Difference between revisions of "2014 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
− | Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the | + | Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the center of the square. |
− | Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the | + | Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the center of the square which I'll label as <math>O</math>. |
Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals. | Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals. | ||
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Consider the area of <math>PFJO</math>. | Consider the area of <math>PFJO</math>. | ||
<cmath>\frac12(PF+OJ)(PO)=65a</cmath> | <cmath>\frac12(PF+OJ)(PO)=65a</cmath> | ||
− | <cmath>(17-\frac{3a}{h})h=65a</cmath> | + | <cmath>\left(17-\frac{3a}{h}\right)h=65a</cmath> |
<cmath>h=4a</cmath> | <cmath>h=4a</cmath> | ||
Thus, <math>KP=1.5</math>. | Thus, <math>KP=1.5</math>. | ||
− | Solving <math>(4a)^2+1.5^2=(\frac{d}{10})^2=13.6a</math>, we get <math>a=\frac58</math>. | + | Solving <math>(4a)^2+1.5^2=\left(\frac{d}{10}\right)^2=13.6a</math>, we get <math>a=\frac58</math>. |
Therefore, the area of <math>ABCD=1360a=\boxed{850}</math> | Therefore, the area of <math>ABCD=1360a=\boxed{850}</math> | ||
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Thus, <math>\boxed{850}</math> is the only valid answer. | Thus, <math>\boxed{850}</math> is the only valid answer. | ||
+ | |||
+ | == Solution 3== | ||
+ | |||
+ | Continue in the same way as solution 1 to get that <math>POK</math> has area <math>3a</math>, and <math>OK = \frac{d}{10}</math>. You can then find <math>PK</math> has length <math>\frac 32</math>. | ||
+ | |||
+ | Then, if we drop a perpendicular from <math>H</math> to <math>BC</math> at <math>L</math>, We get <math>\triangle HLF \sim \triangle OPK</math>. | ||
+ | |||
+ | Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know <math>HL = d</math>, and <math>HF = 34</math>. Thus, we can set up an equation in terms of <math>d</math> using the Pythagorean theorem. | ||
+ | |||
+ | <cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath> | ||
+ | |||
+ | <cmath>d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0</cmath> | ||
+ | |||
+ | <cmath>(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0</cmath> | ||
+ | |||
+ | <math>d^2 = 34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is <math>d^2</math>, we have it is equal to <math>34 \cdot 25 = \boxed{850}</math> | ||
+ | |||
+ | -Alexlikemath | ||
== 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 03:53, 25 February 2021
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 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
Lazy Solution
, 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.