1959 AHSME Problems/Problem 28

Problem 28

In triangle $ABC$ , $AL$ bisects angle $A$ , and $CM$ bisects angle $C$ . Points $L$ and $M$ are on $BC$ and $AB$ , respectively. The sides of $\triangle ABC$ are $a$ , $b$ , and $c$ . Then $\frac{AM}{MB}=k\frac{CL}{LB}$ where $k$ is: $\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a}$

Solution

$[asy] import geometry; point A = (0,0); point B = (3,4); point C = (8,0); point L,M; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Angle Bisectors pair[] l = intersectionpoints(bisector(line(A,B),line(A,C)),B--C); L = l[0]; dot(L); label("L",L,NE); draw(A--L); markscalefactor = 0.15; draw(anglemark(L,A,B)); markscalefactor = 0.17; draw(anglemark(C,A,L)); markscalefactor = 0.18; draw(anglemark(L,A,B)); markscalefactor = 0.2; draw(anglemark(C,A,L)); pair[] m = intersectionpoints(bisector(line(B,C), line(A,C)), A--B); M = m[0]; dot(M); label("M",M,NW); draw(C--M); markscalefactor = 0.15; draw(anglemark(M,C,A)); markscalefactor = 0.17; draw(anglemark(B,C,M)); [/asy]$

By the Angle Bisector Theorem, $\frac{AM}{AB}=\frac{AC}{BC}$ and $\frac{CL}{LB}=\frac{AC}{AB}$ , so by rearranging the given equation and noting $AB=c$ and $BC=a$ , $k=\frac{c}{a}\rightarrow\boxed{\textbf{E}}$ .

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.

Page

Toolbox

Search

1959 AHSME Problems/Problem 28

Problem 28

Solution

See also