Difference between revisions of "2014 AIME I Problems/Problem 13"
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Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the center of the square. | Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the center of the square. | ||
− | 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 <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>. |
Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals. | Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals. |
Revision as of 19:38, 7 July 2020
Contents
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
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 the area of
and the area of
. 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
.
.
The area of , so the area of
.
Let . Then
Consider the area of .
Thus, .
Solving , we get
.
Therefore, the area of
Lazy Solution
, 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.
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.