Difference between revisions of "2019 AMC 10B Problems/Problem 15"
(→Solution 2) |
|||
Line 67: | Line 67: | ||
{{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2019|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | SUB2PEWDS |
Revision as of 16:13, 15 February 2019
Contents
Problem
Two right triangles, and , have areas of 1 and 2, respectively. One side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other. What is the product of the third side lengths of and ?
Solution
First of all, name the two sides which are congruent to be and , where . The only way that the conditions of the problem can be satisfied is if was the shorter leg of and the longer leg of , and is the longer leg of and the hypotenuse of .
Notice that this means the value we are looking for is the square of , which is just .
We have two equations: and .
This means that and that .
Taking the second equation, we get , so since , .
Since , we get .
The value we are looking for is just so the answer is .
Solution 2
First, construct right triangles △ABC and △EDF, with △ABC being the smaller triangle. We are given that one side length of one triangle is congruent to a different side length in the other, and another side length of the first triangle is congruent to yet another side length in the other.
So, ≅ , call this length , and ≅ , call this length
Additionally, call the length , and call the length
Recapping our variables, we have = = , = = , = , and =
We are given that and
Since area = , this gives and
Dividing the two equations, we get =
From this, we get
We see that △EDF is a right triangle, meaning that
In △ABC, and are the legs. By the Pythagorean Theorem,
The question asks for the square of the product of the third side lengths of each triangle, which is
Using substitution, we see that = ) =
We know
Dividing both sides by , we get
Since we want , multiplying both sides by gets us
Squaring this,
.
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
SUB2PEWDS