AIME 2014 I · 第 13 题

Solution (Official Solution, MAA)

Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$.

The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is

$$\frac{275+405}{269+275+405+411}=\frac 12,$$ so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\tfrac 35$, so $RS=\tfrac 35s$ and $RQ=\tfrac{1}{10}s$.

Because $\triangle QSG \cong \triangle RTH$, the area of $DHPG$ is the sum

$$[DHPG]=[DTRS]+[RPQ].$$ Rectangle $DTRS$ has area $RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2$. If $\angle QRP = \theta$ , then $\triangle RPQ$ has area

$$[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.$$ Therefore the area of $[DHPG]$ is $s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)$. Because the area of trapezoid $CDHF$ is $\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$.

Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that

$$\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},$$ from which we get $\sin 2\theta = \tfrac {15}{17}$. Note that $\theta =\angle RHT > \angle QAT = 45^\circ$, so $\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}$. Then

$$[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.$$

Solution 1

Let $s$ be the side length of $ABCD$, let $[ABCD]=1360a$. Let $Q$ and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively; because $269+411=275+405$, $Q$ is also the center of the square. Draw $\overline{IJ} \parallel \overline{HF}$ through $Q$, with $I$ on $\overline{AD}$, $J$ on $\overline{BC}$.

Segments $\overline{EG}$ and $\overline{IJ}$ divide the square into four congruent quadrilaterals, each of area $\tfrac 14 [ABCD]=340a$. Then

$$[HFJI]=[ABJI]-[ABFH]=136a.$$ The fraction of the total area occupied by parallelogram $HFJI$ is $\tfrac 1{10}$, so $RQ=\tfrac{s}{10}$.

Because $[HFJI]= HF\cdot PQ$, with $HF=34$, we get $PQ=4a$. Now

$$[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,$$ and because $[PQR]=\tfrac 12 \cdot PQ\cdot PR$, with $PQ=4a$, we get $PR=\tfrac 32$. By Pythagoras' Theorem on $\triangle PQR$, we get

$$16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,$$ with roots $a=\tfrac 9{40}$ or $a=\tfrac58$. The former leads to a square with diagonal less than $34$, which can't be, since $EG=FH=34$; therefore $a=\tfrac 58$ and $[ABCD]=850$.

Solution 2 (Fakesolve)

$269+275+405+411=1360$, a multiple of $17$. In addition, $EG=FH=34$, which is $17\cdot 2$. Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$. All of these triples are primitive:

$$17=1^2+4^2$$ $$34=3^2+5^2$$ $$51=\emptyset$$ $$68=\emptyset\text{ others}$$ $$85=2^2+9^2=6^2+7^2$$ $$102=\emptyset$$ $$119=\emptyset \dots$$ The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$:

$$\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}$$ $$\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850$$ $$\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}$$ Thus, $\boxed{850}$ is the only valid answer.

Solution 3

Continue in the same way as solution 1 to get that $POK$ has area $3a$, and $OK = \frac{d}{10}$. You can then find $PK$ has length $\frac 32$.

Then, if we drop a perpendicular from $H$ to $BC$ at $L$, We get $\triangle HLF \sim \triangle OPK$.

Thus, $LF = \frac{15\cdot 34}{d}$, and we know $HL = d$, and $HF = 34$. Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.

$$\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2$$ $$d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0$$ $$(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0$$ $d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$. Since the area is $d^2$, we have it is equal to $34 \cdot 25 = \boxed{850}$

-Alexlikemath

Solution 4

saw this at bottom of https://artofproblemsolving.com/community/c5h580667p3428922

Let $O$ be the center of the square. Then let $Q$ be on $HF$ susch that $OQ \perp CD$. Draw $H'F' \parallel HF$ through $O$. Let $s$ be the side length of the square, so that $s^2$ is the desired.

First, we note that as $w + z = x + y$, the area of $[ADGC]$ is half the area of $[ABCD]$, which means $O \in EG$. Similarly, $[CDHF] = y + z = \frac{3}{5}[ABCD]$ because of the ratios. But because $O \in H'F'$, we have $[CDH'F'] = \frac{1}{2}[ABCD]$. Subtracting yields $[H'HFF'] = \frac{1}{10}[ABCD]$. But as it is a parallelogram with base $34$ and height $OP$, we get:

$$340(OP) = s^2$$ Now consider quadrilateral $OEBF'$, which has area $1/4$ of the square. We can write:

$$\frac{1}{4}s^2 = [OEBF'] = [PEBF] + [FPOF'] = x + [FQOF'] - [QOP]$$ However, we know the areas of $x$ and parallelogram $[FQOF'] = 17OP = \frac{s^2}{20}$ by our earlier result. Substituting these yields:

$$[QOP] = \frac{3}{1360}s^2$$ But as we know $OP = \frac{s^2}{340}$, we get $QP = 3/2$. Because $CDHF$ is a trapezoid of height $s$ and area $\frac{3}{5}s^2$, we have $\frac{1}{2}s(CF + DH) = \frac{3}{5}s^2$. This means:

$$CF + DH = \frac{6}{5}s$$ But using the same method on trapezoid $CDH'F'$ of area $\frac{1}{2}s^2$, we get

$$CF' + DH' = s$$ And subtracting yields $HH' + FF' = 2OQ = \frac{1}{5}s$, which gives $OQ = \frac{s}{10}$. At this point, applying Pythagoras on $\triangle QOP$ yields:

$$\left( \frac{s^2}{340} \right)^2 + \left( \frac{3}{2} \right)^2 = \left( \frac{s}{10} \right)^2$$ and the resulting quadratic is easy to solve for $s^2 = 306, 850$. But as $[ABCD] > [EFGH] = 34^2/2 > 306$, we have $\boxed{850}$ as the solution.

Video Solution

https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk

~MathProblemSolvingSkills.com

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=wrxET2c0ZgU