Difference between revisions of "1959 AHSME Problems/Problem 28"
m (formatted fractions) |
(diagram, link) |
||
(One intermediate revision by the same user not shown) | |||
Line 4: | Line 4: | ||
<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 | + | <asy> |
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = (0,0); | ||
+ | point B = (3,4); | ||
+ | point C = (8,0); | ||
+ | point L,M; | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(A--B--C--A); | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | |||
+ | // Angle Bisectors | ||
+ | pair[] l = intersectionpoints(bisector(line(A,B),line(A,C)),B--C); | ||
+ | L = l[0]; | ||
+ | dot(L); | ||
+ | label("L",L,NE); | ||
+ | draw(A--L); | ||
+ | |||
+ | markscalefactor = 0.15; | ||
+ | draw(anglemark(L,A,B)); | ||
+ | markscalefactor = 0.17; | ||
+ | draw(anglemark(C,A,L)); | ||
+ | markscalefactor = 0.18; | ||
+ | draw(anglemark(L,A,B)); | ||
+ | markscalefactor = 0.2; | ||
+ | draw(anglemark(C,A,L)); | ||
+ | |||
+ | pair[] m = intersectionpoints(bisector(line(B,C), line(A,C)), A--B); | ||
+ | M = m[0]; | ||
+ | dot(M); | ||
+ | label("M",M,NW); | ||
+ | draw(C--M); | ||
+ | |||
+ | markscalefactor = 0.15; | ||
+ | draw(anglemark(M,C,A)); | ||
+ | markscalefactor = 0.17; | ||
+ | draw(anglemark(B,C,M)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | 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 15:19, 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.