Difference between revisions of "2021 AIME I Problems/Problem 2"
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− | ==Solution 2 (Coordinate Geometry | + | ==Solution 2 (Coordinate Geometry)== |
+ | I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :) | ||
+ | ===Solution 1 (Slopes)=== | ||
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Clearing fractions | Clearing fractions | ||
+ | ===Solution 2 (Bash)=== | ||
− | I | + | I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :) |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 20:19, 11 March 2021
Contents
Problem
In the diagram below, is a rectangle with side lengths
and
, and
is a rectangle with side lengths
and
as shown. The area of the shaded region common to the interiors of both rectangles is
, where
and
are relatively prime positive integers. Find
.
Solution 1 (Similar Triangles)
Let be the intersection of
and
.
From vertical angles, we know that
. Also, given that
and
are rectangles, we know that
.
Therefore, by AA similarity, we know that triangles
and
are similar.
Let . Then, we have
. By similar triangles, we know that
and
. We have
.
Solving for , we have
.
The area of the shaded region is just
.
Thus, the answer is
.
~yuanyuanC
Solution 2 (Coordinate Geometry)
I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)
Solution 1 (Slopes)
Suppose It follows that
Since
is a rectangle, we have
and
The equation of the circle with center
and radius
is
and the equation of the circle with center
and radius
is
We now have a system of two equations with two variables. Expanding, rearranging, and simplifying respectively give
Subtracting
from
we get
Simplifying and rearranging produce
Substituting
into
gives
which is a quadratic of
Clearing fractions
Solution 2 (Bash)
I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)
~MRENTHUSIASM
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.