Question

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 $s.$

答案

Solution:

Let $AD = CD = a$ and $BD = b,$ and let $E$ be the foot of the perpendicular from $D$ to $\overline{AC}.$ Since $AD = CD,$ point $E$ is the midpoint of $\overline{AC},$ so $AE = 15$ and $BE = 15 - 9 = 6.$ The right triangles $DEA$ and $DEB$ share leg $DE,$ so $a^2 - 15^2 = DE^2 = b^2 - 6^2, \qquad \text{that is} \qquad (a+b)(a-b) = 189.$

The factorizations $189 = 189 \cdot 1 = 63 \cdot 3 = 27 \cdot 7 = 21 \cdot 9$ give $(a, b) = (95, 94),$ $(33, 30),$ $(17, 10),$ and $(15, 6).$ The last is rejected: $b = 6$ would put $D$ on $\overline{AC}.$ Each valid pair gives a triangle with perimeter $2a + 30.$

Therefore $s = (190 + 30) + (66 + 30) + (34 + 30) = 220 + 96 + 64 = 380.$

2003 AIME I Problem 7

Solution:

Problem 7 in Other Years