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 ABP0\triangle ABP_0 be an obtuse isosceles triangle with B=36\angle B=36^{\circ}, P0=108\angle P_0=108^{\circ}, and segment AB=2015\overline{AB}=2015.

Here is a general description of the pattern.

Angle BP0A\angle BP_0A is trisected by segments P0P1\overline{P_0 P_1} and P0D1\overline{P_0 D_1} where P1P_1 and D1D_1 are on side AB\overline{AB} and BP1<BD1\overline{BP_1}<\overline{BD_1}

Next, angle P0P1BP_0P_1B is trisected by segments P1P2\overline{P_1P_2} and P1D2\overline{P_1D_2} where P1P_1 and D1D_1 are on side P0B\overline{P_0B} and P2B<D2B\overline{P_2B}<\overline{D_2B}.

Here is an explicit description

Now, given the pattern above, let the point PnP_n and DnD_n be the points resulting from the trisection of Pn2Pn1B\angle P_{n-2}P_{n-1}B with BPn<BDn\overline{BP_n}<\overline{BD_n}.

If P9P10P6P7\dfrac{\overline{P_9P_{10}}}{\overline{P_6P_7}} can be expressed as (a+bc)k\left(\dfrac{-a+\sqrt{b}}{c}\right)^k where bb is square free and a,ca,c are co-prime positive integers and kk is an integer, what is the value of abckabck?

This is part of the set Trevor's Ten

Details and Assumptions

  • cos36=1+54\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 Pn1Pn\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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