AMC 8 · 2005 · #15
Grade 5 geometry-2dProblem
How many different isosceles triangles have integer side lengths and perimeter 23?
Pick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count the isosceles triangles whose three side lengths are positive integers and whose perimeter is $23$.
Givens: The triangle is isosceles, so two sides are equal; All three side lengths are positive integers; The perimeter (sum of the three sides) is $23$; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $9$, (E) $11$
Unknowns: The number of different isosceles triangles that meet all three conditions
Understand
Restated: Count the isosceles triangles whose three side lengths are positive integers and whose perimeter is $23$.
Givens: The triangle is isosceles, so two sides are equal; All three side lengths are positive integers; The perimeter (sum of the three sides) is $23$; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $9$, (E) $11$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #6 Guess and Check
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 $11 \times 1 = 11$.
Execute — Answer: C
3.MD.D.8 Step 1 - Set up the systematic list.
- An isosceles triangle has sides $(a, a, b)$, so the perimeter equation is $a + a + b = 23$, which gives $b = 23 - 2a$.
- List candidates in order of increasing $a$, starting from the smallest $a$ that keeps $b$ at least $1$.
💡 Grade 3 perimeter says "add the sides." Solving the perimeter equation for $b$ turns the unknown triangle into a one-variable search.
4.OA.A.3 Step 2 - Find the largest $a$ allowed.
- Since $b = 23 - 2a$ must be a positive integer, $23 - 2a \ge 1$, so $a \le 11$.
- The smallest $a$ allowed is $a = 1$ (which gives $b = 21$, an honest positive integer).
- So $a$ ranges over $\{1, 2, \ldots, 11\}$ before checking the triangle rule.
💡 The Grade 4 multi-step move: turn a word constraint ("$b$ is a positive whole number") into a numeric bound on $a$.
5.G.B.3 Step 3 - Test the triangle inequality on each candidate.
- For sides $(a, a, b)$, the only inequality that can fail is $a + a > b$ — the other two ($a + b > a$) are automatic once $b > 0$.
- Check each value of $a$ from $1$ to $11$:
💡 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.
4.OA.A.3 Step 4 - Count the rows that passed.
- The valid triangles are at $a = 6, 7, 8, 9, 10, 11$.
- That is six triangles.
💡 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.
3.MD.D.8 Set up the systematic list. An isosceles triangle has sides $(a, a, b)$, so the 4.OA.A.3 Find the largest $a$ allowed. Since $b = 23 - 2a$ must be a positive integer, $2 5.G.B.3 Test the triangle inequality on each candidate. For sides $(a, a, b)$, the only 4.OA.A.3 Count the rows that passed. The valid triangles are at $a = 6, 7, 8, 9, 10, 11$. Review
Reasonableness: Spot-check the boundary $a = 6$: sides $(6, 6, 11)$ give $6 + 6 = 12 > 11$, just barely a triangle — and dropping to $a = 5$ gives $(5, 5, 13)$ with $5 + 5 = 10 < 13$, which fails. So the cutoff $a \ge 6$ is sharp. On the other end, $a = 11$ gives $(11, 11, 1)$, a very thin triangle but legal since $11 + 11 > 1$. The six valid triangles are $(6,6,11)$, $(7,7,9)$, $(8,8,7)$, $(9,9,5)$, $(10,10,3)$, $(11,11,1)$ — all have perimeter $23$ and all satisfy $2a > b$, confirming answer (C).
Alternative: Tool #13 (Convert to Algebra): combine $b = 23 - 2a$ with the triangle inequality $2a > b$ to get $2a > 23 - 2a$, i.e. $4a > 23$, so $a > 5.75$. Together with $a \le 11$, the integers $a = 6, 7, 8, 9, 10, 11$ — exactly six — give answer (C). The algebra skips the row-by-row check, but the systematic list above also shows the failing cases, which is more honest for a learner who is still building intuition.
CCSS standards used (min grade 5)
3.MD.D.8Solve real world and mathematical problems involving perimeters of polygons (Reading the perimeter condition $a + a + b = 23$ to write $b = 23 - 2a$, turning two unknown sides into one.)4.OA.A.3Solve multistep word problems posed with whole numbers, including problems in which remainders must be interpreted (Combining the perimeter equation and the positive-integer requirement to bound $a$ at $a \le 11$, then tallying the survivors of the triangle-inequality test.)5.G.B.3Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category (Using the defining attribute of an isosceles triangle (two equal sides) and the triangle inequality to decide which side-length triples really form a triangle.)
⭐ When 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.
⭐ When 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.
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