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Difference between revisions of "2006 AMC 10A Problems/Problem 19"

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==Solution 3 (Quick Summation)==
 
==Solution 3 (Quick Summation)==
 
Consider that we have <math>(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n</math>, where <math>n \geq 0</math> and <math>n</math> is an integer. Since <math>a \neq 0</math>, <math>n=0,1,2,3,\cdots, 58</math> which is <math>\boxed{\textbf{(C) }59}</math>.
 
Consider that we have <math>(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n</math>, where <math>n \geq 0</math> and <math>n</math> is an integer. Since <math>a \neq 0</math>, <math>n=0,1,2,3,\cdots, 58</math> which is <math>\boxed{\textbf{(C) }59}</math>.
 +
 
~~QuantumPsiInverted
 
~~QuantumPsiInverted
  

Revision as of 02:09, 22 June 2023

Problem

How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?

$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$

Solution

The sum of the angles of a triangle is $180$ degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it $\frac{180}{3} = 60$ degrees. The minimum possible value for the smallest angle is $1$ and the highest possible is $59$ (since the numbers are distinct), so there are $\boxed{\textbf{(C) }59}$ possibilities.

Solution 2 (Stars and Bars)

Let the first angle be $x$, and the common difference be $d$. The arithmetic progression can now be expressed as $x + (x + d) + (x + 2d) = 180$. Simplifiying, $x + d = 60$. Now, using stars and bars, we have $\binom{61}{1} = 61$. However, we must subtract the two cases in which either $x$ or $d$ equal $0$, so we have $61 - 2$ = $\boxed{\textbf{(C) }59}$.

Solution 3 (Quick Summation)

Consider that we have $(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n$, where $n \geq 0$ and $n$ is an integer. Since $a \neq 0$, $n=0,1,2,3,\cdots, 58$ which is $\boxed{\textbf{(C) }59}$.

~~QuantumPsiInverted

See also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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