Difference between revisions of "2014 AMC 12B Problems/Problem 12"
(→Solution) |
Pi over two (talk | contribs) (→Problem) |
||
(16 intermediate revisions by 3 users not shown) | |||
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? | ||
+ | |||
+ | <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 | + | 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 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 | + | 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?
Solution
Define to be the set of all integral triples such that , , and . Now we enumerate the elements of :
It should be clear that is simply minus the larger "duplicates" (e.g. is a larger duplicate of ). Since is and the number of higher duplicates is , the answer is or .
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.