2019 AMC 12B Problems/Problem 12

Problem

Right triangle $ACD$ with right angle at $C$ is constructed outward on the hypotenuse $AC$ of isosceles right triangle $ABC$ with leg length 1, as shown, so that the two triangles have equal perimeters. What is sin(2 BAD)?

Would you please fix the Latex above? Thanks.

!! Someone with good picture-drawing skills please help !!

$\textbf{(A) } \frac{1}{3} \qquad\textbf{(B) } \frac{\sqrt{2}}{2} \qquad\textbf{(C) } \frac{3}{4} \qquad\textbf{(D) } \frac{7}{9} \qquad\textbf{(E) } \frac{\sqrt{3}}{2}$

Solution 1

Observe that the "equal perimeter" part implies that $BC + BA = 2 = CD + DA$ . A quick Pythagorean chase gives $CD = \frac{1}{2}, DA = \frac{3}{2}$ . Use the sine addition formula on angles $BAC$ and $CAD$ (which requires finding their cosines as well), and this gives the sine of $BAD$ . Now, use $\sin{2x} = 2\sin{x}\cos{x}$ on angle $BAD$ to get $\boxed{\textbf{(D) }\frac{7}{9}}$ .

Feel free to elaborate if necessary.

Solution 2

D 7/9 (SuperWill)

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2019 AMC 12B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also