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 13:38, 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 | ||
