Difference between revisions of "2016 AIME II Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | Letting <math>AI=a</math> and <math>IB=b</math>, we have <math>IJ^{2}=a^{2}+b^{2} \geq 1008</math> by CS inequality. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <math>2016=12^{2} \cdot 14</math>, we have the maximum area is <math>2016 \cdot \dfrac{11}{12} = 1848</math> | + | Letting <math>AI=a</math> and <math>IB=b</math>, we have <math>IJ^{2}=a^{2}+b^{2} \geq 1008</math> by CS inequality. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <math>2016=12^{2} \cdot 14</math>, we have the maximum area is <math>2016 \cdot \dfrac{11}{12} = 1848</math> (the areas of the squares from largest to smallest are <math>12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14</math> forming a geometric progression). |
− | Solution by Shaddoll | + | |
+ | The minimum area is <math>1008</math> (every square is half the area of the square whose sides its vertices touch), so the desired answer is <math>1848-1008=\boxed{840}</math>. | ||
+ | |||
+ | Solution by Shaddoll (edited by ppiittaattoo) | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=6|num-a=8}} | {{AIME box|year=2016|n=II|num-b=6|num-a=8}} |
Revision as of 18:38, 2 April 2018
Problem
Squares and have a common center and . The area of is 2016, and the area of is a smaller positive integer. Square is constructed so that each of its vertices lies on a side of and each vertex of lies on a side of . Find the difference between the largest and smallest positive integer values for the area of .
Solution
Letting and , we have by CS inequality. Also, since , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since , we have the maximum area is (the areas of the squares from largest to smallest are forming a geometric progression).
The minimum area is (every square is half the area of the square whose sides its vertices touch), so the desired answer is .
Solution by Shaddoll (edited by ppiittaattoo)
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |