Difference between revisions of "2021 AIME I Problems/Problem 2"
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MRENTHUSIASM (talk | contribs) m (→Solution 2 (Coordinate Geometry Bash): Omitted one step in factoring.) |
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\left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ | \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ | ||
130y^2-294y-3528&=0 \\ | 130y^2-294y-3528&=0 \\ | ||
− | + | 2(5x+21)(13x-84)&=0 \\ | |
− | (5x+21)(13x-84)&=0 \\ | ||
y&=-\frac{21}{5}, \ \frac{84}{13}. | y&=-\frac{21}{5}, \ \frac{84}{13}. | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 01:54, 12 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
.
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
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 Bash)
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 and rearranging respectively give
Subtracting
from
we get
Simplifying and rearranging produce
Substituting
into
gives
which is a quadratic of
We clear fractions by multiplying both sides by
then solve by factoring:
Since
is in Quadrant IV, we have
It follows that the equation of
is
Let be the intersection of
and
and
be the intersection of
and
Since
is the
-intercept of
we obtain
By symmetry, quadrilateral is a parallelogram. Its area is
and the requested sum is
~MRENTHUSIASM
Solution 3 (Pythagorean Theorem)
Let the intersection of and
be
, and let
, so
.
By the Pythagorean theorem, , so
, and thus
.
By the Pythagorean theorem again, :
Solving, we get , so the area of the parallelogram is
, and
.
~JulianaL25
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.