AMC 10 · 2010 · #16

Grade 7 geometry-2dnumber-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
angle-bisector-theoremratio-proportionpolygon-inequality bound-inequality-then-enumerate ↑ Prerequisites: angle-bisector-theorem
📏 Medium solution 💡 3 insights
Problem
Triangle ABC is nondegenerate and has whole-number side lengths. The bisector of angle B meets side AC at point D, splitting it into AD = 3 and DC = 8. Find the smallest possible perimeter of the triangle.

Pick an answer.

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

AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the Angle Bisector Theorem forces AB and BC into the fixed ratio 3:8, so a single whole-number factor k captures both unknown sides at once and turns the whole problem into one variable. Tool #1 (Draw a Diagram): sketching the bisector keeps the two pieces of AC straight and shows why CA = 11 is fixed. Tool #3 (Eliminate Possibilities): the triangle inequality becomes two inequalities in k that throw out every value except one. Tool #14 (Extreme Principle): the question asks for the smallest perimeter, so I look for the smallest legal value of k.

1STEP 1

Fix the base AC from its two pieces

The bisector from B lands on AC, so the two pieces add back to the whole side: CA = 3 + 8 = 11, locked whatever the other sides do.

CA = AD + DC = 3 + 8 = 11
2STEP 2

Angle Bisector Theorem turns the split into a side ratio

The Angle Bisector Theorem pairs each piece with the side beside it, so AB : BC = 3 : 8, exactly the split AD : DC.

AB/BC = AD/DC = 3/8
3STEP 3

Name both sides with one whole-number factor

Since 3 and 8 share no factor, whole-number sides in that ratio must be equal multiples: AB = 3k, BC = 8k for a positive integer k.

AB = 3k, BC = 8k, k ∈ Z^+
4STEP 4

Triangle inequality pins k to a single value

Triangle inequality with CA = 11: 11k ＞ 11 forces k ＞ 1, and 3k + 11 ＞ 8k forces k ＜ 2.2, leaving k = 2.

11k ＞ 11 → k ＞ 1; 3k + 11 ＞ 8k → k ＜ 2.2; so k = 2
5STEP 5

Add the sides for the smallest perimeter

So AB = 6, BC = 16, CA = 11, the only surviving triangle, and its perimeter is 6 + 16 + 11 = 33, choice (B).

P = 3k + 8k + 11 = 11k + 11 = 11(2) + 11 = 33 → (B) 33
Answer
33
Check the sides 6, 16, 11: 6 + 11 = 17 ＞ 16, 6 + 16 = 22 ＞ 11, and 11 + 16 = 27 ＞ 6, so the triangle is genuinely nondegenerate. The ratio AB : BC = 6 : 16 = 3 : 8 matches the split 3 : 8, and the perimeter 33 is choice (B). Testing k = 1 gives sides 3, 8, 11 with 3 + 8 = 11 — a flat, degenerate triangle, correctly rejected. Testing k = 3 gives 9, 24, 11 with 9 + 11 = 20 ＜ 24, impossible. So 33 is the unique valid perimeter.
💡Key takeaway

An angle bisector splits the far side in the same ratio as the two sides at that corner, so name the sides as equal multiples and let the triangle inequality squeeze out the one value that fits.

  • Fix the base AC from its two pieces
  • Angle Bisector Theorem turns the split into a side ratio
  • Name both sides with one whole-number factor
  • Triangle inequality pins k to a single value
  • Add the sides for the smallest perimeter

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