## Problem 2

The feet of the angle bisectors of the triangle $\mathrm{ABC}$ form a right-angled triangle. If the right-angle is at $\mathrm{X}$, where $\mathrm{AX}$ is the bisector of angle $\mathrm{A}$, find all possible values for angle $\mathrm{A}$.


## Solution

Answer: $120^{\circ}$.

Use vectors origin $A$. Write the vector $A B$ as $\mathbf{B}$ etc. Using the familiar $B X / C X=A B / A C$ etc, we have $\mathbf{Z}=$ $\mathrm{b} \mathbf{B} /(\mathrm{a}+\mathrm{b}), \mathbf{Y}=\mathrm{c} \mathbf{C} /(\mathrm{a}+\mathrm{c}), \mathbf{X}=(\mathrm{b} \mathbf{B}+\mathrm{c} \mathbf{C}) /(\mathrm{b}+\mathrm{c})$. Hence $\mathbf{Z}-\mathbf{X}=\mathrm{b} \mathbf{B}(\mathrm{c}-\mathrm{a}) /((\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c}))-\mathrm{c} \mathbf{C} /(\mathrm{b}+\mathrm{c}), \mathbf{Y}-\mathbf{X}=-$ $b \mathbf{B} /(b+c)+c \mathbf{C}(b-a) /((a+c)(b+c))$

We have (Z-X). $(\mathbf{Y}-\mathbf{X})=0$, so after multiplying through by $(a+b)(a+c)(b+c)^{2}$, we get $b^{2} \mathbf{B}^{2}\left(a^{2}-c^{2}\right)+c^{2} \mathbf{C}^{2}\left(a^{2}-b^{2}\right)$ $+2 b c \mathbf{B} \cdot \mathbf{C}\left(a^{2}+b c\right)=0$. But $\mathbf{B}^{2}=c^{2}, \mathbf{C}^{2}=b^{2}$, so $b c\left(2 a^{2}-b^{2}-c^{2}\right)+2$ B. $C\left(a^{2}+b c\right)=0$.

But $a^{2}=b^{2}+c^{2}-2$ B. $\mathbf{C}$ (cosine rule), so $2 a^{2}-b^{2}-c^{2}=a^{2}-2$ B.C. Hence $b c\left(a^{2}-2\right.$ B.C. $)+2$ B. $\mathbf{C}\left(a^{2}+b c\right)=0$, so $\mathbf{B} \cdot \mathbf{C}=$ -bc/2. Hence $\cos \mathrm{A}=-1 / 2$, so $\angle \mathrm{A}=120^{\circ}$.

