2013 USAJMO Problems/Problem 5

Revision as of 22:29, 28 April 2014 by Suli (talk | contribs) (Solution)


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 \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]


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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 $\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}$.

Solution 2

There should be a geometric solution....

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