Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "== Problem == One of the sides of a triangle is divided into segments of <math>6</math> and <math>8</math> units by the point of tangency of the inscribed circle. If the radius ...")
 
(→‎Solution)
Line 7: Line 7:
 
== Solution ==
 
== Solution ==
  
−
Let the triangle has 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>. 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 has 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>.
  
 
== 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}}

Revision as of 11:30, 6 August 2012

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

Let the triangle has 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 $\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_50&oldid=47834"