AMC 10 · 2011 · #13

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
angle-bisector-theoremisosceles-triangleperimeter physical-representationidentify-subproblems ↑ Prerequisites: isosceles-triangle
📏 Medium solution 💡 2 insights
Problem
A line through the incenter runs parallel to one side, cutting the other two. Find the small triangle's perimeter.

Pick an answer.

(A)
27
(B)
30
(C)
33
(D)
36
(E)
42
How to solve
Strategy Draw a Diagram

The hard-looking quantity is MN, the one length nobody handed us. Tool #1 (Draw a Diagram) shows why it is reachable: the segment MN carries the incenter I on it, and I sits on the bisectors from B and from C. So Tool #7 (Identify Subproblems) splits MN at I into MI and IN and handles each half separately. Each half sits in a triangle where a bisector meets a parallel line, and that pairing always makes two angles equal, hence two sides equal. Tool #15 (Organize Information in More Ways) then finishes: instead of computing three lengths and adding, regroup the four pieces of the perimeter so they re-assemble into whole sides of the original triangle. Notice what the argument really turns on — only that MN ∥ BC and that BI, CI bisect their angles. Nothing needs the incircle, and nothing needs BC = 24. Tool #14 (Extreme Principle) tests that surprising independence at the end.

1STEP 1

Confirm the picture really exists

The triangle really does exist.

12 + 18 = 30 ＞ 24 → △ ABC exists
2STEP 2

Read the alternate angles at I

The parallel makes two angles equal.

MN ∥ BC → ∠ MIB = ∠ IBC
3STEP 3

Bisector plus parallel makes MI = MB

So two little triangles are isosceles.

∠ MBI = ∠ IBC = ∠ MIB → MI = MB, NI = NC
4STEP 4

Split MN at the incenter

The cut segment splits at the incenter.

MN = MI + IN = MB + NC
5STEP 5

Regroup the perimeter into whole sides

Regrouping rebuilds two whole sides, giving 30.

AM + MN + NA = (AM + MB) + (NA + NC) = AB + AC = 12 + 18 = 30 → (B)
Answer
30
Two bounds pin the answer down. The small triangle sits inside the big one, so its perimeter is under 12 + 24 + 18 = 54. It also beats the midline: since △ AMN ∼ △ ABC with ratio k, and the strip between MN and BC has width equal to the inradius r, we get k = (h_a - r)/h_a where h_a is the height from A to BC. Comparing the two area formulas rs = 1/2a h_a with s = 27 and a = 24 gives r/h_a = a/2s = 24/54 = 4/9, so k = 5/9 ＞ 1/2 and the perimeter is 5/9 · 54 = 30, above the midline value 27. That independent route lands on the same number and rules out (A) 27, which is exactly the semiperimeter trap for anyone who assumes MN is the midline. The other traps are wrong side-pairs: (D) 36 = AB + BC and (E) 42 = BC + AC. Finally, a sanity test in the spirit of Tool #14: the general result is perimeter = b + c, with no a in it. Push a toward the degenerate extreme a → b + c; then k → (b+c)/(a+b+c) → 1/2 and the perimeter → 1/2 · 2(b+c) = b+c, still 30. The independence survives the extreme case, which is the strongest evidence that BC = 24 was never needed.
💡Key takeaway

A bisector meeting a parallel line always cuts off an isosceles triangle, so MI = MB and NI = NC; the perimeter of △ AMN is then just AB + AC = 30, and the length of BC never enters at all.

  • Confirm the picture really exists
  • Read the alternate angles at I
  • Bisector plus parallel makes MI = MB
  • Split MN at the incenter
  • Regroup the perimeter into whole sides

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