Difference between revisions of "2010 AMC 8 Problems/Problem 13"
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Since the length of the shortest side is a whole number and is equal to <math>\frac{3}{10}</math> of the perimeter, it follows that the perimeter must be a multiple of <math>10</math>. Adding the two previous integers to each answer choice, we see that <math>11+10+9=30</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct. | Since the length of the shortest side is a whole number and is equal to <math>\frac{3}{10}</math> of the perimeter, it follows that the perimeter must be a multiple of <math>10</math>. Adding the two previous integers to each answer choice, we see that <math>11+10+9=30</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 11}</math> is correct. | ||
| − | == | + | ==Video by MathTalks== |
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| + | https://www.youtube.com/watch?v=6hRHZxSieKc | ||
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| + | ==See Also== | ||
{{AMC8 box|year=2010|num-b=12|num-a=14}} | {{AMC8 box|year=2010|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:54, 22 January 2023
Problem
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 
of the perimeter. What is the length of the longest side?
Solution 1(algebra solution)
Let 
, 
, and 
be the lengths of the sides of the triangle. Then the perimeter of the triangle is 
. Using the fact that the length of the smallest side is of the perimeter, it follows that:
. The longest side is then 
. Thus, answer choice 
is correct.
Solution 2
Since the length of the shortest side is a whole number and is equal to 
of the perimeter, it follows that the perimeter must be a multiple of 
. Adding the two previous integers to each answer choice, we see that 
. Thus, answer choice is correct.
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
