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Difference between revisions of "2005 AIME I Problems/Problem 14"
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== Solution ==
 
== Solution ==
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Let <math> (a,b)</math> denote a normal vector of the side containing <math> A</math>. 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>. With <math>a^2+b^2=1</math> we get <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>936</math>.
+
Let <math> (a,b)</math> denote a normal vector of the side containing <math> A</math>. 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>936</math>.
 +
 
 
== See also ==
 
== See also ==
 
* [[2005 AIME I Problems/Problem 13 | Previous problem]]
 
* [[2005 AIME I Problems/Problem 13 | Previous problem]]

Revision as of 04:35, 31 July 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

Let $(a,b)$ denote a normal vector of the side containing $A$. 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 $936$.

See also

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