Difference between revisions of "2016 AIME II Problems/Problem 7"

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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 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 <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> 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 21:25, 21 December 2016

Squares $ABCD$ and $EFGH$ have a common center at $\overline{AB} || \overline{EF}$. 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 $2016 \cdot \dfrac{11}{12} = 1848$ and the minimum area is $1008$, so the desired answer is $1848-1008=\boxed{840}$.

Solution by Shaddoll

See also

2016 AIME II (ProblemsAnswer KeyResources)
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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