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

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== Problem ==
 
== Problem ==
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How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?  
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How many non-[[similar]] triangles have angles whose degree measures are distinct positive integers in [[arithmetic progression]]?  
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<math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad</math>
  
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<math>\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad</math>
 
 
== Solution ==
 
== Solution ==
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Let us begin by first realizing that the sum of the angles must add up to 180 degrees. Then let us consider the highest and lowest sets of angles that satisfy the conditions of the problem.
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The sum of the angles of a triangle is <math>180</math> 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 <math>\frac{180}{3} = 60</math> degrees. The minimum possible value for the smallest angle is <math>1</math> and the highest possible is <math>59</math> (since the numbers are distinct), so there are <math>\boxed{\textbf{(C) }59}</math> possibilities.
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Highest: 1-60-119
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Lowest: 59-60-61
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==Solution 2 (Stars and Bars)==
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The increment in the highest set is 59, while the increment in the lowest set is 1. Therefore, any increment between 1 and 59 would create a set of angles that work. Therefore, there are 59 possibilities. (c)
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Let the first angle be <math>x</math>, and the common difference be <math>d</math>. The arithmetic progression can now be expressed as <math>x + (x + d) + (x + 2d) = 180</math>. Simplifiying, <math>x + d = 60</math>. Now, using stars and bars, we have <math>\binom{61}{1} = 61</math>.  
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However, we must subtract the two cases in which either <math>x</math> or <math>d</math> equal <math>0</math>, so we have <math>61 - 2</math> = <math>\boxed{\textbf{(C) }59}</math>.
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==Solution 3 (Quick Summation)==
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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> solutions.
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~~QuantumPsiInverted
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== See also ==
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{{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}
  
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== See Also ==
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[[Category:Introductory Geometry Problems]]
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*[[2006 AMC 10A Problems]]
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{{MAA Notice}}

Latest revision as of 02:10, 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 $\boxed{\textbf{(C) }59}$ solutions.

~~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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