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)
(→‎Problem)
 
(13 intermediate revisions by 3 users not shown)
Line 3: Line 3:
 
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?  
 
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?  
  
−
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}}\ 11\qquad\textbf{(E)}\ 12</math>
+
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math>
  
 
==Solution==
 
==Solution==
  
−
Define <math>T</math> to be the set of all triples <math>(a, b, c)</math> such that <math>a \ge b \ge c</math>, <math>b+c > a</math>, and <math>a, b, c \le 5</math>. Now we enumerate the elements of <math>T</math>:
+
Define <math>T</math> to be the set of all integral triples <math>(a, b, c)</math> such that <math>a \ge b \ge c</math>, <math>b+c > a</math>, and <math>a, b, c < 5</math>. Now we enumerate the elements of <math>T</math>:
  
 
<math>(4, 4, 4)</math>
 
<math>(4, 4, 4)</math>
Line 35: 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 <math>13</math> and the number of higher duplicates is <math>4</math>, 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 <math>13</math> and the number of higher duplicates is <math>4</math>, the answer is <math>13 - 4</math> or <math>\boxed{\textbf{(B)}\ 9}</math>.
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2014|ab=B|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 09:13, 3 March 2015

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 integral triples $(a, b, c)$ such that $a \ge b \ge c$, $b+c > a$, and $a, b, c < 5$. 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 $\boxed{\textbf{(B)}\ 9}$.

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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=2014_AMC_12B_Problems/Problem_12&oldid=68159"