Difference between revisions of "2014 AIME I Problems/Problem 13"
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The fraction of the area of the square <math>ABCD</math> which is occupied by trapezoid <math>BCGE</math> is <cmath>\frac{275+405}{269+275+405+411}=\frac 12,</cmath>so <math>Q</math> is the center of <math>ABCD</math>. Thus <math>R</math>, <math>Q</math>, <math>S</math> are collinear, and <math>RT=QS=\tfrac 12 s</math>. Similarly, the fraction of the area occupied by trapezoid <math>CDHF</math> is <math>\tfrac 35</math>, so <math>RS=\tfrac 35s</math> and <math>RQ=\tfrac{1}{10}s</math>. | The fraction of the area of the square <math>ABCD</math> which is occupied by trapezoid <math>BCGE</math> is <cmath>\frac{275+405}{269+275+405+411}=\frac 12,</cmath>so <math>Q</math> is the center of <math>ABCD</math>. Thus <math>R</math>, <math>Q</math>, <math>S</math> are collinear, and <math>RT=QS=\tfrac 12 s</math>. Similarly, the fraction of the area occupied by trapezoid <math>CDHF</math> is <math>\tfrac 35</math>, so <math>RS=\tfrac 35s</math> and <math>RQ=\tfrac{1}{10}s</math>. | ||
− | Because <math>\triangle QSG \cong \triangle RTH</math>, the area of <math>DHPG</math> is the sum <cmath>[DHPG]=[DTRS]+[RPQ].</cmath> Rectangle <math>DTRS</math> has area <math>RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2</math>. If <math>\angle QRP = \theta</math> , then <math>\triangle RPQ</math> has area <cmath>[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.</cmath>Therefore <math>[DHPG] | + | Because <math>\triangle QSG \cong \triangle RTH</math>, the area of <math>DHPG</math> is the sum <cmath>[DHPG]=[DTRS]+[RPQ].</cmath> Rectangle <math>DTRS</math> has area <math>RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2</math>. If <math>\angle QRP = \theta</math> , then <math>\triangle RPQ</math> has area <cmath>[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.</cmath>Therefore the area of <math>[DHPG]</math> is <math>s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)</math>. Because the area of trapezoid <math>CDHF</math> is <math>\tfrac 35 s^2</math>, the area of <math>CGPF</math> is <math>s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)</math>. |
− | Because these areas are in the ratio <math>411:405=(408+3):(408-3)</math>, it follows that <cmath>\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},</cmath>from which we get <math>\sin 2\theta = \tfrac {15}{17}</math>. Note that <math>\theta > 45^\circ</math>, so <math>\cos 2\theta = -\tfrac 8{17}</math> and <math>\sin^2\theta = \tfrac{25}{34}</math>. Then <cmath>[ABCD]=s^2 = EG^2\sin^2\theta = 850.</cmath> | + | Because these areas are in the ratio <math>411:405=(408+3):(408-3)</math>, it follows that <cmath>\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},</cmath>from which we get <math>\sin 2\theta = \tfrac {15}{17}</math>. Note that <math>\theta =\angle RHT > \angle QAT = 45^\circ</math>, so <math>\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}</math> and <math>\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}</math>. Then <cmath>[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.</cmath> |
== Solution 1== | == Solution 1== | ||
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By Pythagoras' Theorem on <math>\triangle PQR</math>, we get <cmath>16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,</cmath> with roots <math>a=\tfrac 9{40}</math> or <math>a=\tfrac58</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 <math>[ABCD]=850</math>. | By Pythagoras' Theorem on <math>\triangle PQR</math>, we get <cmath>16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,</cmath> with roots <math>a=\tfrac 9{40}</math> or <math>a=\tfrac58</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 <math>[ABCD]=850</math>. | ||
− | ==Solution 2 ( | + | ==Solution 2 (Fakesolve)== |
<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: | ||
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-Alexlikemath | -Alexlikemath | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Latest revision as of 21:07, 13 October 2023
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 (Official Solution, MAA)
Let be the side length of
, let
, and
be the midpoints of
and
, respectively, let
be the foot of the perpendicular from
to
, let
be the foot of the perpendicular from
to
.
The fraction of the area of the square
which is occupied by trapezoid
is
so
is the center of
. Thus
,
,
are collinear, and
. Similarly, the fraction of the area occupied by trapezoid
is
, so
and
.
Because , the area of
is the sum
Rectangle
has area
. If
, then
has area
Therefore the area of
is
. Because the area of trapezoid
is
, the area of
is
.
Because these areas are in the ratio , it follows that
from which we get
. Note that
, so
and
. Then
Solution 1
Let be the side length of
, let
. Let
and
be the midpoints of
and
, respectively; because
,
is also the center of the square. Draw
through
, with
on
,
on
.
Segments
and
divide the square into four congruent quadrilaterals, each of area
. Then
The fraction of the total area occupied by parallelogram
is
, so
.
Because , with
, we get
.
Now
and because
, with
, we get
.
By Pythagoras' Theorem on
, we get
with roots
or
. The former leads to a square with diagonal less than
, which can't be, since
; therefore
and
.
Solution 2 (Fakesolve)
, 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
Video Solution
https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk
~MathProblemSolvingSkills.com
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=wrxET2c0ZgU
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 |
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