Competition · AMC preparation · step 4 of 4
AMC 8 · 2005 · #15
Grade 5 geometry-2dPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem asks "how many," the candidate set is small and finite, and each candidate is easy to test — three classic signs that Tool #2 (Systematic List) is the right primary. Order the candidates by the repeated side a: once a is fixed, the perimeter forces the third side b = 23 - 2a, so there is at most one triangle per value of a. Tool #6 (Guess and Check) then handles each candidate the same way — write the three sides and check the triangle inequality a + a ＞ b. No algebra is needed; the longest case to handle is 11 × 1 = 11.
Write b in terms of a
Set up the list: an isosceles triangle has sides (a, a, b) with a + a + b = 23, so b = 23 - 2a. List candidates by increasing a.
Grade 3 perimeter says "add the sides." Solving the perimeter equation for b turns the unknown triangle into a one-variable search.
3.MD.D.8Make A Systematic ListFind the largest a
Bound a: since b = 23 - 2a ≥ 1, we get a ≤ 11, and a ≥ 1, so a runs over 1 to 11 before the triangle test.
The Grade 4 multi-step move: turn a word constraint ("b is a positive whole number") into a numeric bound on a.
4.OA.A.3Make A Systematic ListTest the triangle inequality
Test the triangle inequality a + a ＞ b for each a from 1 to 11: it fails for a ≤ 5 and holds for a ≥ 6.
Grade 5 classifies triangles by side length. The triangle inequality is just the "can these sides actually close into a triangle?" check — for an isosceles, 2a ＞ b is all you need.
The three lengths a, a, b form a real triangle exactly when the two equal sides together are longer than the third side, that is 2a ＞ b.
▸ Why?
A triangle needs the two equal sides to meet at a corner above the third side; the third side is the straight gap between their two lower ends, so the two equal sides together must reach farther than that gap in order to close up.
▸ Why?
The two equal sides make a bent route from one end of the third side to the other, and a bent route between two points is always longer than the straight segment joining them, so their combined length 2a has to be greater than b.
Count the passing rows
Count the passing rows a = 6 to 11: six triangles in all.
Once the systematic list is done, counting the rows that survive is just whole-number tally — the Grade 4 multi-step problem ends with a count.
4.OA.A.3Make A Systematic ListWhen a problem asks "how many," make a systematic list — once the perimeter pins the third side to the repeated side, the only question per row is whether the two equal sides are long enough to close the triangle.
- Write b in terms of a
- Find the largest a
- Test the triangle inequality
- Count the passing rows
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