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Difference between revisions of "1993 AIME Problems/Problem 14"

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== Solution ==
 
== Solution ==
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Answer: 448.
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Solution: Put the rectangle on the coordinate plane so its vertices are at <math>(\pm4,\pm3)</math>, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, <math>O</math>.
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Note that such a rectangle is unstuck iff its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be <math>A(4,y)</math>, <math>B(-x,3)</math>, <math>C(-4,-y)</math> and <math>D(x,-3)</math> for nonnegative <math>x,y</math>. Then this is a rectangle, so <math>OA=OB</math>, or <math>16+y^2=9+x^2</math>, so <math>x^2=y^2+7</math>.
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[[Image:1993 AIME 14 Diagram.png|center]]
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Reflect <math>D</math> across the side of the rectangle containing <math>C</math> to <math>D'(-8-x,-3)</math>. Then <math>BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10</math> is constant, and the perimeter of the rectangle is equal to <math>2(BC+CD')</math>. The midpoint of <math>\overline{BD'}</math> is <math>(-4-x,0)</math>, and since <math>-4>-4-x</math> and <math>-y\le0</math>, <math>C</math> always lies below <math>\overline{BD'}</math>.
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If <math>y</math> is positive, it can be decreased to <math>y'<y</math>. This causes <math>x</math> to decrease as well, to <math>x'</math>, where <math>x'^2=y'^2+7</math> and <math>x'</math> is still positive. If <math>B</math> and <math>D'</math> are held in place as everything else moves, then <math>C</math> moves <math>(y-y')</math> units up and <math>(x-x')</math> units left to <math>C'</math>, which must lie within <math>\triangle BCD'</math>. Then we must have <math>BC'+C'D'<BC+CD'</math>, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with <math>y=0</math>, so <math>x=\sqrt7</math>.
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By the distance formula, this minimum perimeter is <cmath>2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)</cmath> <cmath>=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.</cmath>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1993|num-b=13|num-a=15}}
 
{{AIME box|year=1993|num-b=13|num-a=15}}

Revision as of 01:07, 21 August 2008

Problem

A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$, for a positive integer $N\,$. Find $N\,$.

Solution

Answer: 448.

Solution: Put the rectangle on the coordinate plane so its vertices are at $(\pm4,\pm3)$, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$.

Note that such a rectangle is unstuck iff its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$, $B(-x,3)$, $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$. Then this is a rectangle, so $OA=OB$, or $16+y^2=9+x^2$, so $x^2=y^2+7$.

1993 AIME 14 Diagram.png

Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$. Then $BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$. The midpoint of $\overline{BD'}$ is $(-4-x,0)$, and since $-4>-4-x$ and $-y\le0$, $C$ always lies below $\overline{BD'}$.

If $y$ is positive, it can be decreased to $y'<y$. This causes $x$ to decrease as well, to $x'$, where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$, which must lie within $\triangle BCD'$. Then we must have $BC'+C'D'<BC+CD'$, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$, so $x=\sqrt7$.

By the distance formula, this minimum perimeter is \[2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)\] \[=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.\]

See also

1993 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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