2005 AIME I Problems/Problem 14

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

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 $\boxed{936}$ .

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
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2005 AIME I Problems/Problem 14

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