Difference between revisions of "2021 AIME I Problems/Problem 2"
MRENTHUSIASM (talk | contribs) (→Solution 2 (Coordinate Geometry): More progress. Just save in the middle.) |
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D&=(11,3). | D&=(11,3). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Let <math>G</math> be the intersection of <math>\overline{AD}</math> and <math>\overline{FC},</math> and <math>H</math> be the intersection of <math>\overline{AE}</math> and <math>\overline{BC}.</math> | ||
+ | |||
Two solutions follow from here. | Two solutions follow from here. | ||
Line 65: | Line 67: | ||
y&=-\frac{21}{5}, \ \frac{84}{13}. | y&=-\frac{21}{5}, \ \frac{84}{13}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | Since <math>E</math> is in Quadrant IV, we have <math>E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).</math> It follows that the equation of <math>\overleftrightarrow{AE}</math> is <math>y=-\frac{4}{3}x+3.</math> Since <math>H</math> is the <math>x</math>-intercept of this line, we obtain <math>H=\left(\frac94,0\right).</math> | ||
+ | By symmetry, quadrilateral <math>AGCH</math> is a parallelogram. Its area is <math>HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},</math> and the requested sum is <math>105+4=\boxed{109}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 00:25, 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
.
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)
Suppose It follows that
Let
be the intersection of
and
and
be the intersection of
and
Two solutions follow from here.
Solution 2.1 (Inscribed Angle Theorem)
I WILL BE COMPLETING THE REST RIGHT AFTER TEACHING A CLASS. PLEASE DO NOT EDIT IT. THANKS A LOT! :)
~MRENTHUSIASM
Solution 2 (Circle Equations Bash)
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
Since
is the
-intercept of this line, we obtain
By symmetry, quadrilateral is a parallelogram. Its area is
and the requested sum is
~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.