AMC 8 · 2001 · #23

Grade 8 geometry-2dcounting

Archetype Casework by Position / Vertex Type

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Points $R$ , $S$ and $T$ are vertices of an equilateral triangle, and points $X$ , $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be
drawn using any three of these six points as vertices?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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.)

Givens: $R$, $S$, $T$ are the vertices of an equilateral triangle; $X$ is the midpoint of $RT$, $Y$ is the midpoint of $RS$, $Z$ is the midpoint of $ST$; Triangles are counted up to congruence (same side lengths $=$ same triangle); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $20$

Unknowns: The number of noncongruent (and non-degenerate) triangles formed by any $3$ of the $6$ points

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement

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 $\binom{6}{3} = 20$ 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.

Execute — Answer: D

#1 Draw a Diagram 4.G.A.2 Step 1

Choose a convenient side length and label every distance in the figure.
Let the big triangle $RST$ have side length $2$.
Then each midpoint sits halfway along its side, so each "short" segment (vertex to adjacent midpoint, or midpoint to midpoint) has length $1$.
The medial triangle $XYZ$ is equilateral with side $1$.

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

💡 Marking lengths on the figure turns a geometry problem into a sorting problem. Every triangle we will count is built from segments of length $1$, $\sqrt{3}$, or $2$ — nothing else.

#1 Draw a Diagram 8.G.B.7 Step 2

Find the one remaining length: from a vertex to the midpoint of the opposite side.
By symmetry, the segment from $S$ to $X$ is an altitude of the big equilateral triangle.
Using the right triangle $RXS$ with legs $RX = 1$ and $XS$, and hypotenuse $RS = 2$, apply the Pythagorean theorem to get $XS$.

$1^2 + XS^2 = 2^2 \;\Rightarrow\; XS^2 = 3 \;\Rightarrow\; XS = \sqrt{3}$. By symmetry $RZ = TY = \sqrt{3}$ as well.

💡 The vertex-to-opposite-midpoint segment is the altitude of an equilateral triangle with side $2$, which is the classic $30$-$60$-$90$ leg $\sqrt{3}$.

#16 Change Focus / Count the Complement 4.G.A.2 Step 3

Account for the $\binom{6}{3} = 20$ ways to choose $3$ of the $6$ points and remove the collinear (degenerate) picks.
Each side of the big triangle holds three of our points (two vertices and the midpoint between them), and those three are collinear: $\{R,X,T\}$, $\{R,Y,S\}$, $\{S,Z,T\}$.
That is $3$ degenerate picks, leaving $20 - 3 = 17$ actual triangles.

$$\binom{6}{3} = 20,\quad \text{collinear picks} = 3 \;\Rightarrow\; \text{triangles} = 17$$

💡 Counting the bad cases (collinear) is faster than counting the good ones. We will sort the $17$ triangles into congruence classes next.

#2 Make a Systematic List 4.G.A.2 Step 4

Sort the $17$ triangles by side-length signature.
Every triangle uses only the lengths $1$, $\sqrt{3}$, $2$, so each is one of four types.
Walk through them in order from largest to smallest side.

$$\text{Type 1: }(2,2,2)\;\;\text{Type 2: }(1,1,1)\;\;\text{Type 3: }(1,\sqrt{3},2)\;\;\text{Type 4: }(1,1,\sqrt{3})$$

💡 By SSS, two triangles with the same three side lengths are congruent. So the count of congruence classes is just the count of distinct side-length triples that actually appear.

#2 Make a Systematic List 8.G.A.2 Step 5

Verify each type really does occur (and check the total adds to $17$).
Type 1 is the big triangle $RST$ itself ($1$ triangle).
Type 2 is the small equilaterals — the medial $XYZ$ plus the three corner pieces $SYZ$, $RXY$, $TXZ$ ($4$ triangles).
Type 3 has sides $1$, $\sqrt{3}$, $2$ and is the right triangle formed by a vertex, the midpoint of an adjacent side, and the opposite vertex, e.g., $RXS$ ($6$ triangles, one for each choice of "$\sqrt{3}$ end").
Type 4 has sides $1$, $1$, $\sqrt{3}$ and is the isosceles obtuse triangle formed by a vertex and the two midpoints of the sides meeting at it, e.g., $SYZ$ — wait, $SYZ$ has sides $1,1,1$; the $(1,1,\sqrt{3})$ shape is like $YXR$ with the long side from one midpoint to another vertex, e.g., $SXY$ ($SX = \sqrt{3}$, $SY = 1$, $XY = 1$).
There are $6$ of these as well.

$$1 + 4 + 6 + 6 = 17 \;\checkmark$$

💡 The $17$ non-degenerate triangles split into exactly $4$ side-length signatures, and the totals match — so no class was missed and none was double-counted.

#2 Make a Systematic List 8.G.A.2 Step 6

Read off the answer: the number of congruence classes is the number of distinct side-length triples, which is $4$.

$$\text{Noncongruent triangles} = 4 \;\Rightarrow\; \textbf{(D)}$$

💡 Only choice (D) matches the count of $4$ distinct side-length signatures.

[1] #1 4.G.A.2 Choose a convenient side length and label every distance in the figure. Let the

[2] #1 8.G.B.7 Find the one remaining length: from a vertex to the midpoint of the opposite sid

[3] #16 4.G.A.2 Account for the $\binom{6}{3} = 20$ ways to choose $3$ of the $6$ points and rem

[4] #2 4.G.A.2 Sort the $17$ triangles by side-length signature. Every triangle uses only the l

[5] #2 8.G.A.2 Verify each type really does occur (and check the total adds to $17$). Type 1 is

[6] #2 8.G.A.2 Read off the answer: the number of congruence classes is the number of distinct

Review

Reasonableness: Two quick sanity checks. First, the four side-length triples $(2,2,2)$, $(1,1,1)$, $(1,\sqrt{3},2)$, $(1,1,\sqrt{3})$ are obviously distinct (different multisets), so no two of our types collapse together. Second, the triangle inequality is satisfied for each: $1+1 > 1$, $1+\sqrt{3} > 2$ (since $\sqrt{3} \approx 1.73$ so $1 + \sqrt{3} \approx 2.73 > 2$), and $1+1 > \sqrt{3}$ (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.

Alternative: Tool #16 (Change Focus): instead of counting all $17$ triangles and grouping, exploit the three-fold symmetry of the figure. Anchor one vertex, say $S$, and list only triangles that include $S$ — by symmetry every congruence class has at least one such representative. From $S$ you can reach $R, T$ (distance $2$), $Y, Z$ (distance $1$), and $X$ (distance $\sqrt{3}$). The triangles $SRT$ (gives type $(2,2,2)$), $SYZ$ (gives $(1,1,1)$), $SRX$ (gives $(1,\sqrt{3},2)$), and $SYX$ (gives $(1,1,\sqrt{3})$) cover every congruence class, and no two are congruent. That is again $4$ classes, confirming (D).

CCSS standards used (min grade 8)

8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the length of a vertex-to-opposite-midpoint segment ($\sqrt{3}$ when the big triangle has side $2$) by viewing it as a leg of a right triangle with hypotenuse $2$ and other leg $1$.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another if the second can be obtained by a sequence of rotations, reflections, and translations (Treating two triangles as the same when they have the same three side lengths (SSS congruence), which is what lets us collapse the $17$ actual triangles into congruence classes.)
4.G.A.2 Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or angles of a specified size; recognize triangles (Sorting the $17$ triangles by side-length signature into four types $(2,2,2)$, $(1,1,1)$, $(1,\sqrt{3},2)$, $(1,1,\sqrt{3})$.)

⭐ Mark every distance on the figure first: only $1$, $\sqrt{3}$, 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 — $(2,2,2)$, $(1,1,1)$, $(1,\sqrt{3},2)$, $(1,1,\sqrt{3})$. That is $\textbf{4}$, answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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