Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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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>. | 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 | + | 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 <math>\boxed{\textbf{(D) }\frac{7}{9}}</math>. |
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+ | Feel free to elaborate if necessary. | ||
==Solution 2== | ==Solution 2== |
Revision as of 15:42, 14 February 2019
Contents
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 .
Feel free to elaborate if necessary.
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 |