## What is the distance?

- Ananya Promi
**Posts:**36**Joined:**Sun Jan 10, 2016 4:07 pm**Location:**Naogaon, Bangladesh

### 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**Location:**Naogaon, Bangladesh

### 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, $XI^2=325$

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**Location:**Chittagong, Bangladesh

### Re: What is the distance?

I think you've made a typo here. We've found earlier that $CS=15$, this should be $XS=10$.Ananya Promi wrote: ↑Fri Mar 30, 2018 10:53 pmI'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 $CS=10$

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**Location:**Naogaon, Bangladesh

### Re: What is the distance?

Oh! Yup!!!Atonu Roy Chowdhury wrote: ↑Tue Apr 03, 2018 12:26 pmI think you've made a typo here. We've found earlier that $CS=15$, this should be $XS=10$.Ananya Promi wrote: ↑Fri Mar 30, 2018 10:53 pmI'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 $CS=10$

Also, there's another way to find the value of $PX$ using the area formula $\frac{abc}{4R}$.