2006 AMC 12B Problems/Problem 10

This is an empty template page which needs to be filled. You can help us out by finding the needed content and editing it in. Thanks.

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? $\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$

Solution

$\text {(A) } 43\\$ 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: $\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 \\$ Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43$ .

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

Page

Toolbox

Search

2006 AMC 12B Problems/Problem 10

Problem

Solution

See also