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

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==Solution 2(Stars and Bars)==
 
==Solution 2(Stars and Bars)==
 
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>.  
 
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>.  
However, we must subtract the two cases in which either <math>a</math> or <math>d</math> equal <math>0</math>, so we have <math>61 - 2</math> = C.
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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>C</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:31, 1 June 2021

Problem

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

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

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 $59$ possibilities $\Longrightarrow \mathrm{C}$.

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 $61 - 2$ = $C$.

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
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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