Difference between revisions of "2005 AIME I Problems/Problem 14"
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== Solution == | == Solution == | ||
− | {{ | + | 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>. |
== See also == | == See also == | ||
* [[2005 AIME I Problems/Problem 13 | Previous problem]] | * [[2005 AIME I Problems/Problem 13 | Previous problem]] |
Revision as of 03:33, 31 July 2007
Problem
Consider the points and
There is a unique square
such that each of the four points is on a different side of
Let
be the area of
Find the remainder when
is divided by 1000.
Solution
Let denote a normal vector of the side containing
. The lines containing the sides of the square have the form
,
,
and
. The lines form a square, so the distance between
and the line through
equals the distance between
and the line through
, hence
, or
. With
we get
and
. So the side of the square is
, the area is
, and the answer to the problem is
.