Difference between revisions of "2003 AIME I Problems/Problem 7"
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Denote the height of <math>\triangle ACD</math> as <math>h</math>, <math>x = AD = CD</math>, and <math>y = BD</math>. Using the Pythagorean theorem, we find that <math>h^2 = y^2 - 6^2</math> and <math>h^2 = x^2 - 15^2</math>. Thus, <math>y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189</math>. The LHS is [[difference of squares]], so <math>(x + y)(x - y) = 189</math>. As both <math>x,\ y</math> are integers, <math>x+y,\ x-y</math> must be integral divisors of 189. | Denote the height of <math>\triangle ACD</math> as <math>h</math>, <math>x = AD = CD</math>, and <math>y = BD</math>. Using the Pythagorean theorem, we find that <math>h^2 = y^2 - 6^2</math> and <math>h^2 = x^2 - 15^2</math>. Thus, <math>y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189</math>. The LHS is [[difference of squares]], so <math>(x + y)(x - y) = 189</math>. As both <math>x,\ y</math> are integers, <math>x+y,\ x-y</math> must be integral divisors of 189. | ||
Revision as of 14:38, 16 September 2007
Problem
Point is on
with
and
Point
is not on
so that
and
and
are integers. Let
be the sum of all possible perimeters of
. Find
Solution
Denote the height of as
,
, and
. Using the Pythagorean theorem, we find that
and
. Thus,
. The LHS is difference of squares, so
. As both
are integers,
must be integral divisors of 189.
The divisors of 189 are . This yields the four potential sets for
as
. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of
is equal to
.
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |