AMC 10 · 2010 · #14

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
An angle bisector cuts the opposite side of a whole-number triangle into two known pieces. Find the smallest perimeter.

Pick an answer.

(A)
30
(B)
33
(C)
35
(D)
36
(E)
37
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 two pieces add to a base of 11.

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

Angle Bisector Theorem turns the split into a side ratio

The bisector turns that split into a side ratio.

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

Name both sides with one whole-number factor

One whole-number factor names both other sides.

AB = 3k, BC = 8k, k ∈ Z^+
4STEP 4

Triangle inequality pins k to a single value

The triangle inequality pins that factor to 2.

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

Add the sides for the smallest perimeter

Adding the three sides gives 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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