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

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(→‎Solution)
 
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==Solution==
 
==Solution==
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Listing all possible triangle side lengths satisfying the constraints, we find the following:
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Since we want non-congruent triangles that are neither isosceles nor 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>.
  
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<math>\text{2-3-4}\\
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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>.
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\text{2-4-5}\\
 
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\text{2-5-6}\\
 
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\text{3-4-6}\\
 
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\text{3-5-6}
 
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</math>
 
  
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Thus the answer is <math>\fbox{\textbf{(C)}\; 5}</math>.
+
When <math>a=2</math>, 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 3 triangles <math>(2,3,4), (2,4,5), (2,5,6)</math> within our perimeter constraint.
  
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More explanation:
+
When <math>a=3</math>, <math>c</math> can be <math>b+1</math> or <math>b+2</math>, which gives triangles <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 only 2 triangles.
  
−
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>
+
All in all, this gives us <math>3+2 = \boxed{\textbf{(C)}\; 5}</math> triangles.
−
 
 
−
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}}

Latest revision as of 14:10, 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

Since we want non-congruent triangles that are neither isosceles nor 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 $a=2$, 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 $a=3$, $c$ can be $b+1$ or $b+2$, which gives triangles $(3,4,5), (3,4,6), (3,5,6)$. Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.

All in all, this gives us $3+2 = \boxed{\textbf{(C)}\; 5}$ 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

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