Difference between revisions of "2014 AIME I Problems/Problem 13"
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The former leads to a square with diagonal less than <math>34</math>, which can't be, since <math>EG=FH=34</math>; therefore <math>a=\tfrac 58</math> and the area of <math>ABCD=1360a=\boxed{850}</math> | The former leads to a square with diagonal less than <math>34</math>, which can't be, since <math>EG=FH=34</math>; therefore <math>a=\tfrac 58</math> and the area of <math>ABCD=1360a=\boxed{850}</math> | ||
− | ==Lazy | + | ==Solution 2 (Lazy)== |
<math>269+275+405+411=1360</math>, a multiple of <math>17</math>. In addition, <math>EG=FH=34</math>, which is <math>17\cdot 2</math>. | <math>269+275+405+411=1360</math>, a multiple of <math>17</math>. In addition, <math>EG=FH=34</math>, which is <math>17\cdot 2</math>. | ||
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to <math>EG</math> and <math>FH</math> must be a multiple of <math>17</math>. All of these triples are primitive: | Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to <math>EG</math> and <math>FH</math> must be a multiple of <math>17</math>. All of these triples are primitive: |
Revision as of 19:36, 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
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.