2015 AMC 10B Problems/Problem 19

Problem

In $\triangle{ABC}$ , $\angle{C} = 90^{\circ}$ and $AB = 12$ . Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$ , and $W$ lie on a circle. What is the perimeter of the triangle?

$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32$

Solution

The center of the circle lies on the perpendicular bisectors of both chords $ZW$ and $YX$ . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$ . Draw perpendiculars to $ZW$ and $YX$ from $O$ , and connect $OZ$ and $OY$ . $OY^2=6^2+12^2=180$ . Let $AC=a$ and $BC=b$ . Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$ . Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$ . But by Pythagorean Theorem on $\triangle ABC$ , we know $a^2+b^2=144$ , because $AB=12$ . Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$ . So our equation simplifies further to $a^2+ab=144$ . However $a^2+b^2=144$ , so $a^2+ab=a^2+b^2$ , which means $ab=b^2$ , or $a=b$ . Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$ , and thus the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$ .

 
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2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2015 AMC 10B Problems/Problem 19

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