Difference between revisions of "1988 AHSME Problems/Problem 22"

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==Solution==
 
==Solution==
 
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We first notice that the sides <math>10</math> and <math>24</math>, can be part of <math>2</math> different right triangles, one with sides <math>10,24,26</math>, and the other with a leg
 
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somewhere between <math>21</math> and <math>22</math>. We now notice that if <math>x</math> is less than or equal to <math>21</math>, one of the angles is obtuse, and that the same is the same for
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any value of <math>x</math> above <math>26</math>. Thus the only integer values of <math>x</math> that fit the conditions, are <math>x=22, 23, 24, \text{and }25.</math>
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So, the answer is <math>\boxed{\text{A}}</math>
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1988|num-b=22|num-a=23}}   
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{{AHSME box|year=1988|num-b=21|num-a=23}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:44, 10 August 2018

Problem

For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$


Solution

We first notice that the sides $10$ and $24$, can be part of $2$ different right triangles, one with sides $10,24,26$, and the other with a leg somewhere between $21$ and $22$. We now notice that if $x$ is less than or equal to $21$, one of the angles is obtuse, and that the same is the same for any value of $x$ above $26$. Thus the only integer values of $x$ that fit the conditions, are $x=22, 23, 24, \text{and }25.$ So, the answer is $\boxed{\text{A}}$

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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 26 27 28 29 30
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