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Geometry Difficulty 4.2 AIME Find the answer United States

Problem:

Let $A, B, C, D, E, F$ be 6 points on a circle in that order. Let $X$ be the intersection of $AD$ and $BE$ , $Y$ be the intersection of $AD$ and $CF$ , and $Z$ be 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. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

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

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.