Difference between revisions of "1953 AHSME Problems/Problem 50"
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== Solution 2== | == Solution 2== | ||
− | 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 | + | 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
Contents
[hide]Problem
One of the sides of a triangle is divided into segments of and units by the point of tangency of the inscribed circle. If the radius of the circle is , then the length of the shortest side is
Solution 1
Let the triangle have side lengths and . The area of this triangle can be computed two ways. We have , and , where is the semiperimeter. Therefore, . Solving gives as the only valid solution. This triangle has sides and , so the shortest side is .
Solution 2
Label the tangent points on as respectively. Let , , and The problem is a matter of solving for . To this, we use the fact that if are the angles of a triangle, then We know that , , and so we have the equation Solving this equation yields , so the shortest side has length .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
Followed by Last Question | |
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