AMC 10 · 2010 · #16
Grade 7 geometry-2dNondegenerate △ABC has integer side lengths, BD is an angle bisector, AD=3, and DC=8. What is the smallest possible value of the perimeter?
Pick an answer.
AMC 10 2010 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: Triangle ABC is nondegenerate and has whole-number side lengths. The bisector of angle B meets side AC at point D, splitting it into AD = 3 and DC = 8. Find the smallest possible perimeter of the triangle.
Givens: Triangle ABC is nondegenerate and has integer side lengths.; BD bisects angle B and meets AC at D.; AD = 3 and DC = 8.; Answer choices: (A) $30$, (B) $33$, (C) $35$, (D) $36$, (E) $37$
Unknowns: The smallest possible perimeter $AB + BC + CA$.
Understand
Restated: Triangle ABC is nondegenerate and has whole-number side lengths. The bisector of angle B meets side AC at point D, splitting it into AD = 3 and DC = 8. Find the smallest possible perimeter of the triangle.
Givens: Triangle ABC is nondegenerate and has integer side lengths.; BD bisects angle B and meets AC at D.; AD = 3 and DC = 8.; Answer choices: (A) $30$, (B) $33$, (C) $35$, (D) $36$, (E) $37$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities, #14 Extreme Principle
Tool #4 (Introduce a Variable): the Angle Bisector Theorem forces $AB$ and $BC$ into the fixed ratio $3:8$, so a single whole-number factor $k$ captures both unknown sides at once and turns the whole problem into one variable. Tool #1 (Draw a Diagram): sketching the bisector keeps the two pieces of $AC$ straight and shows why $CA = 11$ is fixed. Tool #3 (Eliminate Possibilities): the triangle inequality becomes two inequalities in $k$ that throw out every value except one. Tool #14 (Extreme Principle): the question asks for the smallest perimeter, so I look for the smallest legal value of $k$.
Execute — Answer: B
7.G.A.2 Step 1 Fix the base AC from its two pieces
- The bisector from B lands on the opposite side AC, so D sits on AC and the two pieces add back to the whole side: $CA = AD + DC = 3 + 8 = 11$.
- This length is locked no matter what the other sides turn out to be.
- Sketch the triangle with the bisector so the pieces stay clear.
💡 A bisector drawn from a vertex always hits the opposite side, so its two pieces just add up to that side.
7.RP.A.2 Step 2 Angle Bisector Theorem turns the split into a side ratio
- The Angle Bisector Theorem says the bisector from B cuts the opposite side into pieces proportional to the two sides meeting at B.
- The piece next to A (length 3) pairs with side AB, and the piece next to C (length 8) pairs with side BC.
- So $AB : BC = AD : DC = 3 : 8$.
💡 The bisector shares the far side in the same proportion as the two sides hugging that corner.
6.NS.B.4 Step 3 Name both sides with one whole-number factor
- The sides $AB$ and $BC$ are in the ratio $3 : 8$.
- Since 3 and 8 share no common factor ($\gcd(3,8) = 1$), the only way to get whole-number sides in this ratio is to scale both by the same positive integer $k$.
- So write $AB = 3k$ and $BC = 8k$.
💡 When a ratio is already in lowest terms, the only whole-number pairs keeping it are equal multiples of the two numbers.
7.EE.B.4 Step 4 Triangle inequality pins k to a single value
- In a real triangle each side is shorter than the sum of the other two.
- Using $CA = 11$: from $AB + BC > CA$ we get $3k + 8k > 11$, so $11k > 11$ and $k > 1$.
- From $AB + CA > BC$ we get $3k + 11 > 8k$, so $11 > 5k$ and $k < 2.2$.
- The third inequality $BC + CA > AB$ always holds.
- The only integer with $1 < k < 2.2$ is $k = 2$.
💡 Two inequalities squeeze $k$ from both sides until only one whole number is left standing.
7.EE.B.3 Step 5 Add the sides for the smallest perimeter
- With $k = 2$ the sides are $AB = 6$, $BC = 16$, and $CA = 11$.
- Because $k = 2$ is the only value that survives, this is both the smallest and the only possible triangle.
- Its perimeter is $6 + 16 + 11 = 33$, so the answer is (B).
💡 Once the smallest legal value of $k$ is fixed, adding the three sides gives the smallest perimeter directly.
7.G.A.2 The bisector from B lands on the opposite side AC, so D sits on AC and the two p 7.RP.A.2 The Angle Bisector Theorem says the bisector from B cuts the opposite side into 6.NS.B.4 The sides $AB$ and $BC$ are in the ratio $3 : 8$. Since 3 and 8 share no common 7.EE.B.4 In a real triangle each side is shorter than the sum of the other two. Using $CA 7.EE.B.3 With $k = 2$ the sides are $AB = 6$, $BC = 16$, and $CA = 11$. Because $k = 2$ i Review
Reasonableness: Check the sides $6$, $16$, $11$: $6 + 11 = 17 > 16$, $6 + 16 = 22 > 11$, and $11 + 16 = 27 > 6$, so the triangle is genuinely nondegenerate. The ratio $AB : BC = 6 : 16 = 3 : 8$ matches the split $3 : 8$, and the perimeter $33$ is choice (B). Testing $k = 1$ gives sides $3, 8, 11$ with $3 + 8 = 11$ — a flat, degenerate triangle, correctly rejected. Testing $k = 3$ gives $9, 24, 11$ with $9 + 11 = 20 < 24$, impossible. So $33$ is the unique valid perimeter.
Alternative: Skip the bounding and use the algebra of the perimeter itself. With $AB = 3k$ and $BC = 8k$, the perimeter is $3k + 8k + 11 = 11k + 11 = 11(k + 1)$, which is always a multiple of $11$. Among the five answer choices, only $33$ is a multiple of $11$, so the answer must be (B) without even solving the inequalities.
CCSS standards used (min grade 7)
7.G.A.2Draw geometric shapes with given conditions including triangles (Placing D on side AC and adding the two pieces to fix $CA = 3 + 8 = 11$.)7.RP.A.2Recognize and represent proportional relationships between quantities (Turning the Angle Bisector Theorem into the fixed side ratio $AB : BC = 3 : 8$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Using $\gcd(3,8) = 1$ to justify writing the integer sides as $AB = 3k$ and $BC = 8k$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Building the triangle-inequality bounds $k > 1$ and $k < 2.2$ and solving for the integer $k = 2$.)7.EE.B.3Solve multi-step real-world problems posed with positive and negative rational numbers (Substituting $k = 2$ into $11k + 11$ to compute the smallest perimeter, $33$.)
⭐ An angle bisector splits the far side in the same ratio as the two sides at that corner, so name the sides as equal multiples and let the triangle inequality squeeze out the one value that fits.
⭐ An angle bisector splits the far side in the same ratio as the two sides at that corner, so name the sides as equal multiples and let the triangle inequality squeeze out the one value that fits.
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