Difference between revisions of "1953 AHSME Problems/Problem 50"
(→Solution) |
|||
| Line 5: | Line 5: | ||
<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>. | ||
| + | |||
| + | == 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, <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
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 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
