In the triangle $ABC$ points $M$ and $N$ are the midpoints of sides $AC$ and $AB$ respectively. $I$ is the incenter of the triangle. It is known that the angle $MIC$ is a right angle. Find the angle $NIB$.
Problem
Source: Belarusian National Olympiad 2023
Tags: geometry
EaZ_Shadow
01.01.2025 02:18
Is the answer $45^{\circ}$?
nAalniaOMliO
01.01.2025 13:35
No, it isn't
CrazyInMath
01.01.2025 22:33
We know that $\angle BIC=90^{\circ}+\frac12\angle C$, so $\angle MIA=\frac12\angle C$
and then $\angle AMI=90^{\circ}+\frac12\angle B$, so $\angle CMI=90^{\circ}-\frac12\angle B$.
Therefore, $\angle B=\angle C$ and $AB=AC$. So $\angle NIB=\angle MIC=90^\circ$.
edit: wrong and I'm not gonna correct it.
Double07
01.01.2025 22:53
$\widehat{AIC}=90^\circ+\dfrac{B}{2}\implies \widehat{AIM}=\dfrac{B}{2}$, so triangles $\Delta AMI$ and $\Delta AIB$ are simmilar, so take $P$ the midpoint of $AI$ and we get $\widehat{NIB}=\widehat{PMI}$.
But $PM\parallel IC$, so $\widehat{IPM}=\dfrac{A+C}{2}$, and since $\widehat{PIM}=\dfrac{B}{2}\implies\widehat{NIB}=\widehat{PMI}=90^\circ$.
nAalniaOMliO
01.01.2025 23:29
CrazyInMath wrote:
We know that $\angle BIC=90^{\circ}+\frac12\angle C$, so $\angle MIA=\frac12\angle C$
and then $\angle AMI=90^{\circ}+\frac12\angle B$, so $\angle CMI=90^{\circ}-\frac12\angle B$.
Therefore, $\angle B=\angle C$ and $AB=AC$. So $\angle NIB=\angle MIC=90^\circ$.
You mixed up all letters everywhere which led to the wrong conclusion.
RagvaloD
02.01.2025 00:20
Let $E$ is midpoint $MC$ and $F$ is point of intersection of $EI$ and $AB$ $ME=EC=IE \to \angle MEI = \angle ACB \to EF \parallel BC$ So $FB=FN$ and $\angle FIB = \angle FBI \to FB=FN=FI$ and so $\angle NIB=90^o$