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== |
Revision as of 13:29, 23 July 2023
Contents
[hide]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 (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
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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