Lines from vertices to some point are perpendicular

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Lines from vertices to some point are perpendicular

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Source: 2012 Indonesia Round 2.5 TST 1 Problem 3

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#1 h May 10, 2012, 1:06 PM • 2 Y

Y by Adventure10, Mango247

The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$ . Prove that $BP$ and $CP$ are perpendicular.

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hadikh

#2 h May 10, 2012, 1:19 PM • 2 Y

Y by Adventure10, Mango247

chaotic_iak wrote:

The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$ . Prove that $BP$ and $CP$ are perpendicular.

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#3 h May 10, 2012, 2:09 PM • 1 Y

Y by Adventure10

There's a reason why this is in Proposed & Own Problems (or otherwise known as "solved by poster, but looking for alternate solutions").

I didn't solve this in the test because I tried on the Algebra one, and when I see this, I immediately think of analytic geometry (hey, one circle, one bisector; use the bisector as an axis and we have a good fact that the gradients of the two sides of the bisected angle are opposites!)...which leads to something messy. So I gave up. Without thinking of simple angle chasing. :headbash:

(Also, I get the one-liner solution probably precisely as you want to say from a friend.)

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#4 h Jul 25, 2023, 12:52 AM

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chaotic_iak wrote:

The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$ . Prove that $BP$ and $CP$ are perpendicular.

$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{Solution 1:}}$
$\color{green}\rule{24cm}{0.3pt}$
Let $I$ be the incenter of $\triangle ABC$
Let $\angle MBI=\angle IBC=\alpha, \angle NCI=\angle ICB=\beta$
$\Rightarrow \angle ABC=2\alpha, \angle ACP=2\beta$ $\Rightarrow \angle BAC=180-2\alpha -2\beta$ $\Rightarrow \angle NAI=90-\alpha -\beta...(I)$ Since $\angle IMA=\angle INA=90$
$\Rightarrow ANIM \text{ is cyclic}$ $\Rightarrow \angle NMI=\angle NAI$ By $(I):$
$\Rightarrow \angle NMI=90-\alpha-\beta$ In $\triangle BMP:$
$\alpha +90+90-\alpha -\beta +\angle MPB=180$ $\Rightarrow \angle MPB=\beta=\angle NCI$ $\Rightarrow PNIC \text{ is cyclic}$ $\Rightarrow \angle IPC=\angle INC=90$ $\Rightarrow \boxed{BP\perp CP}_\blacksquare$ $\color{green}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{Solution 2:}}$
$\color{green}\rule{24cm}{0.3pt}$
Let $T$ be the point of tangency of the incircle to the side $BC$
Working in complex numbers, where $\odot (MNP)$ is the unit circle, let $m=m$ , $n=n$ , $t=1$ and $i=0$
$A$ is the intersection of the tangents to the incircle that pass through $M$ and $N:$
$\Rightarrow a=\frac{2mn}{m+n}$ Analogously we have to:
$b=\frac{2m}{m+1}$ $c=\frac{2n}{n+1}$ $P, B \text{ and }O$ are collinears:
$\Rightarrow \frac{p}{\overline{p}}=\frac{b}{\overline{b}}$ $\Rightarrow \frac{p}{\overline{p}}=\frac{\frac{2m}{m+1}}{\frac{2}{m+1}}=m$ $\Rightarrow \overline{p}=\frac{p}{m}...(\blacktriangle)$ $M, N \text{ and }P$ are collinears:
$\Rightarrow \frac{p-m}{\overline{p}-\overline{m}}=\frac{m-n}{\overline{m}-\overline{n}}=-mn$ $\Rightarrow \overline{p}=\frac{m+n-p}{mn}...(\blacksquare)$ By $(\blacktriangle)$ and $(\blacksquare):$
$\Rightarrow \frac{p}{m}=\frac{m+n-p}{mn}$ $\Rightarrow p=\frac{m+n}{n+1}...(\star)$ $BP\perp CP:$
$\Leftrightarrow \frac{b-p}{\overline{b}-\overline{p}}=-\frac{c-p}{\overline{c}-\overline{p}}$ By $(\star):$
$\Leftrightarrow \frac{\frac{2m}{m+1}-\frac{m+n}{n+1}}{\frac{2}{m+1}-\frac{m+n}{m(n+1)}}=-\frac{\frac{2n}{n+1}-\frac{m+n}{n+1}}{\frac{2}{n+1}-\frac{m+n}{m(n+1)}}$ $\Leftrightarrow \frac{2m(n+1)-(m+n)(m+1)}{\frac{2m(n+1)-(m+n)(m+1)}{m}}=-\frac{n-m}{\frac{m-n}{m}}$ $\Leftrightarrow m=m\checkmark$ $\Rightarrow \boxed{BP\perp CP}_\blacksquare$ $\color{green}\rule{24cm}{0.3pt}$
$\color{blue}\rule{24cm}{0.3pt}$

This post has been edited 1 time. Last edited by hectorleo123, Jul 25, 2023, 1:10 AM

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ismayilzadei1387

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ismayilzadei1387

#5 h Sep 10, 2023, 7:34 PM • 1 Y

Y by FriIzi

It's just Iran Lemma

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