Difference between revisions of "2014 AIME I Problems/Problem 13"
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Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the center of the square which I'll label as <math>O</math>. | Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the center of the square which I'll label as <math>O</math>. | ||
− | <asy> size( | + | <asy> size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,K; |
A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; | A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; | ||
E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); K=H-I; | E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); K=H-I; |
Revision as of 18:36, 5 January 2022
Problem 13
On square , points , and lie on sides and respectively, so that and . Segments and intersect at a point , and the areas of the quadrilaterals and are in the ratio Find the area of square .
Solution 1
Notice that . This means passes through the center of the square.
Draw with on , on such that and intersects at the center of the square which I'll label as . Let the area of the square be . Then and . This is because is perpendicular to (given in the problem), so is also perpendicular to . 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 .
Draw and intersects at . Then Then , so . Let . Then
Consider the area of .
Thus, . Now we solve to get or .
The former leads to a square with diagonal less than , which can't be, since ; therefore and the area of
Solution 2 (Lazy)
, a multiple of . In addition, , which is . Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to and must be a multiple of . All of these triples are primitive:
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting :
Thus, is the only valid answer.
Solution 3
Continue in the same way as solution 1 to get that has area , and . You can then find has length .
Then, if we drop a perpendicular from to at , We get .
Thus, , and we know , and . Thus, we can set up an equation in terms of using the Pythagorean theorem.
is extraneous, so . Since the area is , we have it is equal to
-Alexlikemath
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.