Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (wikified)
m (→‎Solution 3 (Quick Summation))
 
(18 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?  
+
How many non-[[similar]] triangles have angles whose degree measures are distinct positive integers in [[arithmetic progression]]?  
  
−
<math>\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad</math>
+
<math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad</math>
  
 +
== Solution ==
 +
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.
 +
 +
==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>.
 +
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>.
 +
 +
==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> solutions.
  
−
== Solution ==
+
~~QuantumPsiInverted
−
Let us begin by first realizing that the sum of the [[angle]]s must add up to 180 degrees. Then let us consider the highest and lowest sets of angles that satisfy the conditions of the problem.<br>
 
−
Highest: 1-60-119<br>
 
−
Lowest: 59-60-61<br>
 
−
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)
 
  
 +
== See also ==
 +
{{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}
  
−
== See Also ==
+
[[Category:Introductory Geometry Problems]]
−
*[[2006 AMC 10A Problems]]
+
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_19&oldid=194670"