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Difference between revisions of "2021 AIME I Problems/Problem 2"

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</asy>
 
</asy>
  
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==Solution (Similar Triangles)==
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==Solution 1 (Similar Triangles)==
 
Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>.
 
Let <math>G</math> be the intersection of <math>AD</math> and <math>FC</math>.
 
From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>.
 
From vertical angles, we know that <math>\angle FGA= \angle DGC</math>. Also, given that <math>ABCD</math> and <math>AFCE</math> are rectangles, we know that <math>\angle AFG= \angle CDG=90 ^{\circ}</math>.
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Thus, the answer is <math>105+4=\framebox{109}</math>.  
 
Thus, the answer is <math>105+4=\framebox{109}</math>.  
 
~yuanyuanC
 
~yuanyuanC
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==Solution 2 (Coordinate Geometry)==
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Suppose <math>A=(0,0).</math>
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<I will be filling the rest later.>
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 +
~MRENTHUSIASM
  
 
==See also==
 
==See also==

Revision as of 19:04, 11 March 2021

Problem

In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] pair A, B, C, D, E, F; A = (0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]

Solution 1 (Similar Triangles)

Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\angle FGA= \angle DGC$. Also, given that $ABCD$ and $AFCE$ are rectangles, we know that $\angle AFG= \angle CDG=90 ^{\circ}$. Therefore, by AA similarity, we know that triangles $AFG$ and $CDG$ are similar.

Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\frac{7}{3}(11-x)$ and $CG=\frac{3}{7}x$. We have $\frac{7}{3}(11-x)+\frac{3}{7}x=FC=9$.

Solving for $x$, we have $x=\frac{35}{4}$. The area of the shaded region is just $3\cdot \frac{35}{4}=\frac{105}{4}$. Thus, the answer is $105+4=\framebox{109}$. ~yuanyuanC

Solution 2 (Coordinate Geometry)

Suppose $A=(0,0).$

~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

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