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Difference between revisions of "2005 AIME I Problems/Problem 14"

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[[Image:2005_I_AIME-14.png]]
 
[[Image:2005_I_AIME-14.png]]
  
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Let <math> (a,b)</math> denote a [[normal vector]] of the side containing <math> A</math>. Note that <math>\overline{AC}, \overline{BD}</math> intersect and hence must be opposite [[vertex|vertices]]] of the square. The lines containing the sides of the square have the form <math> ax+by=12b</math>, <math> ax+by=8a</math>, <math> bx-ay=10b-9a</math> and <math> bx-ay=-4b-7a</math>. The lines form a square, so the distance between <math>C</math> and the line through <math>A</math> equals the distance between <math>D</math> and the line through <math>B</math>, hence <math> 8a+0b-12b=-4b-7a-10b+9a</math>, or <math>-3a=b</math>. We can take <math>a=-1</math> and <math>b=3</math>. So the side of the square is <math>\frac{44}{\sqrt{10}}</math>, the area is <math>K=\frac{1936}{10}</math>, and the answer to the problem is <math>\boxed{936}</math>.
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Let <math> (a,b)</math> denote a [[normal vector]] of the side containing <math> A</math>. Note that <math>\overline{AC}, \overline{BD}</math> intersect and hence must be opposite [[vertex|vertices]] of the square. The lines containing the sides of the square have the form <math> ax+by=12b</math>, <math> ax+by=8a</math>, <math> bx-ay=10b-9a</math>, and <math> bx-ay=-4b-7a</math>. The lines form a square, so the distance between <math>C</math> and the line through <math>A</math> equals the distance between <math>D</math> and the line through <math>B</math>, hence <math> 8a+0b-12b=-4b-7a-10b+9a</math>, or <math>-3a=b</math>. We can take <math>a=-1</math> and <math>b=3</math>. So the side of the square is <math>\frac{44}{\sqrt{10}}</math>, the area is <math>K=\frac{1936}{10}</math>, and the answer to the problem is <math>\boxed{936}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:14, 2 December 2007

Problem

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$.

Solution

2005 I AIME-14.png

Let $(a,b)$ denote a normal vector of the side containing $A$. Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$, $ax+by=8a$, $bx-ay=10b-9a$, and $bx-ay=-4b-7a$. The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$, hence $8a+0b-12b=-4b-7a-10b+9a$, or $-3a=b$. We can take $a=-1$ and $b=3$. So the side of the square is $\frac{44}{\sqrt{10}}$, the area is $K=\frac{1936}{10}$, and the answer to the problem is $\boxed{936}$.

See also

2005 AIME I (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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