Difference between revisions of "2006 AMC 10A Problems/Problem 19"
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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>61_{ | + | 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>61_{c_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. | 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. | ||
Revision as of 17:40, 4 August 2020
Problem
How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?
Solution
The sum of the angles of a triangle is 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 degrees. The minimum possible value for the smallest angle is and the highest possible is (since the numbers are distinct), so there are possibilities .
Solution 2(Stars and Bars)
Let the first angle be , and the common difference be . The arithmetic progression can now be expressed as . Simplifiying, . Now, using stars and bars, we have . However, we must subtract the two cases in which either or equal , so we have = C.
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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