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 | + | <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)); | ||
| + | |||
| + | // Length Labels | ||
| + | label("$a$", midpoint(B--C), (5,5)); | ||
| + | label("$b$", midpoint(A--C), S); | ||
| + | label("$c$", midpoint(A--B), (-5,5)); | ||
| + | |||
| + | </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}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 14:24, 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
![[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)); // Length Labels label("$a$", midpoint(B--C), (5,5)); label("$b$", midpoint(A--C), S); label("$c$", midpoint(A--B), (-5,5)); [/asy]](http://latex.artofproblemsolving.com/7/4/9/7496248a77b0f59c106c71dbd1f1c143c4437354.png)
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. 
