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Difference between revisions of "2014 AMC 12B Problems/Problem 12"

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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>.
 
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>.
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*Based on the wording of Problem 13 to specifically exclude triangles with zero area: "... triangle with positive area", the definition of a triangle in this test includes degenerate ones. That is, the triangle inequality is not strict.  The following are possible degenerate triangles (excluding duplicates):
 
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<math>(2, 1, 1)</math>
 
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<math>(3, 2, 1)</math>
 
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<math>(4, 3, 1)</math>
 
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As the specifics to the definition of the triangle were not provided, and the only evidence of such (Problem 13) includes degenerates, we must assume the most general case and include these. Our final answer is then <math>9 + 3</math> or <math>\boxed{\textbf{(E)}\ 12}</math>.
 

Revision as of 22:02, 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?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}}\ 11\qquad\textbf{(E)}\ 12$ (Error compiling LaTeX. Unknown error_msg)

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 $\boxed{\textbf{(B)}\ 9}$.

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