Difference between revisions of "2014 AIME I Problems/Problem 13"
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==Strategy== | ==Strategy== | ||
<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 | + | 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: |
<cmath>17=1^2+4^2</cmath> | <cmath>17=1^2+4^2</cmath> | ||
Line 69: | Line 69: | ||
<cmath>119=\emptyset \dots</cmath> | <cmath>119=\emptyset \dots</cmath> | ||
− | The sides of the square | + | The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting <math>EG=FH=34</math>: |
− | <cmath>\sqrt{17}\rightarrow 34\implies | + | <cmath>\sqrt{17}\rightarrow 34\implies 4\sqrt{17}\implies A=\textcolor{red}{1088}</cmath> |
− | <cmath>\sqrt{34}\rightarrow 34\implies | + | <cmath>\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850</cmath> |
− | <cmath>\sqrt{85}\rightarrow 34\implies \{ | + | <cmath>\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}</cmath> |
− | + | Thus, <math>\boxed{850}</math> is the only valid answer. | |
== 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 14:32, 14 February 2016
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 centre of the square.
Draw with
on
,
on
such that
and
intersects at the centre of the square
.
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
Strategy
, 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.