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 "2014 AMC 12B Problems/Problem 12"

(→‎Solution)
(→‎Solution)
Line 1: Line 1:
 +
==Problem==
 +
 +
A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
 +
 +
\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}}\ 11\qquad\textbf{(E)}\ 12
 +
 
==Solution==
 
==Solution==
  
Line 29: Line 35:
 
<math>(1, 1, 1)</math>
 
<math>(1, 1, 1)</math>
  
−
It should be clear that <math>|S|</math> is simply <math>|T|</math> minus the larger "duplicates" (e.g. <math>(2, 2, 2)</math> is a larger duplicate of <math>(1, 1, 1)</math>). Since <math>|T|</math> is 13 and the number of higher duplicates is 4, the answer is <math>13 - 4</math> or <math>9 B</math>
+
It should be clear that <math>|S|</math> is simply <math>|T|</math> minus the larger "duplicates" (e.g. <math>(2, 2, 2)</math> is a larger duplicate of <math>(1, 1, 1)</math>). Since <math>|T|</math> is 13 and the number of higher duplicates is 4, the answer is <math>13 - 4</math> or <math>9 (B)</math>.

Revision as of 22:29, 20 February 2014

Problem

A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?

\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}}\ 11\qquad\textbf{(E)}\ 12

Solution

Define $T$ to be the set of all triples $(a, b, c)$ such that $a \ge b \ge c$, $b+c > a$, and $a, b, c \le 5$. Now we enumerate the elements of $T$:

$(4, 4, 4)$

$(4, 4, 3)$

$(4, 4, 2)$

$(4, 4, 1)$

$(4, 3, 3)$

$(4, 3, 2)$

$(3, 3, 3)$

$(3, 3, 2)$

$(3, 3, 1)$

$(3, 2, 2)$

$(2, 2, 2)$

$(2, 2, 1)$

$(1, 1, 1)$

It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$). Since $|T|$ is 13 and the number of higher duplicates is 4, the answer is $13 - 4$ or $9 (B)$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_12&oldid=60260"