Difference between revisions of "2006 AMC 10B Problems/Problem 10"
m (→Solution) |
Dairyqueenxd (talk | contribs) (→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 1: | Line 1: | ||
== 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? | + | 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 <math>15</math>. What is the greatest possible perimeter of the triangle? |
− | <math> \ | + | <math> \textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47 </math> |
== Solution == | == Solution == | ||
Line 19: | Line 19: | ||
The greatest integer satisfying this inequality is <math>7</math>. | The greatest integer satisfying this inequality is <math>7</math>. | ||
− | So the greatest possible perimeter is <math> 7 + 3\cdot7 + 15 = | + | So the greatest possible perimeter is <math> 7 + 3\cdot7 + 15 =\boxed{\textbf{(A) } 43}</math> |
== See Also == | == See Also == |
Latest revision as of 13:57, 26 January 2022
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 . What is the greatest possible perimeter of the triangle?
Solution
Let be the length of the first side.
The lengths of the sides are: ,
, and
.
By the Triangle Inequality,
The greatest integer satisfying this inequality is .
So the greatest possible perimeter is
See Also
2006 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.