AMC 10 · 2023 · #4

Grade 6 geometry-2d

Archetype Small-Domain Constrained Search

perimeterpolygon-inequalitysystematic-enumeration bound-inequality-then-enumeratecaseworkguess-and-check ↑ Prerequisites: perimetermulti-digit-arithmetic

📏 Medium solution 💡 2 insights

Problem

A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?

$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A quadrilateral has integer side lengths summing to $26$, and one of its four sides is $4$. What is the greatest possible value of any single side?

Givens: Four integer side lengths $a, b, c, d$ with $a + b + c + d = 26$; At least one side equals $4$; set $d = 4$; All side lengths are positive integers; The four lengths must form a genuine (non-degenerate) quadrilateral; Answer choices: (A) $9$, (B) $10$, (C) $11$, (D) $12$, (E) $13$

Unknowns: The largest possible integer side length

Understand

Restated: A quadrilateral has integer side lengths summing to $26$, and one of its four sides is $4$. What is the greatest possible value of any single side?

Plan

Primary tool: #1 Draw a Diagram

Secondary: #6 Guess and Check, #3 Eliminate Possibilities

Geometry with a fixed perimeter cries out for Tool #1 (Draw a Diagram) — a sketch immediately exposes why the longest side cannot be too long: if it equals or exceeds the other three combined, the perimeter "snaps shut" into a line. That visual is the polygon inequality. Once we see the bound "longest $<$ half perimeter," Tool #6 (Guess and Check) walks down from the largest choice (E) $13$ until one works, and Tool #3 (Eliminate Possibilities) finishes by showing the four smaller choices each leave a valid integer pair for the remaining two sides. Algebra (#13) is unnecessary here — the diagram is faster.

Execute — Answer: D

#1 Draw a Diagram 3.MD.D.8 Step 1

Sketch a quadrilateral with one side labelled $4$ and the other three sides unknown.
To make one of the unknown sides as long as possible, push it out; the other three sides must wrap from one end of the long side to the other.

$$a + b + c + d = 26, \quad d = 4 \;\Rightarrow\; a + b + c = 22$$

💡 Drawing the quadrilateral makes the perimeter into a closed loop — picturing the loop is the Grade 3 "perimeter as the sum of the sides" idea.

#1 Draw a Diagram 4.MD.A.3 Step 2

From the picture, if one side $a$ is the longest, the other three sides must close the loop from one end of $a$ to the other.
They can only do this if their total length exceeds $a$; otherwise the figure flattens into a segment.
That gives the polygon inequality.

$$a < b + c + d$$

💡 The triangle-inequality intuition extends: any side of a polygon has to be shorter than the path going around the rest, just like Grade 4 perimeter reasoning about polygons.

#1 Draw a Diagram 6.EE.B.8 Step 3

Replace $b + c + d$ using the perimeter equation, then solve the resulting one-variable inequality.

$$a < 26 - a \;\Longrightarrow\; 2a < 26 \;\Longrightarrow\; a < 13$$

💡 Substituting one expression into another to write a single inequality in one unknown is the Grade 6 "inequality on the number line" move.

#6 Guess and Check 5.OA.A.2 Step 4

Walk down the answer choices from the top.
(E) $13$ fails because the inequality is strict, $a < 13$.
Try $a = 12$ next: the remaining three sides have $b + c + d = 14$ with one of them equal to $4$, so $b + c = 10$.
Many positive integer pairs work (e.g.
$5, 5$), and they satisfy $12 < 5 + 5 + 4 = 14$.
So $a = 12$ is achievable.

$$a = 12: \;\; b = c = 5,\; d = 4 \;\Rightarrow\; 12 + 5 + 5 + 4 = 26, \;\; 12 < 14 \;\checkmark$$

💡 Plugging $12$ into the inequality and finding a concrete legal quadrilateral is the Grade 5 "write a numerical expression and check" move.

#3 Eliminate Possibilities 6.EE.B.8 Step 5

Eliminate the other choices.
Choice (E) $13$ violates $a < 13$ (would force a degenerate flat figure with the other three summing to exactly $13$).
Choices (A)-(C) are smaller than $12$ and clearly leave room, so they are not maximal.
So the greatest possible side is $12$, choice (D).

$$\max a = 12 \;\Rightarrow\; \textbf{(D)}$$

💡 Once the strict bound $a < 13$ is in hand, the only multiple-choice answer that is both legal and largest is (D) — Grade 6 inequality reasoning made the choice.

[1] #1 3.MD.D.8 Sketch a quadrilateral with one side labelled $4$ and the other three sides unkn

[2] #1 4.MD.A.3 From the picture, if one side $a$ is the longest, the other three sides must clo

[3] #1 6.EE.B.8 Replace $b + c + d$ using the perimeter equation, then solve the resulting one-v

[4] #6 5.OA.A.2 Walk down the answer choices from the top. (E) $13$ fails because the inequality

[5] #3 6.EE.B.8 Eliminate the other choices. Choice (E) $13$ violates $a < 13$ (would force a de

Review

Reasonableness: Double-check the boundary. If $a = 13$, then $b + c + d = 13$, which equals $a$ — the polygon inequality is violated (the loop would lie flat on top of the long side), so $13$ truly fails. If $a = 12$, the explicit quadrilateral with sides $12, 5, 5, 4$ has perimeter $26$ and satisfies $12 < 14$, so it really exists. The answer $12$ matches choice (D). Magnitude sanity: the bound is "less than half the perimeter," and half of $26$ is $13$ — the answer $12$ is the largest integer below that.

Alternative: Tool #13 (Convert to Algebra): set up $a + b + c + 4 = 26$ and impose $a < b + c + 4$ (polygon inequality for the largest side). Substituting $b + c = 22 - a$ gives $a < 26 - a$, hence $a < 13$. The largest integer below $13$ is $12$. Same answer (D).

CCSS standards used (min grade 6)

3.MD.D.8 Solve real-world problems involving perimeters of polygons (Reading the perimeter $a + b + c + d = 26$ as the closed loop around the quadrilateral and using $d = 4$ to get $a + b + c = 22$.)
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Extending the perimeter-of-a-polygon picture to the polygon-inequality intuition: the longest side cannot equal or exceed the path of the others without collapsing the figure.)
5.OA.A.2 Write simple expressions that record calculations with numbers (Constructing the concrete witness quadrilateral $(12, 5, 5, 4)$ and verifying $12 + 5 + 5 + 4 = 26$ and $12 < 5 + 5 + 4$.)
6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ and graph on a number line (Solving the substituted inequality $a < 26 - a$ to get $a < 13$, then picking the largest integer that satisfies it.)

⭐ This AMC 10 problem only needs Grade 6 "$x < c$" inequality reasoning — one side can't be more than half the perimeter, so the cap is $\tfrac{26}{2} = 13$ exclusive, giving $12$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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