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Difference between revisions of "2016 AIME II Problems/Problem 7"
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==Solution==
 
==Solution==
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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).  
+
Letting <math>AI=a</math> and <math>IB=b</math>, we have <math>IJ^{2}=a^{2}+b^{2} \geq 1008</math> 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 <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).  
  
  

Revision as of 22:44, 15 March 2021

Problem

Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest positive integer values for the area of $IJKL$.

Solution

Letting $AI=a$ and $IB=b$, we have $IJ^{2}=a^{2}+b^{2} \geq 1008$ by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since $2016=12^{2} \cdot 14$, we have the maximum area is $2016 \cdot \dfrac{11}{12} = 1848$ (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).


The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is $1848-1008=\boxed{840}$.

[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

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

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