Difference between revisions of "2006 AMC 12B Problems/Problem 10"
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== Problem == | == Problem == | ||
| − | {{ | + | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? |
| + | <math> | ||
| + | \text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47 | ||
| + | </math> | ||
== Solution == | == Solution == | ||
| + | <math> | ||
| + | \text {(A) } 43 | ||
| + | </math> | ||
| + | If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: x+15>3x, so 2x<15 and x<7.5. Now, since we want the greatest perimeter, we want the greatest integer x, which is 7. Then, the first side has length 3*7=21, the second side has length 7 and the third side has length 15, and so the perimeter is 21+7+15=43. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}} | ||
Revision as of 09:47, 26 February 2011
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Problem
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
Solution
If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: x+15>3x, so 2x<15 and x<7.5. Now, since we want the greatest perimeter, we want the greatest integer x, which is 7. Then, the first side has length 3*7=21, the second side has length 7 and the third side has length 15, and so the perimeter is 21+7+15=43.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
