AMC 10 · 2006 · #10
Grade 7 geometry-2dIn a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
Pick an answer.
AMC 10 2006 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 whole-number side lengths. One side is three times the length of a second side, and the third side is $15$. Find the largest perimeter such a triangle can have.
Givens: All three side lengths are integers; One side is exactly three times as long as a second side; The remaining (third) side has length $15$; Answer choices: (A) $43$, (B) $44$, (C) $45$, (D) $46$, (E) $47$
Unknowns: The greatest possible perimeter of the triangle
Understand
Restated: A triangle has whole-number side lengths. One side is three times the length of a second side, and the third side is $15$. Find the largest perimeter such a triangle can have.
Givens: All three side lengths are integers; One side is exactly three times as long as a second side; The remaining (third) side has length $15$; Answer choices: (A) $43$, (B) $44$, (C) $45$, (D) $46$, (E) $47$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The question asks for the greatest possible perimeter, and since the perimeter is $4x+15$, that means pushing $x$ as large as the rules allow — a textbook use of Tool #14 (Extreme Principle): the answer lives at the top boundary. To find where that boundary is, Tool #4 (Introduce a Variable) names the short related side $x$, writes the sides as $x$, $3x$, $15$, and turns the triangle condition into inequalities. Tool #3 (Eliminate Possibilities) then rejects any $x$ that is too big to close into a real triangle.
Execute — Answer: A
6.EE.B.6 Step 1 Name the sides with one letter
- Let $x$ be the length of the shorter of the two related sides.
- Then the side that is three times as long is $3x$, and the third side is $15$.
- Every side is now written in terms of $x$, and the perimeter is $x+3x+15=4x+15$.
💡 Writing all three sides from a single unknown makes the perimeter a plain function of that one number.
7.G.A.2 Step 2 Write the triangle condition
- Three lengths form a real triangle only when each side is shorter than the sum of the other two.
- Checking each side against the other two gives $x+3x>15$, $x+15>3x$, and $3x+15>x$.
- The last one, $3x+15>x$, is always true, so only the first two matter.
💡 If one side were as long as the other two combined, the triangle would flatten into a straight line instead of closing up.
7.EE.B.4 Step 3 Solve the two inequalities for x
- Simplify each condition.
- From $x+3x>15$: $4x>15$, so $x>3.75$.
- From $x+15>3x$: $15>2x$, so $x<7.5$.
- Together, $3.75<x<7.5$.
- Because $x$ must be a whole number, the allowed values are $x=4,5,6,7$.
💡 The two triangle rules trap $x$ between a floor and a ceiling, leaving only a short list of legal integers.
4.OA.A.3 Step 4 Take the largest x and add up the sides
- For the greatest perimeter, take the biggest allowed value, $x=7$.
- The sides are $7$, $21$, and $15$.
- Check the tight condition: $7+15=22>21$, so it really is a triangle.
- The perimeter is $7+21+15=43$.
- (Trying $x=8$ would give sides $8,24,15$, but $8+15=23<24$, so it collapses — $x=7$ is the true ceiling.) The answer is $43$, choice (A).
💡 Since perimeter grows with $x$, the biggest legal $x$ gives the biggest perimeter — no need to test the smaller ones.
6.EE.B.6 Let $x$ be the length of the shorter of the two related sides. Then the side tha 7.G.A.2 Three lengths form a real triangle only when each side is shorter than the sum o 7.EE.B.4 Simplify each condition. From $x+3x>15$: $4x>15$, so $x>3.75$. From $x+15>3x$: $ 4.OA.A.3 For the greatest perimeter, take the biggest allowed value, $x=7$. The sides are Review
Reasonableness: The winning triangle has sides $7$, $21$, $15$, all whole numbers, with $21=3\times 7$ and a third side of $15$, so it fits every condition. The two short sides just clear the long one, $7+15=22>21$, meaning the triangle is barely valid — exactly what you expect at the maximum. All the larger answer choices $44,45,46,47$ would need $x>7$, which fails the triangle inequality, so $43$ is the only reachable one.
Alternative: Skip solving inequalities symbolically and just test integers from the top down. $x=8$ gives $8,24,15$ and $8+15=23<24$ (fails); $x=7$ gives $7,21,15$ and $7+15=22>21$ (works). The first success from above is the maximum, giving perimeter $7+21+15=43$, choice (A).
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $x$ be the short related side and writing the sides as $x$, $3x$, $15$ with perimeter $4x+15$.)7.G.A.2Draw geometric shapes with given conditions including triangles (Applying the rule that three lengths form a triangle only when each side is less than the sum of the other two.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Building and solving $4x>15$ and $15>2x$ to trap $x$ in $3.75<x<7.5$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding $7+21+15$ to get the perimeter $43$ at the extreme value $x=7$.)
⭐ When a problem asks for the biggest possible answer, find the boundary the rules allow and push right up to it — here the triangle inequality caps the short side at $7$, giving perimeter $43$.
⭐ When a problem asks for the biggest possible answer, find the boundary the rules allow and push right up to it — here the triangle inequality caps the short side at $7$, giving perimeter $43$.
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