Printable

2010 AMC 10A Problem 16

2010 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Triangle Centers & Cevians · difficulty 3 of 5 · about 4 min · position P16-20

Nondegenerate ${\triangle ABC}$ has integer side lengths, ${\overline{BD}}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?

A)
$30$
B)
$33$
C)
$35$
D)
$36$
E)
$37$

Answer

Key insight

The angle bisector theorem forces AB : BC = 3 : 8, so sides are 3m, 8m, 11; m = 1 is degenerate, m = 2 gives 6, 16, 11.

Solution

Since $BD$ bisects $\angle B$ , the angle bisector theorem gives
$\frac{AB}{BC} = \frac{AD}{DC} = \frac{3}{8}.$
With integer sides and $\gcd(3, 8) = 1$ , this means $AB = 3m$ and $BC = 8m$ for some positive integer $m$ . Also $AC = AD + DC = 11$ .

For a nondegenerate triangle, the two shorter sides must exceed the longest. With $m = 1$ the sides are $3, 8, 11$ and $3 + 8 = 11$ , which is degenerate. With $m = 2$ the sides are $6, 16, 11$ : the longest is $16$ , and $6 + 11 = 17 > 16$ , so the triangle exists.

Its perimeter is $6 + 16 + 11 = 33$ .

The answer is $\boxed{\textbf{(B)}\ 33}$ .

Why this works

The angle bisector theorem converts a cevian condition into a ratio of sides, and "integer side lengths" plus a ratio in lowest terms forces both sides to be multiples of a common integer. The triangle inequality then rules out the tempting smallest multiple; always check it when minimizing.

The trap

Accepting m = 1 (sides 3, 8, 11) and answering 22, or forgetting the triangle inequality check and picking the wrong multiple.

Common mistakes

Accepting m = 1 (sides 3, 8, 11) and answering 22, or forgetting the triangle inequality check and picking the wrong multiple.
Applying the bisector ratio to the wrong pair of sides (e.g. $AB/AC = 3/8$ ), which no longer uses $AC = 11$ correctly.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed