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 (2u3w,v+4w)(2u-3w, v+4w) with 0u10\le u\le1, 0v1,0\le v\le1, and 0w10\le w\le1?

(A) 103(B) 13(C) 12(D) 18(E) 16\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16

Pick an answer.

(A)
$10\sqrt{3}$
(B)
13
(C)
12
(D)
18
(E)
16
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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 $R$.

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) $16$

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).

⭐ 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).