Difference between revisions of "2016 AIME II Problems/Problem 7"
m (I don't know if this is right, but I think it should be AM-GM inequality instead of Cauchy-Schwarz) |
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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 | + | 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> and the minimum area is <math>1008</math>, so the desired answer is <math>1848-1008=\boxed{840}</math>. |
Solution by Shaddoll | Solution by Shaddoll |
Revision as of 20:26, 21 December 2016
Squares and
have a common center at
. 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
and the minimum area is
, so the desired answer is
.
Solution by Shaddoll
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 |