Difference between revisions of "2016 AIME II Problems/Problem 7"
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− | Squares <math>ABCD</math> and <math>EFGH</math> have a common center | + | ==Problem== |
+ | Squares <math>ABCD</math> and <math>EFGH</math> have a common center and <math>\overline{AB} || \overline{EF}</math>. The area of <math>ABCD</math> is 2016, and the area of <math>EFGH</math> is a smaller positive integer. Square <math>IJKL</math> is constructed so that each of its vertices lies on a side of <math>ABCD</math> and each vertex of <math>EFGH</math> lies on a side of <math>IJKL</math>. Find the difference between the largest and smallest positive integer values for the area of <math>IJKL</math>. | ||
==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 <cmath>1848-1008=\boxed{840}</cmath> | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D,E,F,G,H,I,J,K,L; | ||
+ | A=(0,0); | ||
+ | B=(2016,0); | ||
+ | C=(2016,2016); | ||
+ | D=(0,2016); | ||
+ | I=(1008,0); | ||
+ | J=(2016,1008); | ||
+ | K=(1008,2016); | ||
+ | L=(0,1008); | ||
+ | E=(504,504); | ||
+ | F=(1512,504); | ||
+ | G=(1512,1512); | ||
+ | H=(504,1512); | ||
+ | draw(A--B--C--D--A); | ||
+ | draw(I--J--K--L--I); | ||
+ | draw(E--F--G--H--E); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NW); | ||
+ | label("$E$",E,SW); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,NE); | ||
+ | label("$H$",H,NW); | ||
+ | label("$I$",I,S); | ||
+ | label("$J$",J,NE); | ||
+ | label("$K$",K,N); | ||
+ | label("$L$",L,NW); | ||
+ | </asy> | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=II|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:10, 1 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.