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 $\sin(2 BAD)$ ?

!! 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