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 "1966 AHSME Problems/Problem 11"

(Created page with "== Problem == The sides of triangle <math>BAC</math> are in the ratio <math>2:3:4</math>. <math>BD</math> is the angle-bisector drawn to the shortest side <math>AC</math>, dividi...")
 
(→‎Solution)
 
(3 intermediate revisions by one other user not shown)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
 +
By the [[Angle Bisector Theorem]], we have <math>\frac{BA}{AD}=\frac{BC}{CD}</math> which implies <math>\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}</math>. So <math>AC=10=AD+DC=\frac{7}{4}DC</math>, and thus <math>DC=\frac{40}{7}=5\frac{5}{7}</math>.
 +
Hence
 +
<math>\fbox{C}</math>
 +
 +
== See also ==
 +
{{AHSME box|year=1966|num-b=10|num-a=12}} 
 +
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:50, 25 June 2018

Problem

The sides of triangle $BAC$ are in the ratio $2:3:4$. $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$. If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac{1}{2} \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac{5}{7} \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac{1}{2}$

Solution

By the Angle Bisector Theorem, we have $\frac{BA}{AD}=\frac{BC}{CD}$ which implies $\frac{AD}{DC}=\frac{BA}{BC}=\frac{3}{4}$. So $AC=10=AD+DC=\frac{7}{4}DC$, and thus $DC=\frac{40}{7}=5\frac{5}{7}$. Hence $\fbox{C}$

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_11&oldid=95625"