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Difference between revisions of "2005 AMC 8 Problems/Problem 15"

(Created page with "==Problem== How many different isosceles triangles have integer side lengths and perimeter 23? <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}...")
 
 
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==Solution==
 
==Solution==
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Let <math>b</math> be the base of the isosceles triangles, and let <math>a</math> be the lengths of the other legs. From this, <math>2a+b=23</math> and <math>b=23-2a</math>. From triangle inequality, <math>2a>b</math>, then plug in the value from the previous equation to get <math>2a>23-2a</math> or <math>a>5.75</math>. The maximum value of <math>a</math> occurs when <math>b=1</math>, in which from the first equation <math>a=11</math>. Thus, <math>a</math> can have integer side lengths from <math>6</math> to <math>11</math>, and there are <math>\boxed{\textbf{(E)}\ 6}</math> triangles.
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Let <math>b</math> be the base of the isosceles triangles, and let <math>a</math> be the lengths of the other legs. From this, <math>2a+b=23</math> and <math>b=23-2a</math>. From triangle inequality, <math>2a>b</math>, then plug in the value from the previous equation to get <math>2a>23-2a</math> or <math>a>5.75</math>. The maximum value of <math>a</math> occurs when <math>b=1</math>, in which from the first equation <math>a=11</math>. Thus, <math>a</math> can have integer side lengths from <math>6</math> to <math>11</math>, and there are <math>\boxed{\textbf{(C)}\ 6}</math> triangles.
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==Video Solution==
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https://youtu.be/DGUZn2BMuiE  Soo, DRMS, NM
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=14|num-a=16}}
 
{{AMC8 box|year=2005|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 23:11, 1 July 2023

Problem

How many different isosceles triangles have integer side lengths and perimeter 23?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution

Let $b$ be the base of the isosceles triangles, and let $a$ be the lengths of the other legs. From this, $2a+b=23$ and $b=23-2a$. From triangle inequality, $2a>b$, then plug in the value from the previous equation to get $2a>23-2a$ or $a>5.75$. The maximum value of $a$ occurs when $b=1$, in which from the first equation $a=11$. Thus, $a$ can have integer side lengths from $6$ to $11$, and there are $\boxed{\textbf{(C)}\ 6}$ triangles.

Video Solution

https://youtu.be/DGUZn2BMuiE Soo, DRMS, NM

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions

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