2019 AMC 12B Problems/Problem 12

Revision as of 15:42, 14 February 2019 by Expilncalc (talk | contribs) (Added new solution.)

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}}.

Solution 2

D 7/9 (SuperWill)

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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