AMC 8 · 2010 · #13

Grade 6 geometry-2d
perimetersequences-arithmeticpercentagelinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: perimeterpercentagelinear-equations-one-var
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Problem

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30%30\% of the perimeter. What is the length of the longest side?

Pick an answer.

(A)
7
(B)
8
(C)
9
(D)
10
(E)
11
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Toolkit + CCSS Solution

Understand

Restated: A triangle has three side lengths (in inches) that are three consecutive integers. The shortest side is exactly $30\%$ of the perimeter. Find the length of the longest side.

Givens: Side lengths are three consecutive integers, e.g. $n,\;n+1,\;n+2$; Shortest side $= 30\% = \tfrac{3}{10}$ of the perimeter; Perimeter $=$ sum of the three side lengths; Answer choices for the longest side: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$

Unknowns: The length of the longest side (one of the five choices)

Understand

Restated: A triangle has three side lengths (in inches) that are three consecutive integers. The shortest side is exactly $30\%$ of the perimeter. Find the length of the longest side.

Givens: Side lengths are three consecutive integers, e.g. $n,\;n+1,\;n+2$; Shortest side $= 30\% = \tfrac{3}{10}$ of the perimeter; Perimeter $=$ sum of the three side lengths; Answer choices for the longest side: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

The reference solution uses Tool #13 (Algebra): set up $n = 0.3(3n+3)$ and solve. But the answer choices give us the longest side directly, and each choice pins down all three sides. So Tool #6 (Guess and Check) is faster and needs no variable: for each choice $L \in \{7,8,9,10,11\}$, the three sides are $L-2,\;L-1,\;L$, the perimeter is $3L - 3$, and we just check whether $\text{shortest} = 0.3 \times \text{perimeter}$. Tool #2 (Systematic List) keeps the five cases organized; Tool #3 (Eliminate Possibilities) crosses off any choice that misses the $30\%$ target. This path uses only multiplication and basic decimals — no algebra.

Execute — Answer: E

#2 Make a Systematic List 4.OA.C.5 Step 1
  • Set up the systematic list.
  • If the longest side is $L$, the three consecutive integer sides are $L-2,\;L-1,\;L$, so the perimeter is $(L-2) + (L-1) + L = 3L - 3$.
  • The shortest side is $L - 2$, and the $30\%$ test is whether $L - 2 = 0.3 \times (3L - 3)$.
$$\text{sides} = (L-2,\;L-1,\;L),\quad P = 3L - 3,\quad \text{test: } L - 2 \;\stackrel{?}{=}\; 0.3\,(3L - 3)$$

💡 Writing all five cases in the same template $(L-2, L-1, L)$ is the systematic-list move — same shape, only $L$ changes.

#6 Guess and Check 5.NBT.B.7 Step 2
  • Try (A) $L = 7$: sides $(5, 6, 7)$, perimeter $5 + 6 + 7 = 18$.
  • Check: is $5 = 0.3 \times 18$?
  • $0.3 \times 18 = 5.4$, not $5$.
  • Reject (A).
$$0.3 \times 18 = 5.4 \neq 5\;\;\text{(A) rejected}$$

💡 Multiplying $18$ by $0.3$ is the same as $18 \times 3 \div 10 = 54 \div 10 = 5.4$ — a Grade 5 decimal-times-whole-number move.

#3 Eliminate Possibilities 5.NBT.B.7 Step 3
  • Try (B) $L = 8$: sides $(6, 7, 8)$, perimeter $6 + 7 + 8 = 21$.
  • Check: $0.3 \times 21 = 6.3 \neq 6$.
  • Reject (B).
  • Try (C) $L = 9$: sides $(7, 8, 9)$, perimeter $24$.
  • Check: $0.3 \times 24 = 7.2 \neq 7$.
  • Reject (C).
  • Try (D) $L = 10$: sides $(8, 9, 10)$, perimeter $27$.
  • Check: $0.3 \times 27 = 8.1 \neq 8$.
  • Reject (D).
$$\begin{aligned}L=8:&\;0.3\times 21 = 6.3 \neq 6\\L=9:&\;0.3\times 24 = 7.2 \neq 7\\L=10:&\;0.3\times 27 = 8.1 \neq 8\end{aligned}$$

💡 Each candidate misses by a small amount — the gap between shortest and $30\%$ of perimeter keeps shrinking, so we are on the right track.

#6 Guess and Check 6.RP.A.3 Step 4
  • Try (E) $L = 11$: sides $(9, 10, 11)$, perimeter $9 + 10 + 11 = 30$.
  • Check: $0.3 \times 30 = 9$, which exactly equals the shortest side $9$.
  • The $30\%$ condition is satisfied, so the longest side is $11$.
$$L = 11:\;\text{sides }(9,10,11),\;P=30,\;0.3\times 30 = 9 = \text{shortest}\;\checkmark\;\Rightarrow\;\textbf{(E)}$$

💡 Finding the percent of a number ($30\%$ of $30$) and matching it to a target is core Grade 6 percent reasoning.

[1] #2 4.OA.C.5 Set up the systematic list. If the longest side is $L$, the three consecutive in
[2] #6 5.NBT.B.7 Try (A) $L = 7$: sides $(5, 6, 7)$, perimeter $5 + 6 + 7 = 18$. Check: is $5 = 0
[3] #3 5.NBT.B.7 Try (B) $L = 8$: sides $(6, 7, 8)$, perimeter $6 + 7 + 8 = 21$. Check: $0.3 \tim
[4] #6 6.RP.A.3 Try (E) $L = 11$: sides $(9, 10, 11)$, perimeter $9 + 10 + 11 = 30$. Check: $0.3

Review

Reasonableness: If the three sides were equal, each side would be $\tfrac{1}{3} \approx 33.3\%$ of the perimeter. Here the shortest side is only $30\%$ — a bit less than a third — so the sides must be a little spread apart, not tightly bunched. The winning triple $(9, 10, 11)$ has a shortest side $9$ out of a perimeter $30$, and $9/30 = 0.30$ exactly. Triangle inequality also holds: $9 + 10 = 19 > 11$. Everything checks out.

Alternative: Tool #13 (Algebra): let the shortest side be $n$, so the sides are $n,\;n+1,\;n+2$ and the perimeter is $3n+3$. The condition $n = 0.3(3n+3)$ becomes $n = 0.9n + 0.9$, then $0.1n = 0.9$, so $n = 9$ and the longest side is $n+2 = 11$. Same answer, but it requires setting up and solving a linear equation — Tool #6's table is more elementary.

CCSS standards used (min grade 6)

  • 4.OA.C.5 Generate a number or shape pattern following a given rule (Setting up the template $(L-2,\;L-1,\;L)$ so that each answer choice gives a complete side triple with the same shape.)
  • 5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing $0.3 \times P$ for each candidate perimeter ($5.4,\;6.3,\;7.2,\;8.1,\;9.0$) to test the $30\%$ condition.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Interpreting "shortest side $= 30\%$ of the perimeter" as a percent relationship and verifying it for the winning triple $(9,10,11)$.)

⭐ This AMC 8 problem only needs Grade 6 percent reasoning — "$30\%$ of the perimeter" — that you already know!

⭐ This AMC 8 problem only needs Grade 6 percent reasoning — "$30\%$ of the perimeter" — that you already know!