# Angle trisection: A Beautiful problem

Geometry Level 5

Note: I give two explanations for this problem (both are the same thing) to make the problem more understandable.

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.

###### By the way, this problem is VERY beautiful and is without a doubt my new favorite problem that I've made.
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