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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 at <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>.
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==Problem==
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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==
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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 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>.
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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).  
  
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Solution by Shaddoll
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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>
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pair A,B,C,D,E,F,G,H,I,J,K,L;
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A=(0,0);
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B=(2016,0);
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C=(2016,2016);
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D=(0,2016);
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I=(1008,0);
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J=(2016,1008);
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K=(1008,2016);
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L=(0,1008);
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E=(504,504);
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F=(1512,504);
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G=(1512,1512);
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H=(504,1512);
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draw(A--B--C--D--A);
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draw(I--J--K--L--I);
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draw(E--F--G--H--E);
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,NE);
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label("$D$",D,NW);
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label("$E$",E,SW);
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label("$F$",F,SE);
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label("$G$",G,NE);
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label("$H$",H,NW);
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label("$I$",I,S);
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label("$J$",J,NE);
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label("$K$",K,N);
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label("$L$",L,NW);
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</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 $ABCD$ 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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