JEE-Advanced 2015 (20.A/40)

As $a^2-b^2=\frac{c^2}{2} \gt 0$ , side $ZY$ has an internal point $W$ such that $\triangle WXY$ is isosceles in $W$ . Set $w=|ZW|$ . Now from sine theorem $\frac{\sin(X-Y)}{w} =\frac{\sin Z}{a-w}$ we get $\frac{\sin(X-Y)}{\sin Z} = \frac{w}{a-w}$ Now cosine formula yields $(a-w)^2 = w^2+b^2 - 2bw\cos Z$ , whence $w(2b\cos Z - 2a) = b^2-a^2.$

Now note (reverse cosine formula) that $\cos Z = \frac{a^2+b^2-c^2}{2ab} = \frac{a^2+b^2-2a^2+2b^2}{2ab} = \frac{3b^2 - a^2}{2ab}$ Substitution in the previous equation gives $w\left(\frac{3b^2-a^2}{a} - 2a\right) = b^2 - a^2$ whence $w= \frac{a}{3},$ that is $3w=a$ , then $a-w = 2w$ , and finally $\frac{w}{a-w} = \frac{1}{2}$ At last one has to remember that cosine function is zero exactly on odd multiples of $\frac{\pi}{2}$ .