Difference between revisions of "2014 AIME I Problems/Problem 13"

Revision as of 03:53, 25 February 2021

Problem 13

On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$ .

$[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]$

Solution

Notice that $269+411=275+405$ . This means $\overline{EG}$ passes through the center of the square.

Draw $\overline{IJ} \parallel \overline{HF}$ with $I$ on $\overline{AD}$ , $J$ on $\overline{BC}$ such that $\overline{IJ}$ and $\overline{EG}$ intersects at the center of the square which I'll label as $O$ .

Let the area of the square be $1360a$ . Then the area of $HPOI=71a$ and the area of $FPOJ=65a$ . This is because $\overline{HF}$ is perpendicular to $\overline{EG}$ (given in the problem), so $\overline{IJ}$ is also perpendicular to $\overline{EG}$ . These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.

Let the side length of the square be $d=\sqrt{1360a}$ .

Draw $\overline{OK}\parallel \overline{HI}$ and intersects $\overline{HF}$ at $K$ . $OK=d\cdot\frac{[HFJI]}{[ABCD]}=\frac{d}{10}$ .

The area of $HKOI=\frac12\cdot HFJI=68a$ , so the area of $POK=3a$ .

Let $\overline{PO}=h$ . Then $KP=\frac{6a}{h}$

Consider the area of $PFJO$ . $\frac12(PF+OJ)(PO)=65a$ $\left(17-\frac{3a}{h}\right)h=65a$ $h=4a$

Thus, $KP=1.5$ .

Solving $(4a)^2+1.5^2=\left(\frac{d}{10}\right)^2=13.6a$ , we get $a=\frac58$ .

Therefore, the area of $ABCD=1360a=\boxed{850}$

Lazy Solution

$269+275+405+411=1360$ , 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

@@ Line 78: / Line 78: @@
 == Solution 3==
-Continue in the same way as solution 1 to get that POK has area 3a, and OK = d/10. You can then find PK has length 3/2.
+Continue in the same way as solution 1 to get that <math>POK</math> has area <math>3a</math>, and <math>OK = \frac{d}{10}</math>. You can then find <math>PK</math> has length <math>\frac 32</math>.
-Then, if we drop a perpendicular from H to BC get L, We get HLF ~ OPK.
+Then, if we drop a perpendicular from <math>H</math> to <math>BC</math> at <math>L</math>, We get <math>\triangle HLF \sim \triangle OPK</math>.
-Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know HL = d, and HF = 34. Thus, we can set up an equation in terms of d using the Pythagorean theorem.
+Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know <math>HL = d</math>, and <math>HF = 34</math>. Thus, we can set up an equation in terms of <math>d</math> using the Pythagorean theorem.
 <cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath>
@@ Line 90: / Line 90: @@
 <cmath>(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0</cmath>
-<math>d^2 = 34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is d^2, we have it is equal to <math>34 \cdot 25 = \boxed{850}</math>
+<math>d^2 = 34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is <math>d^2</math>, we have it is equal to <math>34 \cdot 25 = \boxed{850}</math>
 -Alexlikemath
 == See also ==
 {{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 {{MAA Notice}}

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