Competition · AMC preparation · step 4 of 4
AMC 8 · 2010 · #13
Grade 6 geometry-2dPick an answer.
AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 ∈ {7,8,9,10,11}, the three sides are L-2, L-1, L, the perimeter is 3L - 3, and we just check whether shortest = 0.3 × 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.
Set up the systematic list
If the longest side is L, the three sides are L-2, L-1, L and the perimeter is 3L-3; test whether L-2 = 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.
Labelling the sides downward from the longest side L as L-2, L-1, L, the perimeter is 3L-3, and the condition "the shortest side is 30% of the perimeter" becomes exactly the test L-2 = 0.3 (3L-3).
▸ Why?
The perimeter is the whole boundary made by laying the three sides end to end, so it equals their sum (L-2)+(L-1)+L, which tidies to 3L-3.
▸ Why?
Laying the three sides end to end joins them into one length with nothing missing or counted twice, so the perimeter equals the three parts added together, (L-2)+(L-1)+L.
▸ Why?
In that sum the three L's may be gathered together first and the constants -2 and -1 added separately, because regrouping which numbers you add first never changes the total.
▸ Why?
The gathered L+L+L is three equal groups of L, which is what 3L means, while the leftover -2 and -1 combine to -3, giving the perimeter 3L-3.
▸ Why?
"30% of the perimeter" means 30 out of every hundred of it, and 30 per hundred is the fraction 30/100=0.3, so it equals 0.3 (3L-3); the word "is" in "the shortest side is 30% of the perimeter" sets the shortest side L-2 equal to that amount, giving the test L-2 = 0.3 (3L-3).
Test 7 as the longest side
Try (A) L=7: sides (5,6,7), perimeter 18, but 0.3 × 18 = 5.4 ≠ 5, so reject (A).
Multiplying 18 by 0.3 is the same as 18 × 3 ÷ 10 = 54 ÷ 10 = 5.4 — a Grade 5 decimal-times-whole-number move.
5.NBT.B.7Guess And CheckTest 8, 9, and 10
Try L=8, 9, 10: perimeters 21, 24, 27 give 0.3 × = 6.3, 7.2, 8.1, none matching the shortest side, so reject (B), (C), (D).
Each candidate misses by a small amount — the gap between shortest and 30% of perimeter keeps shrinking, so we are on the right track.
5.NBT.B.7Eliminate PossibilitiesTest 11 and confirm
Try (E) L=11: sides (9,10,11), perimeter 30, and 0.3 × 30 = 9 exactly matches the shortest side, so the longest side is 11.
Finding the percent of a number (30% of 30) and matching it to a target is core Grade 6 percent reasoning.
6.RP.A.3Guess And CheckThis AMC 8 problem only needs Grade 6 percent reasoning — "30% of the perimeter" — that you already know!
- Set up the systematic list
- Test 7 as the longest side
- Test 8, 9, and 10
- Test 11 and confirm
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