Difference between revisions of "2015 AMC 12B Problems/Problem 10"
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==Problem== | ==Problem== | ||
+ | How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles? | ||
+ | <math>\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7</math> | ||
+ | ==Solution== | ||
+ | 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>. | ||
+ | |||
+ | 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>, 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. | ||
− | = | + | 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. |
+ | All in all, this gives us <math>3+2 = \boxed{\textbf{(C)}\; 5}</math> 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?
Solution
Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths with . Furthermore, "positive area" tells us that and the perimeter constraints means .
There are no triangles when because then must be less than , implying that , contrary to .
When , similar to above, must be less than , so this leaves the only possibility . This gives 3 triangles within our perimeter constraint.
When , can be or , which gives triangles . Note that is a right triangle, so we get rid of it and we get only 2 triangles.
All in all, this gives us 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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