Competition · AMC preparation · step 4 of 4

AMC 8 · 2001 · #23

Grade 8 geometry-2dcounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

combinations-basicsimilar-figuresspatial-visualizationsystematic-enumeration caseworksystematic-enumerationcomplementary-counting ↑ Prerequisites: combinations-basicsimilar-figures

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An equilateral triangle has vertices R, S, T and midpoints X, Y, Z on its three sides — six points in all. Choosing any three of these six points as vertices, how many noncongruent triangles can be drawn? (Three chosen points lying on one line do not form a triangle and are not counted.)

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only finitely many distances between the six points, and any triangle is fixed (up to congruence) by its three side lengths. So Tool #2 (Make a Systematic List) is the natural lead: label every possible distance, then list each side-length triple that actually occurs and count distinct ones. Tool #1 (Draw a Diagram) on the given figure lets us read every distance straight off the picture using one short Pythagorean step. Tool #16 (Count the Complement) handles the bookkeeping: out of all C(6, 3) = 20 ways to pick 3 of the 6 points, three picks are collinear (degenerate); the remaining 17 are real triangles, and we sort those into congruence classes.

1STEP 1

Label every length

Set the big triangle RST to side 2; then each short segment (vertex to adjacent midpoint, or midpoint to midpoint) has length 1.

RS = ST = RT = 2, RX = XT = RY = YS = SZ = ZT = 1, XY = YZ = XZ = 1

2STEP 2

Find the missing length

Segment SX from a vertex to the opposite midpoint is the altitude; the Pythagorean theorem in right triangle RXS gives its length as √(3).

1² + XS² = 2² → XS² = 3 → XS = √(3). By symmetry RZ = TY = √(3) as well.

3STEP 3

Count the triangle picks

Of the C(6, 3) = 20 point-triples, the three collinear ones {R,X,T}, {R,Y,S}, {S,Z,T} form no triangle, leaving 17 real triangles.

C(6, 3) = 20, collinear picks = 3 → triangles = 17

4STEP 4

Sort by side lengths

Every side is 1, √(3), or 2, so the 17 triangles have just four side-length signatures: (2,2,2), (1,1,1), (1,√(3),2), (1,1,√(3)).

Type 1: (2,2,2) Type 2: (1,1,1) Type 3: (1,√(3),2) Type 4: (1,1,√(3))

5STEP 5

Check each type occurs

Each type occurs: RST (1), four small equilaterals, right triangles like RXS (6), isosceles like SXY (6) — 1 + 4 + 6 + 6 = 17 checks out.

1 + 4 + 6 + 6 = 17 ✓

6STEP 6

Count the distinct types

The number of congruence classes equals the number of distinct side-length triples, which is four — choice (D).

Noncongruent triangles = 4 → (D)

Answer

Two quick sanity checks. First, the four side-length triples (2,2,2), (1,1,1), (1,√(3),2), (1,1,√(3)) are obviously distinct (different multisets), so no two of our types collapse together. Second, the triangle inequality is satisfied for each: 1+1 ＞ 1, 1+√(3) ＞ 2 (since √(3) ≈ 1.73 so 1 + √(3) ≈ 2.73 ＞ 2), and 1+1 ＞ √(3) (since 2 ＞ 1.73). All four shapes really exist. Finally, choice (E) 20 would be the count of all 3-point picks — but that ignores both collinear picks and the congruence collapses, so it is a classic trap; the correct answer (D) 4 is much smaller and matches our enumeration.

💡Key takeaway

Mark every distance on the figure first: only 1, √(3), and 2 appear. Two triangles with the same three side lengths are the same triangle, so the answer is just the number of distinct side-length triples — (2,2,2), (1,1,1), (1,√(3),2), (1,1,√(3)). That is 4, answer (D).

Label every length
Find the missing length
Count the triangle picks
Sort by side lengths
Check each type occurs
Count the distinct types

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