RMO 2012 Kar Region

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krishnamurthy95

16 posts

krishnamurthy95

#1 h Dec 16, 2012, 3:56 AM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle. Let $BE$ and $CF$ be internal angle bisectors of $\angle B$ and $\angle C$
respectively with $E$ on $AC$ and $F$ on $AB$ . Suppose $X$ is a point on the segment $CF$
such that $AX$ perpendicular $CF$ ; and $Y$ is a point on the segment $BE$ such that $AY$ perpendicular $BE$ . Prove
that $XY = (b + c-a)/2$ where $BC = a, CA = b$ and $AB = c$ .

This post has been edited 2 times. Last edited by krishnamurthy95, Dec 18, 2012, 6:40 AM

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krishnamurthy95

16 posts

krishnamurthy95

#2 h Dec 16, 2012, 4:34 AM • 3 Y

Y by math_aspirant, Adventure10, Mango247

Although a smaller proof has been given here http://olympiads.hbcse.tifr.res.in/. I give another proof
which is bit lengthier.
In $ABC, BE$ and $CF$ intersect at $I$ the incenter. Thus $\angle BIC = 90^{\circ}+A/2$ .
Also $AI$ is the diameter of circle passing though $A,X,I,Y$ . By sine formula in triangle $XIY$
we get $XY= AI cos(A/2)$ .
Now in $AIC$ by sine formula $AI= b \frac{sin(C/2)} {cos(B/2)}$
By sine formula in $ABC$ itself gives $b+c-a = b+ b\frac{sin(C)}{sin(B)}-b\frac {sin(A)}{sin(B)}$
Just after few elementary trigonometric manipulation you end up with
$(b+c-a)/2 = AI cos(A/2)$ which is same as $XY$ and we are done.

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underzero

318 posts

underzero

#3 h Dec 19, 2012, 7:09 PM • 3 Y

Y by math_aspirant, Adventure10, Mango247

We know $BIC=90+A/2$ and we have $XY/sin(90+A/2)=AI$ (I is the incenter of triangle ABC)
$T$ is a point on $AB$ such that $IT$ is perpendicular to $AB$ so we have $AT=b+c-a/2=cosA/2.AI$
we know $sin(90+A/2)=cosA/2$ thus $XY=AT=b+c-a/2$
QED

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Geoclid

11 posts

Geoclid

#4 h Jul 25, 2021, 4:48 AM

Y by

My solution without trigonometry HERE, as well as many other geometry materials.

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HoRI_DA_GRe8

590 posts

HoRI_DA_GRe8

#5 h Aug 12, 2021, 6:23 AM

Y by

krishnamurthy95 wrote:

Let $ABC$ be a triangle. Let $BE$ and $CF$ be internal angle bisectors of $\angle B$ and $\angle C$
respectively with $E$ on $AC$ and $F$ on $AB$ . Suppose $X$ is a point on the segment $CF$
such that $AX$ perpendicular $CF$ ; and $Y$ is a point on the segment $BE$ such that $AY$ perpendicular $BE$ . Prove
that $XY = (b + c-a)/2$ where $BC = a, CA = b$ and $AB = c$ .

I used ptolemy and cosine rule again and again

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trying_to_solve_br

191 posts

trying_to_solve_br

#6 h Aug 12, 2021, 11:40 AM

Y by

Mehhhh solution without trig:

First notice that if $AX,AY \cap BC=J,K$ , then $X$ is the midpoint of $AJ$ , because it lies on the bissector and $CX\perp AJ$ . Similarly for $Y$ , we just need to prove $JK=2XY$ , but notice $BK=BA$ and $CJ=CA$ : then $JK=BC-KC-JB=a-(a-b)-(a-c)=b+c-a$ , which is what we wanted.

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INMO2021

2 posts

INMO2021

#8 h Jan 30, 2022, 7:57 PM

Y by

Let $I$ be the incenter of $\triangle ABC$ . Note that $\angle BIC =90+\angle \frac{A}{2}$
Since AXIY are concyclic, $\angle XAY = 90- \frac{A}{2}$
The circle passing through AXIY has $AI=rcosec(A/2)$ as diameter
Applying sine rule in $\triangle XAY$ , $\frac {XY}{sin(\angle XAY)} = 2R = AI$
$XY=rcosec(\frac{A}{2})cos(\frac{A}{2})$ $\Rightarrow XY=rcot(\frac{A}{2})$ $\Rightarrow \fbox {XY = s-a}$

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bal09

21 posts

bal09

#9 h May 31, 2024, 7:13 AM

Y by

here is my synthetic solution

Attachments:

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