Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 12B Problems/Problem 10"

(→‎Solution)
(→‎Solution)
Line 15: Line 15:
  
 
Thus the answer is <math>\fbox{\textbf{(C)}\; 5}</math>.
 
Thus the answer is <math>\fbox{\textbf{(C)}\; 5}</math>.
 +
 +
More explanation:
 +
 +
Since we want non-congruent triangles that are not isosceles or equilateral, we can just list side lengths <math>a-b-c</math> with <math>a<b<c</math>. Furthermore, "positive area" tells us that <math>c < a + b</math> and the perimeter constraints means <math>a+b+c < 15</math>
 +
 +
There are no triangles when <math>a = 1</math> because then <math>c</math> must be less than <math>b+1</math>, implying that <math>b \geq c</math>, contrary to <math>b < c</math>
 +
 +
When <math>a=2</math> then, similar to above, <math>c</math> must be less than <math>b+2</math>, so this leaves the only possibility <math>c = b+1</math>. This gives <math>3</math> triangles (<math>2-3-4, 2-4-5, 2-5-6</math>) within our perimeter constraint.
 +
 +
When <math>a=2</math> then <math>c</math> can be <math>b+1</math> or <math>b+2</math>, which gives <math>3-4-5, 3-4-6, 3-5-6</math>. Note that <math>3-4-5</math> is a right triangle, so we get rid of it and we get <math>2</math> triangles.
 +
 +
All in all this gives us 5 triangles.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=11|num-b=9}}
 
{{AMC12 box|year=2015|ab=B|num-a=11|num-b=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:19, 5 March 2015

Problem

How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?

$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$

Solution

Listing all possible triangle side lengths satisfying the constraints, we find the following:

$\text{2-3-4}\\ \text{2-4-5}\\ \text{2-5-6}\\ \text{3-4-6}\\ \text{3-5-6}$

Thus the answer is $\fbox{\textbf{(C)}\; 5}$.

More explanation:

Since we want non-congruent triangles that are not isosceles or equilateral, we can just list side lengths $a-b-c$ with $a<b<c$. Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$

There are no triangles when $a = 1$ because then $c$ must be less than $b+1$, implying that $b \geq c$, contrary to $b < c$

When $a=2$ then, similar to above, $c$ must be less than $b+2$, so this leaves the only possibility $c = b+1$. This gives $3$ triangles ($2-3-4, 2-4-5, 2-5-6$) within our perimeter constraint.

When $a=2$ then $c$ can be $b+1$ or $b+2$, which gives $3-4-5, 3-4-6, 3-5-6$. Note that $3-4-5$ is a right triangle, so we get rid of it and we get $2$ triangles.

All in all this gives us 5 triangles.

See Also

2015 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_10&oldid=68481"