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 < | + | Letting <math>AI=a</math> and <math>IB=b</math>, we have <cmath>IJ^{2}=a^{2}+b^{2} \geq 1008</cmath> by [[AM-GM 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 <cmath>2016=12^{2} \cdot 14</cmath> we have the maximum area is <cmath>2016 \cdot \dfrac{11}{12} = 1848</cmath> (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). |
− | 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 < | + | 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 <cmath>1848-1008=\boxed{840}</cmath> |
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<asy> | <asy> | ||
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label("$H$",H,NW); | label("$H$",H,NW); | ||
label("$I$",I,S); | label("$I$",I,S); | ||
− | label("$J$",J, | + | label("$J$",J,NE); |
label("$K$",K,N); | label("$K$",K,N); | ||
− | label("$L$",L, | + | label("$L$",L,NW); |
</asy> | </asy> | ||
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== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:10, 2 September 2021
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 AM-GM 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.