Difference between revisions of "1959 AHSME Problems/Problem 28"
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a} </math> | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a} </math> | ||
== Solution == | == Solution == | ||
| − | By the angle bisector theorem, <math>\frac{AM}{AB}=\frac{AC}{BC}</math> and <math>\frac{CL}{LB}=\frac{AC}{AB}</math>, so by rearranging the given equation and noting <math>AB=c</math> and <math>BC=a</math>, <math>k=c | + | By the angle bisector theorem, <math>\frac{AM}{AB}=\frac{AC}{BC}</math> and <math>\frac{CL}{LB}=\frac{AC}{AB}</math>, so by rearranging the given equation and noting <math>AB=c</math> and <math>BC=a</math>, <math>k=\frac{c}{a}\rightarrow\boxed{\textbf{E}}</math>. |
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=27|num-a=29}} | {{AHSME 50p box|year=1959|num-b=27|num-a=29}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:55, 21 July 2024
Problem 28
In triangle 
, 
bisects angle 
, and 
bisects angle 
. Points 
and 
are on 
and 
, respectively. The sides of 
are 
, 
, and 
. Then 
where 
is:
Solution
By the angle bisector theorem, 
and 
, so by rearranging the given equation and noting 
and 
, 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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. 
