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2021 AMC Fall 10B Problem 23

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2021 AMC Fall 10B Problem 23AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 7 min · position P21-25

Each of the 55{ } sides and the 55{ } diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?

  1. A)

    23\frac23

  2. B)

    105128\frac{105}{128}

  3. C)

    125128\frac{125}{128}

  4. D)

    253256\frac{253}{256}

  5. E)

    11

Answer

D

Key insight

With no one-color triangle, every vertex has exactly 2 red and 2 blue edges, so the red edges form a 5-cycle; only 12 of 1024 colorings.

Solution

There are 210=10242^{10} = 1024 colorings of the ten segments. Count the colorings with no single-color triangle.

Fix a vertex VV; four segments leave it. If three of them, say to A,B,CA, B, C, were red, then to avoid red triangles VABVAB, VBCVBC, VCAVCA the segments AB,BC,CAAB, BC, CA would all be blue, forming a blue triangle. So no vertex has three edges of one color: every vertex has exactly 22 red and 22 blue edges.

The red edges then form a graph on 55 vertices in which every vertex has degree 22. Such a graph is a union of cycles covering all 55 vertices; since a cycle needs at least 33 vertices, 55 cannot be split, so the red graph is a single 55-cycle. Its complement (the blue edges) is then also a 55-cycle, and a 55-cycle contains no triangle, so every such coloring works.

The number of 55-cycles on 55 labeled vertices is 4!2=12\frac{4!}{2} = 12 (arrange the vertices around a cycle, up to rotation and reflection).

Therefore the probability of at least one monochromatic triangle is
1−121024=1−3256=253256.1 - \frac{12}{1024} = 1 - \frac{3}{256} = \frac{253}{256}.

The answer is (D) 253256\boxed{\textbf{(D)}\ \frac{253}{256}}.

Why this works

This is the Ramsey-theory fact that K5K_5 can be 2-colored without monochromatic triangles, but only barely: the local pigeonhole condition at each vertex is so restrictive that it determines the global structure (two complementary pentagons). Whenever "at least one" is nearly certain, count the rare complement.

Alternative approach

Answer-choice reasoning: the probability is 1−k10241 - \frac{k}{1024} for some integer kk that is small but positive (the two-pentagon coloring shows k≥1k \ge 1). Only (D), with k=12k = 12, and (C), with k=24k = 24, have the right shape; a quick count of the pentagon colorings (1212) selects (D).

The trap

Trying to count colorings that contain a monochromatic triangle directly with inclusion-exclusion over the 10 triangles, which is long and error-prone.

Common mistakes

  • Trying to count colorings that contain a monochromatic triangle directly with inclusion-exclusion over the 10 triangles, which is long and error-prone.
  • Counting the 55-cycles as 5!=1205! = 120 or 4!=244! = 24 by not dividing out rotations and reflections, or forgetting that the blue edges must also be checked for triangles.

Techniques

Count the complement and subtract from the total · Pigeonhole principle