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Difference between revisions of "1953 AHSME Problems/Problem 50"

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== Solution 2==
 
== Solution 2==
  
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Label the tangent points on <math>BC, CA, AB</math> as <math>D, E, F</math> respectively. Let <math>AF=AE=6</math>, <math>BF=BD=8</math>, and <math>CE=CD=x.</math> The problem is a matter of solving for <math>x</math>. To this, we use the fact that if <math>A,B,C</math> are the angles of a triangle, then <math>\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.</math> We know that <math>\tan{\frac{A}{2}} = \frac{2}{3}</math>, <math>\tan{\frac{B}{2}} = \frac{1}{2}</math>, and <math>\tan{\frac{C}{2}} = \frac{4}{x},</math> so we have the equation <math>\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.</math> Solving, <math>x=7</math>, so the shortest side has length <math>\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}</math>.
+
Label the tangent points on <math>BC, CA, AB</math> as <math>D, E, F</math> respectively. Let <math>AF=AE=6</math>, <math>BF=BD=8</math>, and <math>CE=CD=x.</math> The problem is a matter of solving for <math>x</math>. To this, we use the fact that if <math>A,B,C</math> are the angles of a triangle, then <math>\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.</math> We know that <math>\tan{\frac{A}{2}} = \frac{2}{3}</math>, <math>\tan{\frac{B}{2}} = \frac{1}{2}</math>, and <math>\tan{\frac{C}{2}} = \frac{4}{x},</math> so we have the equation <math>\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.</math> Solving this equation yields <math>x=7</math>, so the shortest side has length <math>\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1953|num-b=49|after=Last Question}}
 
{{AHSME 50p box|year=1953|num-b=49|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:32, 31 December 2023

Problem

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$, then the length of the shortest side is

$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$

Solution 1

Let the triangle have side lengths $14, 6+x,$ and $8+x$. The area of this triangle can be computed two ways. We have $A = rs$, and $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\sqrt{(14+x)(x)(8)(6)}$. Solving gives $x = 7$ as the only valid solution. This triangle has sides $13,14$ and $15$, so the shortest side is $\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}$.

Solution 2

Label the tangent points on $BC, CA, AB$ as $D, E, F$ respectively. Let $AF=AE=6$, $BF=BD=8$, and $CE=CD=x.$ The problem is a matter of solving for $x$. To this, we use the fact that if $A,B,C$ are the angles of a triangle, then $\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.$ We know that $\tan{\frac{A}{2}} = \frac{2}{3}$, $\tan{\frac{B}{2}} = \frac{1}{2}$, and $\tan{\frac{C}{2}} = \frac{4}{x},$ so we have the equation $\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.$ Solving this equation yields $x=7$, so the shortest side has length $\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 49 		Followed by
Last Question
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All AHSME Problems and Solutions

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