Difference between revisions of "1966 AHSME Problems/Problem 11"
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== 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 == | == See also == | ||
{{AHSME box|year=1966|num-b=10|num-a=12}} | {{AHSME box|year=1966|num-b=10|num-a=12}} | ||
| − | [[Category:]] | + | [[Category:Introductory Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:50, 25 June 2018
Problem
The sides of triangle 
are in the ratio 
. 
is the angle-bisector
drawn to the shortest side 
, dividing it into segments 
and 
.
If the length of is 
, then the length of the longer segment of is:
Solution
By the Angle Bisector Theorem, we have 
which implies 
. So 
, and thus 
.
Hence
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. 
