Consider all triangles ABC satisfying the following conditions: AB = AC, D is a point on AC for which BD ⊥ AC, AD and CD are integers, and BD^2 = 57. Among all such triangles, the smallest possible value of AC is

1999 AMC 12 Problem 19

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19.

Consider all triangles

ABC

satisfying the following conditions:

AB = AC,

D

is a point on

\overline{AC}

for which

\overline{BD} \perp \overline{AC},

AD

and

CD

are integers, and

BD^2 = 57.

Among all such triangles, the smallest possible value of

AC

9

10

11

12

13

Right triangle

ADB

gives

AB^2 = AD^2 + BD^2

With

AB = AC = AD + CD,

reduce to

CD(CD + 2\,AD) = 57

Let

AD = n

and

CD = m.

Since

\triangle ADB

is right-angled at

D,

AB^2 = n^2 + 57.

Also

AB = AC = m + n,

(m + n)^2 = n^2 + 57,

which simplifies to

m(m + 2n) = 57.

The positive integer solutions are

m = 1, n = 28

(giving

AC = 29

) and

m = 3, n = 8

(giving

AC = 11

). The smallest possible value of

AC

11.

Thus, the correct answer is C.