Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 33: Line 33:
  
 
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>.
 
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>.
 +
 +
Draw line <math>\overline {IJ} \parallel \overline{FH}</math>, with <math>I</math> on <math>\overline{BC}</math> and <math>J</math> on <math>\overline{AD}</math> and <math>\overline {IJ}</math> intersects <math>\overline{EG}</math> at its midpoint <math>O</math>. Then the area of parallelogram <math>FIJH</math> is <math>136k</math>.  Let <math>\overline{IK} \vertical \overline{AD}</math> at <math>K</math> and <math>\overline{FL} \vertical \overline{IJ}</math> at <math>L</math>. <math>\triangle FIL \sim \triangle HFK</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 00:01, 21 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 $w + x = 544k = \frac 2 3 (x + y)$.

Draw line $\overline {IJ} \parallel \overline{FH}$, with $I$ on $\overline{BC}$ and $J$ on $\overline{AD}$ and $\overline {IJ}$ intersects $\overline{EG}$ at its midpoint $O$. Then the area of parallelogram $FIJH$ is $136k$. Let $\overline{IK} \vertical \overline{AD}$ (Error compiling LaTeX. Unknown error_msg) at $K$ and $\overline{FL} \vertical \overline{IJ}$ (Error compiling LaTeX. Unknown error_msg) at $L$. $\triangle FIL \sim \triangle HFK$.

See also

2014 AIME I (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_13&oldid=61139"