2019 AMC 12B Problems/Problem 12: Difference between revisions
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<math>\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}</math> | <math>\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}</math> | ||
==Solution== | ==Solution 1== | ||
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>. | |||
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get $\boxed{\textbf{(D) }\frac{7}{9}}. | |||
==Solution 2== | |||
D 7/9 (SuperWill) | D 7/9 (SuperWill) | ||
Revision as of 15:42, 14 February 2019
Problem
Right triangle 
with right angle at 
is constructed outward on the hypotenuse 
of isosceles right triangle 
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 !!
Solution 1
Observe that the "equal perimeter" part implies that 
. A quick Pythagorean chase gives 
.
Use the sine addition formula on angles 
and 
(which requires finding their cosines as well), and this gives the sine of 
. Now, use 
on angle to get $\boxed{\textbf{(D) }\frac{7}{9}}.
Solution 2
D 7/9 (SuperWill)
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
