Difference between revisions of "2014 AIME I Problems/Problem 13"
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Therefore, the area of <math>ABCD=1360a=\boxed{850}</math> | Therefore, the area of <math>ABCD=1360a=\boxed{850}</math> | ||
+ | |||
+ | ==Strategy== | ||
+ | <math>269+275+405+411=1360</math>, a multiple of <math>17</math>. | ||
+ | Therefore the <math>\text{square}</math> of the "hypotenuse" of a triangle with slope <math>m</math> must be a multiple of <math>17</math>. All of these triples are primitive: | ||
+ | |||
+ | <cmath>17=1^2+4^2</cmath> | ||
+ | <cmath>34=3^2+5^2</cmath> | ||
+ | <cmath>51=\emptyset</cmath> | ||
+ | <cmath>85=2^2+9^2=6^2+7^2</cmath> | ||
+ | <cmath>102=\emptyset</cmath> | ||
+ | <cmath>119=\emptyset \dots</cmath> | ||
+ | |||
+ | The sides of the square could either be the longer leg or the shorter leg. Substituting <math>EG=FH=34</math>: | ||
+ | <cmath>\sqrt{17}\rightarrow 34\implies \{2\sqrt{17},4\sqrt{17}\}\implies 272,\textcolor{red}{1088}</cmath> | ||
+ | <cmath>\sqrt{34}\rightarrow 34\implies \{3\sqrt{34},5\sqrt{34}\}\implies 306,850</cmath> | ||
+ | <cmath>\sqrt{85}\rightarrow 34\implies \{4\sqrt{85}/5,18\sqrt{85}/5,12\sqrt{85}/5,14\sqrt{85}/5\}\implies\textcolor{red}{54.4,1101.6,489.6,666.4}</cmath> | ||
+ | |||
+ | The three viable answers here are <math>272,306,850</math>. Since the areas add up to <math>1360</math> parts, probably the answer should be <math>\boxed{850}</math>. And perhaps MAA wants a high answer at the end of the AIME :). | ||
== 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 14:05, 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
.
Therefore the
of the "hypotenuse" of a triangle with slope
must be a multiple of
. All of these triples are primitive:
The sides of the square could either be the longer leg or the shorter leg. Substituting :
The three viable answers here are . Since the areas add up to
parts, probably the answer should be
. And perhaps MAA wants a high answer at the end of the AIME :).
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.