AMC 10 · 2006 · #19
Grade 8 arithmeticHow many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?
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: Count the triangles whose three angle measures are all different positive whole numbers of degrees and form an arithmetic progression (each angle is a fixed step larger than the one before). Two triangles that have the same three angles count as the same shape (similar), so only differently-angled triangles are counted.
Givens: The three angles of the triangle are in arithmetic progression.; Each angle is a positive integer number of degrees.; The three angle measures are distinct (all different).; The three interior angles of any triangle add up to $180^\circ$.; Answer choices: (A) $0$, (B) $1$, (C) $59$, (D) $89$, (E) $178$.
Unknowns: How many triangles that are not similar to each other satisfy all the conditions.
Understand
Restated: Count the triangles whose three angle measures are all different positive whole numbers of degrees and form an arithmetic progression (each angle is a fixed step larger than the one before). Two triangles that have the same three angles count as the same shape (similar), so only differently-angled triangles are counted.
Givens: The three angles of the triangle are in arithmetic progression.; Each angle is a positive integer number of degrees.; The three angle measures are distinct (all different).; The three interior angles of any triangle add up to $180^\circ$.; Answer choices: (A) $0$, (B) $1$, (C) $59$, (D) $89$, (E) $178$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #14 Extreme Principle, #2 Make a Systematic List
The angles are tied together by two rules — they form an arithmetic progression and they sum to $180^\circ$ — so the natural first move is to name them with a variable (Tool #4). Writing the three angles as $60-d,\;60,\;60+d$ makes the common difference $d$ the single thing that decides the whole triangle. Then the only question is: which values of $d$ are allowed? That is a boundary question, so the Extreme Principle (Tool #14) pins down the smallest and largest legal $d$ from the 'positive integer' and 'distinct' rules. Finally, because each allowed $d$ gives one differently-angled (non-similar) triangle, counting the whole-number values of $d$ in that range is a clean systematic count (Tool #2).
Execute — Answer: C
6.EE.A.2 Step 1 Name the three angles
- An arithmetic progression steps up by the same amount each time.
- Call the smallest angle $a$ and the common step $d$ (with $d>0$ so the angles really do increase and stay distinct).
- Then the three angles are $a$, $a+d$, and $a+2d$.
💡 One starting value plus one common step describes any arithmetic progression, so two letters capture all three angles.
8.G.A.5 Step 2 Use the triangle angle-sum rule
- The three interior angles of a triangle always add to $180^\circ$.
- Adding the three expressions gives $a+(a+d)+(a+2d) = 3a+3d = 3(a+d)$.
- Setting this equal to $180$ isolates the middle angle.
💡 In any 3-term arithmetic progression the sum is three times the middle term, so the angle-sum rule locks the middle term.
6.EE.B.7 Step 3 The middle angle is fixed at 60
- Dividing $3(a+d)=180$ by $3$ gives $a+d = 60$.
- So the middle angle is always $60^\circ$, no matter which triangle we pick.
- Rewriting the three angles around this center, they become $60-d$, $60$, and $60+d$.
💡 Because the average of the three angles is $180/3 = 60$, the middle angle must be exactly $60^\circ$.
7.EE.B.4 Step 4 Find the allowed values of d
- Every angle must be a positive integer, so $d$ must be a whole number, and the smallest angle $60-d$ must be at least $1$.
- That forces $60-d \ge 1$, i.e.
- $d \le 59$.
- Distinctness needs $d \ge 1$ (if $d=0$ all three angles equal $60$).
- The largest angle $60+d$ stays under $180$ automatically.
- So $d$ can be any integer from $1$ to $59$.
💡 Push the smallest angle to its limit: it can shrink to $1^\circ$ but no further, which caps how big the step $d$ can be.
7.EE.B.4 Step 5 Count the triangles
- Each whole-number $d$ from $1$ to $59$ gives one set of angles $60-d,\,60,\,60+d$, and different values of $d$ give different angle sets — hence non-similar triangles.
- Using $-d$ instead of $d$ just relabels the same three angles, so it is not a new triangle.
- Counting the integers $1,2,\dots,59$ gives $59$ triangles, which is choice (C).
💡 One legal step size equals one triangle shape, so the answer is simply how many step sizes fit between $1$ and $59$.
6.EE.A.2 An arithmetic progression steps up by the same amount each time. Call the smalle 8.G.A.5 The three interior angles of a triangle always add to $180^\circ$. Adding the th 6.EE.B.7 Dividing $3(a+d)=180$ by $3$ gives $a+d = 60$. So the middle angle is always $60 7.EE.B.4 Every angle must be a positive integer, so $d$ must be a whole number, and the s 7.EE.B.4 Each whole-number $d$ from $1$ to $59$ gives one set of angles $60-d,\,60,\,60+d Review
Reasonableness: Spot-check the endpoints. With $d=1$ the angles are $59,60,61$ — three distinct positive integers summing to $180$, valid. With $d=59$ they are $1,60,119$ — still valid. With $d=60$ the smallest angle would be $0^\circ$, which is not a real angle, so $59$ is correctly the last one. There is exactly one triangle per step size, so the total is $59$, matching choice (C). The tempting wrong answers check out as traps: $89$ would come from mistakenly allowing the middle angle to vary, and $178$ from double-counting $d$ and $-d$ as separate triangles.
Alternative: List the smallest angle directly. Since the middle angle is $60$, the smallest angle $60-d$ can be any integer from $1$ up to $59$ (it must be below $60$ to keep the angles increasing and above $0$ to be a real angle). That is $59$ possible smallest angles, and each one fixes the whole triangle, giving $59$ non-similar triangles.
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Writing the three arithmetic-progression angles as the expressions $a$, $a+d$, $a+2d$ using a starting value and a common step.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using the fact that a triangle's three interior angles sum to $180^\circ$ to set up $3(a+d)=180$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $3(a+d)=180$ to get the fixed middle angle $a+d=60$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Setting up and solving $60-d \ge 1$ and $d \ge 1$ to bound the common difference, then counting the integer values $1 \le d \le 59$.)
⭐ Because three angles in arithmetic progression must average $60^\circ$, the middle angle is always $60$; the step size can be any whole number from $1$ to $59$, giving $59$ different triangles.
⭐ Because three angles in arithmetic progression must average $60^\circ$, the middle angle is always $60$; the step size can be any whole number from $1$ to $59$, giving $59$ different triangles.
More like this
Same archetype — closest grade level first.