Difference between revisions of "1959 AHSME Problems/Problem 28"
Clarkculus (talk | contribs) (Created page with "== Problem 28== In triangle <math>ABC</math>, <math>AL</math> bisects angle <math>A</math>, and <math>CM</math> bisects angle <math>C</math>. Points <math>L</math> and <math>...") |
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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
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 | |
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