# Angle trisection: A Beautiful problem

**Geometry**Level 5

Let triangle \(\triangle ABP_0\) be an obtuse isosceles triangle with \(\angle B=36^{\circ}\), \(\angle P_0=108^{\circ}\), and segment \(\overline{AB}=2015\).

Here is a general description of the pattern.

Angle \(\angle BP_0A\) is trisected by segments \(\overline{P_0 P_1}\) and \(\overline{P_0 D_1}\) where \(P_1\) and \(D_1\) are on side \(\overline{AB}\) and \(\overline{BP_1}<\overline{BD_1}\)

Next, angle \(P_0P_1B\) is trisected by segments \(\overline{P_1P_2}\) and \(\overline{P_1D_2}\) where \(P_1\) and \(D_1\) are on side \(\overline{P_0B}\) and \(\overline{P_2B}<\overline{D_2B}\).

Here is an explicit description

Now, given the pattern above, let the point \(P_n\) and \(D_n\) be the points resulting from the trisection of \(\angle P_{n-2}P_{n-1}B\) with \(\overline{BP_n}<\overline{BD_n}\).

If \(\dfrac{\overline{P_9P_{10}}}{\overline{P_6P_7}}\) can be expressed as \(\left(\dfrac{-a+\sqrt{b}}{c}\right)^k\) where \(b\) is square free and \(a,c\) are co-prime positive integers and \(k\) is an integer, what is the value of \(abck\)?

This is part of the set Trevor's Ten

**Details and Assumptions**

\(\cos{36}=\dfrac{1+\sqrt{5}}{4}\). (And I know what you're thinking, the answer is not 20. Lol, if only it were that simple)

Hint: make an equation for \(\overline{P_{n-1}P_{n}}\). Don't work with numbers.