Difference between revisions of "2014 AIME I Problems/Problem 13"
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+ | Let <math>w, x, y, z</math> be <math>269k, 275k, 405k</math> and <math>411k</math> respectively. Then the total area of <math>ABCD</math> is <math>1360k</math>. We have <math>w + z = x + y = 680</math>, and <math>w + x = 544k = \frac 2 3 (x + y)</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 23:51, 20 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
Let be and respectively. Then the total area of is . We have , and .
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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