AMC 10 · 2023 · #24

Grade 8 geometry-2d

vector-parameterizationcoordinate-geometryperimeterminkowski-sumspatial-visualization physical-representationidentify-subproblems ↑ Prerequisites: coordinate-geometryperimeter

📏 Long solution 💡 4 insights

Problem

What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ , $0\le v\le1,$ and $0\le w\le1$ ?

Pick an answer.

(A)

$10\sqrt{3}$

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Let $R$ be the set of points $(2u - 3w,\;v + 4w)$ as $u, v, w$ each range over $[0, 1]$. Find the perimeter of the boundary of $R$.

Givens: Parameters $u, v, w$ each independent in $[0, 1]$; Point $P(u, v, w) = (2u - 3w,\;v + 4w) = u(2, 0) + v(0, 1) + w(-3, 4)$; Three direction vectors $\vec{a} = (2, 0)$, $\vec{b} = (0, 1)$, $\vec{c} = (-3, 4)$; Answer choices: (A) $10\sqrt{3}$, (B) $13$, (C) $12$, (D) $18$, (E) $16$

Unknowns: The perimeter of the boundary of $R$

Understand

Restated: Let $R$ be the set of points $(2u - 3w,\;v + 4w)$ as $u, v, w$ each range over $[0, 1]$. Find the perimeter of the boundary of $R$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #7 Identify Subproblems, #10 Create a Physical Representation

The region is built from three direction-vectors added in scaled amounts. Tool #1 (Diagram) — plot the three vectors $(2, 0), (0, 1), (-3, 4)$ and sketch the shape they sweep out. Tool #2 (Systematic List) enumerates the $2^{3} = 8$ corner images by trying $u, v, w \in \{0, 1\}$. Tool #7 (Subproblems) breaks 'find perimeter' into 'identify the boundary vertices' then 'add up six edge lengths.' Tool #10 (Physical) is the Minkowski-sum picture: $R$ is the unit segment in direction $\vec{a}$ plus the unit segment in direction $\vec{b}$ plus the unit segment in direction $\vec{c}$ — a hexagon whose six edges come in pairs $\pm \vec{a}, \pm \vec{b}, \pm \vec{c}$.

Execute — Answer: E

#10 Create a Physical Representation 8.G.A.1 Step 1

Recognize the shape.
The point $P = u(2, 0) + v(0, 1) + w(-3, 4)$ is the sum of three independent segments: $[0, (2, 0)] + [0, (0, 1)] + [0, (-3, 4)]$.
This Minkowski sum of three segments in the plane is a centrally symmetric hexagon whose six sides come in parallel pairs — one pair parallel to each direction vector.

$$R = [0, \vec{a}] + [0, \vec{b}] + [0, \vec{c}],\quad \vec{a} = (2, 0),\;\vec{b} = (0, 1),\;\vec{c} = (-3, 4)$$

💡 Grade 8 'rigid motions and translations' — adding a unit segment to a shape just translates it; doing this three times sweeps out a hexagonal patch.

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

List the $8$ corner images by plugging $(u, v, w) \in \{0, 1\}^{3}$ into $P$.
Compute the $(x, y)$ pairs: $(0,0,0)\!\to\!(0,0)$; $(1,0,0)\!\to\!(2,0)$; $(0,1,0)\!\to\!(0,1)$; $(0,0,1)\!\to\!(-3,4)$; $(1,1,0)\!\to\!(2,1)$; $(1,0,1)\!\to\!(-1,4)$; $(0,1,1)\!\to\!(-3,5)$; $(1,1,1)\!\to\!(-1,5)$.

$$\text{corners}: (0,0),(2,0),(0,1),(-3,4),(2,1),(-1,4),(-3,5),(-1,5)$$

💡 Grade 5 'graph points on the coordinate plane' — plot all eight and look at the outline.

#1 Draw a Diagram 6.G.A.3 Step 3

Plot and identify the outer hexagon.
The points $(0, 1)$ and $(-1, 4)$ sit strictly inside the convex hull of the others, so the boundary is the hexagon with vertices $A = (0, 0),\; B = (2, 0),\; C = (2, 1),\; D = (-1, 5),\; E = (-3, 5),\; F = (-3, 4)$, listed counter-clockwise.
Each consecutive pair of vertices differs by one of $\pm \vec{a}, \pm \vec{b}, \pm \vec{c}$.

$$\text{hexagon vertices (CCW): } A(0,0),\;B(2,0),\;C(2,1),\;D(-1,5),\;E(-3,5),\;F(-3,4)$$

💡 Grade 6 'draw polygons from given vertices' — six points outline the hexagon; two points are interior decorations.

#7 Identify Subproblems 8.G.B.8 Step 4

Compute each side length using the distance formula or by reading off $\vec{a}, \vec{b}, \vec{c}$.
$AB = (2,0) - (0,0) = \vec{a}$, length $|\vec{a}| = 2$.
$BC = (2,1) - (2,0) = \vec{b}$, length $1$.
$CD = (-1,5) - (2,1) = (-3, 4) = \vec{c}$, length $\sqrt{9 + 16} = 5$.
$DE = (-3,5) - (-1,5) = -\vec{a}$, length $2$.
$EF = (-3,4) - (-3,5) = -\vec{b}$, length $1$.
$FA = (0,0) - (-3,4) = (3, -4) = -\vec{c}$, length $5$.

$$|\vec{a}| = 2,\;|\vec{b}| = 1,\;|\vec{c}| = \sqrt{3^{2} + 4^{2}} = 5$$

💡 Grade 8 'Pythagoras for distance between coordinate points' — $\vec{c} = (-3, 4)$ is the classic $3$-$4$-$5$ triangle, so $|\vec{c}| = 5$.

#7 Identify Subproblems 3.MD.D.8 Step 5

Add the six lengths.
Each direction contributes twice (one positive, one negative), so the perimeter is $2(|\vec{a}| + |\vec{b}| + |\vec{c}|) = 2(2 + 1 + 5) = 2 \cdot 8 = 16$.

$$\text{Perimeter} = 2(2 + 1 + 5) = 16 \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 3 'perimeter is the sum of side lengths' — six sides, three lengths each appearing twice.

[1] #10 8.G.A.1 Recognize the shape. The point $P = u(2, 0) + v(0, 1) + w(-3, 4)$ is the sum of

[2] #2 5.G.A.2 List the $8$ corner images by plugging $(u, v, w) \in \{0, 1\}^{3}$ into $P$. Co

[3] #1 6.G.A.3 Plot and identify the outer hexagon. The points $(0, 1)$ and $(-1, 4)$ sit stric

[4] #7 8.G.B.8 Compute each side length using the distance formula or by reading off $\vec{a},

[5] #7 3.MD.D.8 Add the six lengths. Each direction contributes twice (one positive, one negativ

Review

Reasonableness: Cross-check by tracing $ABCDEFA$ as a closed loop: walk $\vec{a}$ to $B$, $\vec{b}$ to $C$, $\vec{c}$ to $D$, $-\vec{a}$ to $E$, $-\vec{b}$ to $F$, $-\vec{c}$ back to $A$. Net displacement is $\vec{a} + \vec{b} + \vec{c} - \vec{a} - \vec{b} - \vec{c} = \vec{0}$, confirming the polygon closes. The two interior points $(0, 1)$ and $(-1, 4)$ correspond to corners where only one or two parameters are at $1$ — geometrically they're tucked inside because their direction vectors don't 'stick out' compared to the convex hull. Perimeter $16$ is choice (E).

Alternative: Tool #16 (Change Focus): instead of finding the hexagon, use the general fact that the Minkowski sum of segments $\vec{v}_{1}, \ldots, \vec{v}_{k}$ in the plane (a zonotope) has perimeter $2(|\vec{v}_{1}| + \cdots + |\vec{v}_{k}|)$ provided no two are parallel. Here $(2, 0), (0, 1), (-3, 4)$ are pairwise non-parallel, so the perimeter is $2(2 + 1 + 5) = 16$ — a one-line computation.

CCSS standards used (min grade 8)

3.MD.D.8 Solve real-world and mathematical problems involving perimeters of polygons (Adding up the six side lengths to report the perimeter.)
5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Plotting the eight corner images on the coordinate plane.)
6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Tracing the outer hexagon $ABCDEF$ from the plotted points.)
8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Recognizing that the region is a Minkowski sum of three segments — each segment translates the previous shape, sweeping out a hexagon.)
8.G.B.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Computing $|\vec{c}| = \sqrt{3^{2} + 4^{2}} = 5$ for the slanted sides.)

⭐ This AMC 10 problem only needs Grade 8 Pythagoras you already know — the region is built by adding three unit segments $(2,0), (0,1), (-3,4)$, sweeping out a hexagon whose six sides come in pairs of those three lengths. Side lengths $2, 1, 5$ each twice, so perimeter $= 2(2+1+5) = 16$, choice (E).