2013 USAJMO Problems/Problem 5

Problem

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$ . Segments $AY$ and $BX$ meet at $P$ . Point $Z$ is the foot of the perpendicular from $P$ to line $XY$ . Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$ . Let $Q$ be the intersection of segments $AY$ and $XC$ . Prove that $\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.$

Problem

Please copy and paste the problem here!

Solution 1

Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants $a$ and $b$ set A $(\cos a, \sin a)$ and B $(\cos b, \sin b)$ . Now, let's use our coordinate tools. It is easily derived that the equation of $BX$ is $y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)$ and the equation of $AY$ is $y = \frac{\sin a}{1 - cos a}(x - 1) = u(x-1)$ , where $u$ and $v$ are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, $P$ , is $(\frac{u-v}{u+v}, \frac{2uv}{u+v})$ . Also, $Z(\frac{u-v}{u+v}, 0)$ . It shall be left to the reader to find the slope of $AZ$ , the coordinates of Q and C, and use the distance formula to verify that $\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}$ .