Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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~IceMatrix | ~IceMatrix | ||
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+ | == Video Solution 2== | ||
+ | https://youtu.be/4_x1sgcQCp4?t=4245 | ||
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+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:41, 24 January 2021
Contents
Problem
In with a right angle at , point lies in the interior of and point lies in the interior of so that and the ratio . What is the ratio
Solution 1
Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a triangle with .
Then , and is a triangle.
In isosceles triangles and , drop altitudes from and onto ; denote the feet of these altitudes by and respectively. Then by AAA similarity, so we get that , and . Similarly we get , and .
Solution 2
Let , and . (For this solution, is above , and is to the right of ). Also let , so , which implies . Similarly, , which implies . This further implies that .
Now we see that . Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ). Therefore (by definition), , and . Hence (by the double angle formula), giving .
By the Law of Cosines in , if , we have Now . Thus the answer is .
Solution 3
WLOG, let , and . . Because of this, is a 3-4-5 right triangle. Draw the altitude of . is by the base-height triangle area formula. is similar to (AA). So . is of . Therefore, is .
~Thegreatboy90
Video Solution 1
~IceMatrix
Video Solution 2
https://youtu.be/4_x1sgcQCp4?t=4245
~ pi_is_3.14
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
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