AMC 8 · 2015 · #8
Grade 6 geometry-2dProblem
What is the smallest whole number larger than the perimeter of any triangle with a side of length 5 and a side of length 19?
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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