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.
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.
Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.