Difference between revisions of "2005 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
| − | Consider the | + | Consider the [[point]]s <math> A(0,12), B(10,9), C(8,0),</math> and <math> D(-4,7). </math> There is a unique [[square]] <math> S </math> such that each of the four points is on a different side of <math> S. </math> Let <math> K </math> be the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by <math>1000</math>. |
== 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>. 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>. | + | [[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>. | ||
== See also == | == See also == | ||
Revision as of 12:12, 2 December 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 
.
Solution
Let 
denote a normal vector of the side containing 
. Note that 
intersect and hence must be opposite vertices] of the square. 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 
. We can take 
and 
. So the side of the square is 
, the area is 
, and the answer to the problem is 
.
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 | ||
