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

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== Solution ==
 
== Solution ==
  
Let <math>w, x, y, z</math> be <math>269k, 275k, 405k</math> and <math>411k</math> respectively. Then the total area of <math>ABCD</math> is <math>1360k</math>. We have <math>w + z = x + y = 680</math>, and <math>w + x = 544k = \frac 2 3 (x + y)</math>.
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Let <math>w, x, y, z</math> be <math>269k, 275k, 405k</math> and <math>411k</math> respectively. Then the total area of <math>ABCD</math> is <math>1360k</math>. We have <math>w + z = x + y = 680k</math>, and <math>w + x = 544k = \frac 2 3 (x + y)</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:52, 20 March 2014

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

Let $w, x, y, z$ be $269k, 275k, 405k$ and $411k$ respectively. Then the total area of $ABCD$ is $1360k$. We have $w + z = x + y = 680k$, and $w + x = 544k = \frac 2 3 (x + y)$.

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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