AMC 8 · 2005 · #15
Easy mode Grade 5Problem
Picture an isosceles triangle. That means two of its sides have the same length.
The three side lengths are all whole numbers, and the three sides add up to 23.
How many different isosceles triangles like this are there?
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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