AMC 10 · 2023 · #4

Grade 6 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

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

📏 Medium solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Sketch the quadrilateral with one side labelled 4; to maximize another side, push it out while the other three close the loop.

a + b + c + d = 26, d = 4 → a + b + c = 22

2STEP 2

If a is the longest side, the other three must total more than a to close the loop instead of flattening — the polygon inequality.

a ＜ b + c + d

3STEP 3

Replace b + c + d with 26 − a, giving a ＜ 26 − a, which simplifies to a ＜ 13.

a ＜ 26 - a ⟹ 2a ＜ 26 ⟹ a ＜ 13

4STEP 4

Walk down the choices: (E) 13 fails the strict bound; a = 12 works with sides 12, 5, 5, 4 since 12 ＜ 14.

a = 12: b = c = 5, d = 4 → 12 + 5 + 5 + 4 = 26, 12 ＜ 14 ✓

5STEP 5

(E) 13 violates a ＜ 13; the smaller choices leave room but aren't maximal, so the greatest side is 12 — choice (D).

max a = 12 → (D)

Answer

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.

💡Key takeaway

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 $\frac{26}{2}$ = 13 exclusive, giving 12.

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