Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Angles and Triangles.
Let triangle ABC be a right angled triangle with right angle at C. Let D and E be points on AB with D between A and E such that CD and CE trisect angle C. If \(\frac{DE}{BE}\)=\(\frac{8}{15}\), then tan B can be written as \(\frac{mp^\frac{1}{2}}{n}\) where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime , find m+n+p.
Angles
Algebra
Triangles
Answer: is 18.
AIME I, 2012, Question 12
Geometry Vol I to Vol IV by Hall and Stevens
Let CD=2a,then with angle bisector theorem of triangle we have for triangle CDB \(\frac{2a}{8}\)=\(\frac{CB}{15}\) then \(CB=\frac{15a}{4}\)
DF drawn perpendicular to BC gives CF=a, FD=\(a \times 3^\frac{1}{2}\), FB= \(\frac{11a}{4}\)
then tan B = \(\frac{a \times 3^\frac{1}{2}}{\frac{11a}{4}}\)=\(\frac{4 \times 3^\frac{1}{2}}{11}\) then m+n+p=4+3+11=18.
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