Interpreting congruent angles!

Geometry Level 5

Q = ( 1 B A 1 + 1 B A n + 1 ) ( 1 B A 1 + 1 B A 2 + + 1 B A n + 1 B A n + 1 ) \large{ Q= \dfrac{\left(\dfrac{1}{|BA_1|} + \dfrac{1}{|BA_{n+1}|} \right)}{\left( \dfrac{1}{|BA_1|} + \dfrac{1}{|BA_2|} + \ldots + \dfrac{1}{|BA_n|} + \dfrac{1}{|BA_{n+1}|} \right)} }

Given an A B C = ϕ \angle ABC = \phi and rays l 1 , l 2 , , l n 1 l_1, l_2, \ldots, l_{n-1} dividing the angle into n n congruent angles. For a line l l intersecting sides A B AB and B C BC at two distinct points, denote l ( A B ) = A 1 l \cap (AB) = A_{1} , l ( B C ) = A n + 1 l \cap (BC) = A_{n+1} and l l i = A i + 1 l \cap l_i = A_{i+1} for 1 i < n 1 \leq i < n . Show that the value of Q Q is a constant which doesn't depend on l l , and find the value of Q Q in terms of n n and ϕ \phi .

If n = 5 , ϕ = 7 5 n=5, \ \phi = 75^\circ , then Q Q can be expressed as A B C , \large{A - \dfrac{B}{\sqrt{C}}},

where A , B A,B and C C are positive integers, and C C square-free, find the value of A + B + C A+B+C .


The answer is 4.

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1 solution

This problem can we done by using polar coordinate system. After a lot of (simple) calculations, we get that Q = 1 sin ( n 1 2 n ϕ ) sin ( n + 1 2 n ϕ ) Q=1-\frac{\sin \left( \frac{n-1}{2n} \phi \right)}{\sin \left( \frac{n+1}{2n} \phi \right)}

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