# Prove that the intersection of $BM$ and $CN$ is on the circumcircle of triangle $ABC.$

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB$ = $\angle BCA$ and $\angle CAQ = \angle ABC$.Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$.Prove that the intersection of $BM$ and $CN$ is on the circumcircle of triangle $ABC.$

• What have you attempted to solve this problem? Jan 20 '15 at 10:29

Let $\angle PBM= \alpha$ and $\angle NCQ=\beta$. Then, as $BP$ is the median of $\triangle ABM$, we have that $\cot \alpha +\cot (\pi-A)=\cot B+\cot A$, which gives $\cot \alpha = 2\cot A+\cot B$ and similarly $\cot \beta = 2\cot A+\cot C$. For convenience, let us take $\cot A=c_1$, $\cot B=c_2$ and $\cot C=c_3$.

By the addition formula we have:

$$\cot(\alpha+\beta)=\frac{\cot \alpha \cot\beta-1}{\cot\alpha+\cot\beta}$$ $$=\frac{(2c_1+c_2)(2c_1+c_3)-1}{4c_1+c_2+c_3}=\frac{4c_1^2+2c_1c_2+2c_1c_3+c_2c_3-1}{4c_1+c_2+c_3}$$

To prove that the intersection of $BM$ and $CN$ (call it $X$) is on the circumcircle, we need to prove that $\angle AXC=180^{\circ}-A$, or $\cot(\alpha+\beta)=\cot A$.

To do so, we use the identity, $\tan A+\tan B+\tan C= \tan A\tan B \tan C$ when $A+B+C=\pi$, which gives $c_1c_2+c_1c_3+c_2c_3=1$. Thus,

$$\cot(\alpha+\beta)=\frac{4c_1^2+c_1c_2+c_1c_3+(c_1c_2+c_1c_3+c_2c_3)-1}{4c_1+c_2+c_3}=c_1=\cot A$$

$\text{QED}$.

• (+1) Despite being a trigonometric solution, it is rather clean, I like it. Jan 20 '15 at 14:43
• @Sawarnik Congrats! You have solved the IMO 2014 question!I'll make sure to ask the doubts sir. Jan 20 '15 at 15:36
• @SNEHILSINHA I am stunned :O Jan 20 '15 at 15:38
• @SNEHILSINHA Can you come to the chat? chat.stackexchange.com/rooms/13523/the-root-of-math :D We can discuss it there :) Jan 20 '15 at 15:48
• @SNEHILSINHA sure come to chat.. Jan 24 '15 at 10:01