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 centre 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{EC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the centre of the square <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>. | ||
+ | |||
+ | Let the side side length be <math>d=\sqrt{1360a}</math>. | ||
+ | |||
+ | Draw <math>\overline{OK}\parallel \overline{HI}</math> and intersects <math>\overline{HF}</math> at <math>K</math>. <math>OK=d\cdot\frac{HFJI}{ABCD}=\frac{d}{10}</math>. | ||
+ | |||
+ | The area of <math>HKOI=\frac12\cdot HFJI=68a</math>, so the area of <math>POK=3a</math>. | ||
+ | |||
+ | Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math> | ||
+ | |||
+ | Consider the area of <math>PFJO</math>. | ||
+ | <cmath>\frac12(PF+OJ)(PO)=65a</cmath> | ||
+ | <cmath>(17-\frac{3a}{h})h=65a</cmath> | ||
+ | <cmath>h=4a</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Therefore, the area of <math>ABCD=1360a=\boxed{850}</math> | ||
== 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 22:50, 23 March 2014
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 .
Let the side side length 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
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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