Difference between revisions of "1953 AHSME Problems/Problem 50"
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<math> \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} </math> | <math> \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} </math> | ||
− | == Solution == | + | == Solution 1== |
Let the triangle have side lengths <math>14, 6+x,</math> and <math>8+x</math>. The area of this triangle can be computed two ways. We have <math>A = rs</math>, and <math>A = \sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s = 14+x</math> is the semiperimeter. Therefore, <math>4(14+x)=\sqrt{(14+x)(x)(8)(6)}</math>. Solving gives <math>x = 7</math> as the only valid solution. This triangle has sides <math>13,14</math> and <math>15</math>, so the shortest side is <math>\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}</math>. | Let the triangle have side lengths <math>14, 6+x,</math> and <math>8+x</math>. The area of this triangle can be computed two ways. We have <math>A = rs</math>, and <math>A = \sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s = 14+x</math> is the semiperimeter. Therefore, <math>4(14+x)=\sqrt{(14+x)(x)(8)(6)}</math>. Solving gives <math>x = 7</math> as the only valid solution. This triangle has sides <math>13,14</math> and <math>15</math>, so the shortest side is <math>\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}</math>. | ||
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+ | == 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>. | ||
== 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 02:24, 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, , 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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