Difference between revisions of "2016 AIME II Problems/Problem 7"
m (teh -> the) |
|||
(16 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
− | 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 <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 Cauchy-Schwarz 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). |
− | + | ||
+ | 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>. | ||
+ | |||
+ | <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}} |
Revision as of 19:57, 18 October 2020
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 Cauchy-Schwarz 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.