Let $\vartriangle ABC$ be a triangle such that $\angle ABC = 30^o$, $\angle ACB = 15^o$. Let $M$ be midpoint of segment $BC$ and point $N$ lies on segment $MC$, such that the length of $NC$ is equal to length of $AB$. Proce that $AN$ is the bisector of the angle $\angle MAC$.
Problem
Source: 2023 Chile NMO L2 P6
Tags: geometry, angle bisector
joeym2011
19.05.2024 23:33
WLOG by law of sines, let $AC=\frac12$, $BC=\frac{\sqrt2}2$, and $AB=\frac{\sqrt6-\sqrt2}4$. This gives $CN=\frac{\sqrt6-\sqrt2}4$ and $MN=\frac{2\sqrt2-\sqrt6}4$. Then, by stewart's theorem,
$$\frac{\sqrt2}2\left(\left(\frac{\sqrt2}4\right)^2+AM^2\right)=\frac{\sqrt2}4\left(\left(\frac{\sqrt6-\sqrt2}4\right)^2+\left(\frac12\right)^2\right),$$$$\frac{\sqrt2}2\left(\frac18+AM^2\right)=\frac{\sqrt2}4\left(\frac{3-\sqrt3}2\right),$$$$AM=\frac{\sqrt3-1}4.$$By the angle bisector theorem, $AN$ is an angle bisector because
$$\frac{MN}{AM}=\frac{CN}{AC}=\frac{\sqrt6-\sqrt2}2.$$
sunken rock
20.05.2024 19:52
Let $O$ be the circumcenter of $\triangle ABC\implies\triangle AOC$ equilateral, $\triangle BOC$ right-angled and isosceles, so $OM=CM$ ( 1 ), that is, $AM$ is bisector of $\widehat{OAC}=60^\circ\implies AN$ bisector of $\widehat{MAC}=30^\circ$. Best regards, sunken rock
Krishijivi
15.09.2024 08:36
Let angle MAC =2x So, angle BAM=135°-2x sin(135°-2x)/BM=sin30°/AM BM/AM= √2{sin2x+cos2x}..(1) BM=MC MC/AM=sin2x/sin15° ..(2) So, equating the 2 equations (√3-1)(√3 sin2x-cos2x)=0 So, tan 2x=1/√3 2x=30° Let angle MAN=y, so angle NAC=30°-y sin(30°-y)/NC=sin(45°-y)/ AC NC/AC=sin(30°-y)/sin(45°-y) AB/AC=sin15°/sin30°= 1/2cos 15° NC=AB So, 2sin(30°-y) cos15°= sin(45°-y) sin(15°-y)=0=sin 0° So,y=15° So, angle MAN= angle NAC=15° AN bisects angle MAC