2019 AMC 10B Problems/Problem 16
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
~ By Little Mouse
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 .
Alternatively, once finding the length of one could use the Pythagorean Theorem to find and consequently , and then compute the ratio.
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 .
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 .
Solution 4 (a bit long)
WLOG, and . Notice that in , we have . Since and , we find that and , so and is right. Therefore, by 3-4-5 triangle, and . Define point F such that is an altitude; we know the area of the whole triangle is and we know the hypotenuse is , so . By the geometric mean theorem, . Solving the quadratic we get , so . For now, assume . Then . splits into two parts (quick congruence by Leg-Angle) so and . . Now we know and , we can find or .
Solution 5 (Short with Trig)
Let , then . Since , . Similarly, . Then, . Therefore is right. Let and , then . Let . We know that so we can apply the Law of Cosines on to find . Doing Pythagorean for , we get . Then, so the requested ratio is .
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