Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

Let $A, B, C, D, E, F$ be 6 points on a circle in that order. Let $X$ be the intersection of $AD$ and $BE$ , $Y$ is the intersection of $AD$ and $CF$ , and $Z$ is the intersection of $CF$ and $BE$ . $X$ lies on segments $BZ$ and $AY$ and $Y$ lies on segment $CZ$ . Given that $AX=3, BX=2, CY=4, DY=10, EZ=16$ , and $FZ=12$ , find the perimeter of triangle $XYZ$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $XY=z, YZ=x$ , and $ZX=y$ . By Power of a Point, we have that $3(z+10)=2(y+16), 4(x+12)=10(z+3), \text{ and } 12(x+4)=16(y+2)$ . Solving this system gives $XY=\frac{11}{3}$ and $YZ=\frac{14}{3}$ and $ZX=\frac{9}{2}$ . Therefore, the answer is $XY+YZ+ZX=\frac{77}{6}$ .

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