## What is the distance?

For discussing Olympiad level Geometry Problems
Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### What is the distance?

Let $ABC$ be a triangle with $AB=26, AC=28, BC=30$. Let $X, Y, Z$ be the midpoints of arcs $BC, CA, AB$ (not containg the opposite vertics) respectively on the circumcircle of triangle $ABC$. Let $P$ be the midpoint of arc $BC$ containing point A. Suppose, lines $BP$ and $XZ$ meet at $M$, while lines $CP$ and $XY$ meet at $N$. Find the square of the distance from $X$ to $MN$.

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm

### Re: What is the distance?

We use Pascal's theorem on $YXZCPB$ and get $M, N, I$ collinear where $I$ is the intersection point of $ZC$ and $BY$.
In fact, $I$ is the incenter of triangle $ABC$.
So, $XC=XI$
Again, $$\angle{CXY}=\angle{AXY}=\angle{CXN}=\angle{IXN}$$
And $$\angle{PCX}=\angle{NCX}=90^o$$
Ultimately we get $CNIX$ cyclic.
So, $XI$ is perpendicular on $NM$.
We need to measure $XI^2=CX^2$.
Let $CB$ intersects with $PX$ at $S$.
We get $CS=SB=15$
I'm avoiding the calculations how I got that $PX=2R=65/2$ (By Heron's formula we got the area of the triangle $ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.)
Using the fact that, $CS/PS=XS/CS$ we get $CS=10$
So, $CX^2=SX^2+CS^2=10^2+15^2=325$
So, $XI^2=325$

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: What is the distance?

Ananya Promi wrote:
Fri Mar 30, 2018 10:53 pm
I'm avoiding the calculations how I got that $PX=2R=65/2$ (By Heron's formula we got the area of the triangle $ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.)
Using the fact that, $CS/PS=XS/CS$ we get $CS=10$
I think you've made a typo here. We've found earlier that $CS=15$, this should be $XS=10$.
Also, there's another way to find the value of $PX$ using the area formula $\frac{abc}{4R}$.
This was freedom. Losing all hope was freedom.

Ananya Promi
Posts: 36
Joined: Sun Jan 10, 2016 4:07 pm
I'm avoiding the calculations how I got that $PX=2R=65/2$ (By Heron's formula we got the area of the triangle $ABC$, then found out the height from on vertix to oppsite side, and then applied Brahmagupta's theorem.)
Using the fact that, $CS/PS=XS/CS$ we get $CS=10$
I think you've made a typo here. We've found earlier that $CS=15$, this should be $XS=10$.
Also, there's another way to find the value of $PX$ using the area formula $\frac{abc}{4R}$.