AoPSWiki

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

Personal tools

Search

Toolbox

2003 AMC 12A Problems/Problem 7

From AoPSWiki

Problem

How many non- congruent triangles with perimeter $7$ have integer side lengths?

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$

Solution

By the triangle inequality, no one side may have a length greater than half the perimeter which is $\frac{1}{2}\cdot7=3.5$ .

Since all sides must be integers, the largest possible length of a side is $3$

Therefore, all such triangles must have all sides of length $1$ , $2$ , or $3$ .

Since $2+2+2=6<7$ , atleast one side must have a length of $3$

Thus, the remaining two sides have a combined length of $7-3=4$ .

So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ .

Therefore, the number of triangles is $2 \Longrightarrow \mathrm{(B)}$ .

See Also

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2003_AMC_12A_Problems/Problem_7"

Category: Introductory Geometry Problems

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us