2003 AIME I Problems/Problem 7

Problem

Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$ . Find $s.$

Solution

$[asy] size(220); pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle); D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2); [/asy]$

Denote the height of $\triangle ACD$ as $h$ , $x = AD = CD$ , and $y = BD$ . Using the Pythagorean theorem, we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$ . Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$ . The LHS is difference of squares, so $(x + y)(x - y) = 189$ . As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$ .

The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$ . This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$ . The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{252}$ .

Solution 2

Using Stewart's Theorem, letting the side length be c, and the cevian be d, then we have $30d^2+21*9*30=9c^2+21c^2=30c^2$ . Dividing both sides by thirty leaves $d^2+189=c^2$ . The solution follows as above.

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2003 AIME I Problems/Problem 7

Contents

Problem

Solution

Solution 2

See also