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

(→‎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>
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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>
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Two solutions follow from here.
 
Two solutions follow from here.
  
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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 23:25, 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 $B=(0,0).$ It follows that \begin{align*} A&=(0,3), \\ C&=(11,0), \\ D&=(11,3). \end{align*} Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$

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 $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$

We now have a system of two equations with two variables. Expanding and rearranging respectively give \begin{align*} x^2+y^2-6y&=72, \ &(1) \\ x^2+y^2-22x&=-72. \ &(2) \end{align*} Subtracting $(2)$ from $(1),$ we get $22x-6y=144.$ Simplifying and rearranging produce \[x=\frac{3y+72}{11}. \ \ \ \ \ \ \ \ \ (*)\] Substituting $(*)$ into $(1)$ gives \[\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \begin{align*} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 65y^2-147y-1764&=0 \\ (5x+21)(13x-84)&=0 \\ y&=-\frac{21}{5}, \ \frac{84}{13}. \end{align*} Since $E$ is in Quadrant IV, we have $E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).$ It follows that the equation of $\overleftrightarrow{AE}$ is $y=-\frac{4}{3}x+3.$ Since $H$ is the $x$-intercept of this line, we obtain $H=\left(\frac94,0\right).$

By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},$ and the requested sum is $105+4=\boxed{109}.$

~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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