Difference between revisions of "2019 AMC 12B Problems/Problem 12"

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==Problem==
 
==Problem==
Right triangle <math>ACD</math> with right angle at <math>C</math> is constructed outward on the hypotenuse <math>AC</math> of isosceles right triangle <math>ABC</math> with leg length 1, as shown, so that the two triangles have equal perimeters. What is sin(2 BAD)?
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Right triangle <math>ACD</math> with right angle at <math>C</math> is constructed outward on the hypotenuse <math>AC</math> of isosceles right triangle <math>ABC</math> with leg length 1, as shown, so that the two triangles have equal perimeters. What is <math>\sin(2 BAD)</math>?
 
 
Would you please fix the Latex above? Thanks.
 
  
 
!! Someone with good picture-drawing skills please help !!
 
!! Someone with good picture-drawing skills please help !!

Revision as of 15:44, 14 February 2019

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)

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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