AMC 8 · 2015 · #8
Easy mode Grade 6Problem
Picture a triangle. One of its sides has length 5, and another side has length 19. The third side could be many different lengths.
The perimeter of the triangle is the total of all three sides. Depending on the third side, the perimeter can be many different numbers.
What is the smallest whole number that is bigger than every possible perimeter?
Pick an answer.
AMC 8 2015 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: A triangle has two sides of length $5$ and $19$. The third side can be any length that still forms a real triangle. Find the smallest whole number that is strictly greater than the perimeter, no matter which valid third side is chosen.
Givens: Side $a = 5$; Side $b = 19$; Third side $x$ can be any value that produces a valid triangle; Answer choices: (A) $24$, (B) $29$, (C) $43$, (D) $48$, (E) $57$
Unknowns: The smallest whole number larger than every possible perimeter $P = 5 + 19 + x = 24 + x$
Understand
Restated: A triangle has two sides of length $5$ and $19$. The third side can be any length that still forms a real triangle. Find the smallest whole number that is strictly greater than the perimeter, no matter which valid third side is chosen.
Givens: Side $a = 5$; Side $b = 19$; Third side $x$ can be any value that produces a valid triangle; Answer choices: (A) $24$, (B) $29$, (C) $43$, (D) $48$, (E) $57$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems, #6 Guess and Check
The hidden gate is figuring out which values of $x$ are even allowed. Tool #1 (Draw a Diagram) shows it directly: lay the side of length $19$ flat, then swing the side of length $5$. The far tip can reach at most $19 + 5 = 24$ (when both are stretched in the same direction) and at least $19 - 5 = 14$ (when folded back), so $x$ must satisfy $14 < x < 24$ — those are the limits where the triangle flattens into a line. Tool #9 (Solve an Easier Related Problem) confirms the same rule with a friendly case (sides $3$ and $4$ allow $1 < x < 7$). Once we know $x < 24$, the perimeter $P = 24 + x < 48$, and Tool #6 (Guess and Check) on the choices picks the smallest whole number that stays above every legal $P$.
Execute — Answer: D
4.G.A.2 Step 1 - Sketch the situation.
- Draw the side of length $19$ as a flat segment on the ground.
- Attach the side of length $5$ at one end and let it swing like a hand on a clock.
- The third side $x$ runs from the free tip of the $5$ to the far end of the $19$.
💡 Drawing a picture turns the abstract "is this a triangle?" question into something you can see: when the $5$ points straight out along the $19$, the tip lands at $24$; when it folds back along the $19$, the tip lands at $14$.
4.G.A.2 Step 2 - Read the limits off the diagram.
- As the side of length $5$ swings, the third side $x$ can be as close to $24$ as you like (almost-flat triangle pointing out) or as close to $14$ as you like (almost-flat triangle folded in), but it can never reach either value, because then the three sides would lie on one line instead of forming a triangle.
💡 This is the Triangle Inequality discovered visually: the third side has to be longer than the difference of the other two and shorter than their sum.
4.G.A.2 Step 3 - Double-check the rule on an easier related problem.
- Sides $3$ and $4$: a swing argument gives $4 - 3 < x < 4 + 3$, so $1 < x < 7$.
- Try $x = 7$: $3 + 4 = 7$ exactly, the triangle collapses to a line — confirmed.
💡 Testing the rule on small numbers — Tool #9's job — makes the inequality feel solid before we trust it on the real numbers.
6.EE.B.8 Step 4 - Write the perimeter as a function of $x$, then apply the bound on $x$.
- Since $x$ is strictly less than $24$, the perimeter is strictly less than $24 + 24 = 48$.
- It can get as close to $48$ as we want (e.g., $x = 23.99$ gives $P = 47.99$), but it never actually reaches $48$.
💡 Adding the same number to both sides of the inequality $x < 24$ keeps it true — a Grade 6 inequality move.
6.EE.B.8 Step 5 - Find the smallest whole number that beats every possible perimeter.
- Guess each choice in turn.
- $47$ fails because $P$ could be $47.5$.
- $48$ works because $P < 48$ always.
- Anything larger ($57$) works too but isn't the smallest.
- So $48$ is the answer.
💡 Checking the choices against the bound is the Tool #6 (Guess and Check) finish: the right answer is the smallest whole number sitting just above the open boundary.
4.G.A.2 Sketch the situation. Draw the side of length $19$ as a flat segment on the grou 4.G.A.2 Read the limits off the diagram. As the side of length $5$ swings, the third sid 4.G.A.2 Double-check the rule on an easier related problem. Sides $3$ and $4$: a swing a 6.EE.B.8 Write the perimeter as a function of $x$, then apply the bound on $x$. Since $x$ 6.EE.B.8 Find the smallest whole number that beats every possible perimeter. Guess each c Review
Reasonableness: Sanity-test the bound. Pick a legal $x$, say $x = 20$: the sides $5, 19, 20$ satisfy $5 + 19 = 24 > 20$, so it is a real triangle, and $P = 44 < 48$. Push $x$ toward the limit: $x = 23.9$ gives $P = 47.9 < 48$. Push past: $x = 24$ gives sides $5, 19, 24$ with $5 + 19 = 24$ (collapsed line, not a triangle). The cutoff sits exactly at $48$, confirming the answer (D).
Alternative: Use the Triangle Inequality algebraically (Tool #13). For sides $5, 19, x$: each side must be less than the sum of the other two. The binding case is $x < 5 + 19 = 24$ (the other two cases give $x > 14$ and the trivial $19 + x > 5$). So $P = 24 + x < 48$, and the smallest whole number above this open bound is $48$.
CCSS standards used (min grade 6)
4.G.A.2Classify two-dimensional figures based on properties (Recognizing when three given side lengths can form an actual triangle versus collapsing into a straight line — a Grade 4 geometry-properties argument done visually with a swinging-side diagram.)6.EE.B.8Write inequalities to represent constraints, and recognize their solutions (Translating the geometric bound $14 < x < 24$ into the perimeter inequality $P = 24 + x < 48$, and reading off the smallest whole number above an open upper bound.)
⭐ This AMC 8 problem only needs Grade 6 inequality reasoning — once you see why the third side has to stay under $5 + 19 = 24$, the perimeter cap of $48$ falls right out!
⭐ This AMC 8 problem only needs Grade 6 inequality reasoning — once you see why the third side has to stay under $5 + 19 = 24$, the perimeter cap of $48$ falls right out!
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