AIME 2021 I · 第 2 题

Solution 1 (Similar Triangles)

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

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 $105+4=\boxed{109}$.

~yuanyuanC

Solution 2 (Similar Triangles)

Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\triangle AFG \sim \triangle CDG$ 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 $33 - \frac{27}{4} = \frac{105}{4}$.

Finally, the answer is $105+4=\boxed{109}$.

~mathboy100

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 $3\cdot\left(11-\frac{9}{4}\right)=\frac{105}{4}$, and $105+4=\boxed{109}$.

~JulianaL25

Solution 4 (Pythagorean Theorem)

Let $P = AD \cap FC$, and $K = AE \cap BC$. Also let $AP = x$.

$CK$ also has to be $x$ by parallelogram properties. 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:

$$\begin{aligned} DC^2 + PD^2 &= PC^2\\ 9 + (11-x)^2 &= PC^2 \end{aligned}$$ It follows shortly that $PC = \sqrt{x^2-22x+30}$.

Also, $FC = 9$, and $FP + PC = 9$. We can then say that $PC = \sqrt{x^2-22x+30}$, so $FP = 9 - \sqrt{x^2-22x+30}$.

Now we can apply the Pythagorean Theorem to $\triangle AFP$.

$$\begin{aligned} AF^2 + FP^2 = AP^2\\ 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2 \end{aligned}$$ This simplifies (not-as-shortly) to $x = \dfrac{35}{4}$. 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.

From the area of a parallelogram (we know that the base is $\dfrac{35}{4}$ and the height is $3$), it is clear that the area is $\dfrac{105}{4}$, giving an answer of $\boxed{109}$.

~ishanvannadil2008 (Solution Sketch)

~Tuatara (Rephrasing and $\LaTeX$)

Solution 5 (Coordinate Geometry)

Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$

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

$$\begin{aligned} x^2+y^2-6y&=72, &(1) \\ x^2+y^2-22x&=-72. &(2) \end{aligned}$$ Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce

$$x=\frac{3y+72}{11}. \hspace{34.5mm} (*)$$ 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:

$$\begin{aligned} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 2(5y+21)(13y-84)&=0 \\ y&=-\frac{21}{5},\frac{84}{13}. \end{aligned}$$ 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 get $H=\left(\frac94,0\right).$

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

~MRENTHUSIASM

Solution 6 (Trigonometry)

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 $\angle CAD = \alpha$. Let $\angle CAE = \beta$. Note, $\alpha+\beta=\angle DAE$.

$\alpha=\tan^{-1}\left(\frac{3}{11}\right)$ $\beta=\tan^{-1}\left(\frac{7}{9}\right)$ $\tan(\angle DAE) = \tan\left(\tan^{-1}\left(\frac{3}{11}\right)+\tan^{-1}\left(\frac{7}{9}\right)\right) = \frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}} = \frac{\frac{104}{99}}{\frac{78}{99}} = \frac{4}{3}$

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

~twotothetenthis1024

Solution 7

A self-admittedly worse solution 1

Construct a point at line $\overline{BC}$, and name it $G$. Let $a$ = $\overline{AG}$ Let $b$ = $\overline{GE}$ Let $c$ = $\overline{BG}$ Let $d$ = $\overline{AB}$ Let $e$ = $\overline{GC}$ Let $f$ = $\overline{AF}$

We know $\overline{AB} = 3$, so let $d$ = $3$. We also know that $\overline{AF} = 7$, so let $f$ = $7$.

Triangle $\triangle ABG \sim \triangle ECG$ due to all the angles being equal from alternate interior angles at point G and both of the triangles having right angles. Dividing $d$ and $f$ yields the ratio of the side lengths to be $\frac{3}{7}$

We will now create several equations with these variables. $d = 3$ $c + e = 11$ $f = 7$ $a + b = 9$ $\frac{d}{f} = \frac{3}{7}$ $\frac{a}{e} = \frac{3}{7}$ $\frac{c}{b} = \frac{3}{7}$

We can find $b$ to be $9 - a$ from the fourth equation, and then find $c$ to be $7c = 3b$ from the seventh equation. Then substitute $b$ for $9 - a$ to find that $c = \frac{3}{7}\left(9 - a \right)$. Now we just substitute $c$ into our second equation, $c + e = 11$ to yield that $e$ is $e = 11 - \frac{3}{7}\left(9 - a \right)$. That's not useful right now, but will be once we look at our sixth equation, which shows that $\frac{a}{e} = \frac{3}{7}$, and we yield that $e$ is equal to $3e = 7a$ which shows that when simplified we get $e = \frac{7a}{3}$. Now we can see that we have 2 equations with 2 variables. When substituting $e$ into the first equation, which is $\frac{a}{e} = \frac{3}{7}$, we get

$$\frac{7a}{3} = 11 - \frac{3}{7}\left(9 - a \right)$$ We then yield that $15 = 4a$, which gives us $a = \frac{15}{4}$. We then substitute $a$ into our fourth equation, which is $a + b = 9$, and then we get that $b = 9 - \frac{15}{4}$, and then we get that $b$ is equal to $\frac{21}{4}$. Then we use our seventh equation, which is $\frac{c}{b} = \frac{3}{7}$. We then get $7c = 3b$. We substitute $b$ in to get that $c$ is $\frac{9}{4}$ Then we use our second equation which is $c + e = 11$ and then we substitute $c$ into it to get that $e$ is $11 - \frac{9}{4}$. Then we get that $e = \frac{35}{4}$. Then to find the area we multiply $e$ by the height, $d$ to get $\frac{35}{4} \cdot 3$ to get $\frac{105}{4}$, which is $\frac{m}{n}$, and we get that $m = 105$ and $n = 4$, so

$$m + n = 105 + 4 = \boxed{109}$$ ~imhappy262789

Video Solution by Punxsutawney Phil

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

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

Video Solution by Steven Chen (in Chinese)

https://youtu.be/eaS5gRLSqgY

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https://www.youtube.com/watch?v=BinfKrc5bWo

Video Solution by Power of Logic

https://youtu.be/WS6X1MQ37jg