$. Then
$IA.IR = IB^2$ and $IA.IU =IC^2$. Multiplying and using (1), we get $IB.IC = IA^2$. So $V$ coincides with $I$.

Fm Jakaria

Posts: 77
Joined: Thu Feb 28, 2013 11:49 pm