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Difference between revisions of "2021 AIME I Problems/Problem 2"
(→‎Solution 1 (Similar Triangles))
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Solving for <math>x</math>, we have <math>x=\frac{35}{4}</math>.
 
Solving for <math>x</math>, we have <math>x=\frac{35}{4}</math>.
 
The area of the shaded region is just <math>3\cdot \frac{35}{4}=\frac{105}{4}</math>.
 
The area of the shaded region is just <math>3\cdot \frac{35}{4}=\frac{105}{4}</math>.
 +
 
Thus, the answer is <math>105+4=\framebox{109}</math>.  
 
Thus, the answer is <math>105+4=\framebox{109}</math>.  
 
~yuanyuanC
 
~yuanyuanC

Revision as of 08:53, 13 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 Bash)

Suppose $B=(0,0).$ It follows that \begin{align*} A&=(0,3), \\ C&=(11,0), \\ D&=(11,3). \end{align*}

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 \\ 2(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 $y=-\frac{4}{3}x+3.$

Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$ Since $H$ is the $x$-intercept of $\overleftrightarrow{AE},$ 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

Solution 3 (Pythagorean Theorem)

Let the intersection of $AE$ and $BC$ be $G$, and let $BG=x$, so $CG=11-x$.

By the Pythagorean theorem, ${AG}^2={AB}^2+{BG}^2$, so $AG=\sqrt{x^2+9}$, and thus $EG=9-\sqrt{x^2+9}$.

By the Pythagorean theorem again, ${CG}^2={EG}^2+{CE}^2$: \[11-x=\sqrt{7^2+(9-\sqrt{x^2+9})^2}.\]

Solving, we get $x=\frac{9}{4}$, so the area of the parallelogram is $3\cdot(11-\frac{9}{4})=\frac{105}{4}$, and $105+4=\framebox{109}$.

~JulianaL25

Solution 4 (Similar triangles and area)

Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $AFG \sim CDG$ with a $\frac{7}{3}$ ratio. Define $x$ as $\frac{[CDG]}{9}$. Because of similar triangles, $[AFG] = 49x$. Using $ABCD$, the area of the parallelogram is $33-18x$. Using $AECF$, the area of the parallelogram is $63-98x$. These equations are equal, so we can solve for $x$ and obtain $x = \frac{3}{8}$. Thus, $18x = \frac{27}{4}$, so the area of the parallelogram is $33 - \frac{27}{4} = \frac{105}{4}$.

~mathboy100

Solution 5

The intersection of AD and FC = P. The intersection of AE and BC = K. Let's set AP to x. CK also has to be x because of the properties of a parallelogram. Then PD and BK must be 11-x because the sum of the segments has to be 11. We can easily solve for PC by the Pythagorean Theorem. DC^2 + PD^2 = PC^2. 9 + (11-x)^2 = PC^2. After 10 seconds of simplifying, we get that PC = sqrt(x^2-22x+30).

FC = 9, and FP + PC = 9. PC = sqrt(x^2-22x+30), so FP = 9 - sqrt(x^2-22x+30).

Now we can apply the Pythagorean Theorem to triangle AFP. AF^2 + FP^2 = AP^2. 49 + (9 - (sqrt(x^2-22x+30)))^2 = x^2. After simplifying (took me 2 minutes on the test), we get that x = 35/4. If you don't believe me, then plug it into WolframAlpha.

Now we have to solve for the area of APCK. We know that the height is 3 because the height of the parallelogram is the same as the height of the smaller rectangle.

The area of a parallelogram is base * height. The base is x (or 35/4), and the height is 3.

Multiplying, (35/4)*3 = 105/4. m+n = 105+4 = 109. ~ishanvannadil2008


Solution 6 (Trigbash)

Let the intersection of $AE$ and $BC$ be $G$. It is useful to find $tan(\angle DAE)$, because $tan(\angle DAE)=\frac{3}{BG}$ and $\frac{3}{tan(\angle DAE)}=BG$.

From there, subtracting the areas of the two triangles from the larger rectangle, we get  Area = $33-3BG=33-\frac{9}{tan(\angle DAE)}$.

let $\angle CAD$ = α. Let $\angle CAE$ = β. Note, α+β=$\angle DAE$.

α =  arctan($\frac{3}{11}$)

β  = arctan($\frac{7}{9}$)

tan($\angle DAE$)= tan(arctan($\frac{3}{11}$)+arctan($\frac{7}{9}$)) =

$\frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}}$ = $\frac{\frac{104}{99}}{\frac{78}{99}}$ = $\frac{4}{3}$

Area= $33-\frac{9}{\frac{4}{3}}$ = $33-\frac{27}{4 }= \frac{105}{4}$. The answer is $105+4=109$. ~ twotothetenthis1024

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=H17E9n2nIyY&t=289s

Video Solution

https://youtu.be/M3DsERqhiDk?t=275

See also

2021 AIME I (Problems • Answer Key • Resources)
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