Problem:
Triangle ABC has AB=2β
AC. Let D and E be on AB and BC, respectively, such that β BAE=β ACD. Let F be the intersection of segments AE and CD, and suppose that β³CFE is equilateral. What is β ACB?
Answer Choices:
A. 60β
B. 75β
C. 90β
D. 105β
E. 120β
Solution:
Let Ξ±=β BAE=β ACD=β ACF. Because β³CFE is equilateral, it follows that β CFA=120β and then
β FAC=180ββ120βββ ACF=60ββΞ±.
Therefore
β BAC=β BAE+β FAC=Ξ±+(60ββΞ±)=60β
Because AB=2β
AC, it follows that β³BAC is a 30β60β90β triangle, and thus β ACB=90ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions
