Difference between revisions of "2019 AMC 10B Problems/Problem 15"
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The value we are looking for is just <math>y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}</math> so the answer is <math>\boxed{\textbf{(A) }\frac{28}{3}}</math>. | The value we are looking for is just <math>y^{4}-x^{4} = \frac{64-36}{3} = \frac{28}{3}</math> so the answer is <math>\boxed{\textbf{(A) }\frac{28}{3}}</math>. | ||
− | Solution by Invoker | + | Solution by <math>\underline{\textbf{Invoker}}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 20:49, 20 February 2019
Problem
Right triangles and
, have areas of 1 and 2, respectively. A side of
is congruent to a side of
, and a different side of
is congruent to a different side of
. What is the square of the product of the lengths of the other (third) side of
and
?
Solution 1
First of all, let the two sides which are congruent be and
, where
. The only way that the conditions of the problem can be satisfied is if
is 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
.
The area conditions give us 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 by
Solution 2
Like in Solution 1, we have and
.
Squaring both equations yields and
.
Let and
. Then
, and
, so
.
We are looking for the value of , so the answer is
.
Solution 3
Firstly, let the right triangles be and
, with
being the smaller triangle. As in Solution 1, let
and
. Additionally, let
and
.
We are given that and
, so using
, we have
and
. Dividing the two equations, we get
=
, so
.
Thus is a
right triangle, meaning that
. Now by the Pythagorean Theorem in
,
.
The problem requires the square of the product of the third side lengths of each triangle, which is . By substitution, we see that
=
. We also know
.
Since we want , multiplying both sides by
gets us
. Now squaring gives
.
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.