Angle trisection: A Beautiful problem

We notice a bunch of $36-36-108$ triangles; in particular, $P_kP_{k+1}B$ is a triangle of this type.

We also know, from our repertoire of not really useful knowledge, that the ratio of the shorter side to the longer side of a $36-36-108$ triangle is $\dfrac{1}{\varphi}$

With this in mind, it is easy to see that $P_kP_{k+1}=\dfrac{2015}{\varphi^{k+2}}$

Finally, this means $\dfrac{P_9P_{10}}{P_6P_7} =\dfrac{\dfrac{2015}{\varphi^{11}}}{\dfrac{2015}{\varphi^{8}}} =\dfrac{1}{\varphi^3}=\left(\dfrac{-1+\sqrt{5}}{2}\right)^3$ so our answer is $1\times 5\times 2\times 3 = \boxed{30}$

Get ready for the mother of all solutions.

Well, to be exact, $\phi-1$

Incase you're wondering why this is my new favorite problem.

Now for the solution

Important Note: when we trisect the angle, two of the three smaller triangles are similar to the larger one. And when the angle of the smaller triangle is trisected, the two of the three smaller smaller triangles are similar, this continues for every trisection.

Let the "double leg" (is there a term for the leg that's used twice in an isosceles triangle) be x, the angle trisector segment be a, and the angle be 36. I'm not using a variable for $\cos(36)$ because for this problem, the angle 36 is unique. But for just one trisection, any angle can be used. This is because for any angle besides 36, when trisected, the triangles aren't similar.

By law of sines, we equate the area of the three smaller triangles to the area of the larger one. Using $\frac{1}{2} ab\sin(c)$ .

$\frac{1}{2}x^2\sin(3\cdot36)=\frac{1}{2}xa \sin(36)+\frac{1}{2}xa\sin(36)+\frac{1}{2}a^2 \sin(36)$

$x^2\sin(3\cdot36)=2xa \sin(36)+a^2 \sin(36)$

Let's derive the triple angle formula, shall we?

$\cos(t)+i\sin(t)$ ^3 looking at the imaginary part.

$\sin(3t)=3\cos^2(t)\sin(t)-\sin^3(t)$

$\sin(3t)=3\sin(t)(1-\sin^2(t))-\sin^3(t)$

$\sin(3t)=3\sin(t)-3\sin^3(t)-\sin^3(t)$

$\sin(3t)=3\sin(t)-4\sin^3(t)$

Using this identity

$x^2(3\sin(36)-4\sin^3(36))=2xa \sin(36)+a^2 \sin(36)$

Making this into a quadratic with a as the variable

$0=(\sin(36))a^2+(2x\sin(36))a-x^2(3\sin(36)-4\sin^3(36))$

Ugly as this may look, it's quite beautiful. Using the quadratic equation.

$a=\dfrac{-2x\sin(36)\pm\sqrt{4x^2\sin^2(36)-4(\sin(36)(-3x^2\sin(36)+4x^2\sin(36))}}{2\sin(36)}$

$a=\dfrac{-2x\sin(36)\pm\sqrt{4x^2\sin^2(36)+12x^2\sin^2(36)-16x^2\sin^2(36)}}{2\sin(36)}$

$a=\dfrac{-2x\sin(36)\pm\sqrt{16x^2\sin^2(36)(1-\sin^2(36)}}{2\sin(36)}$

Using the identity $\cos^2(s)=1-\sin^2(s)$

$a=\dfrac{-2x\sin(36)\pm4x\cos(36)\sin(36)}{2\sin(36)}$

$a=-x\pm x\cos(36)$

Since the side length can't be negative, we disregard the - case.

$a=(2\cos(36)-1)x$

If you replace the 36 with $\theta$ , then this is the explicit formula for the length of an angle trisector of an isosceles triangle with side length x.

Now, $\textbf{since there is no constant term here}$

$\textbf{this is a ratio, aka: a scale factor since the triangles are similar}$

This means the ratio of $\overline{P_0B}:\overline{P_0P_1}=1:(2\cos(36)-1)$

It follows that $\overline{P_0P_1}:\overline{P_1P_2}=1:(2\cos(36)-1)$

And that

$\overline{P_{n-1}P_{n}}:\overline{P_n P_{n+1}}=1:(2\cos(36)-1)$

And finally

$\overline{P_{n-1}P_{n}}:\overline{P_{n+k-1} P_{n+k}}=1:(2\cos(36)-1)^k$

Thus it follows that

$\dfrac{\overline{P_{9}P_{10}}}{\overline{P_6P_7}}=(2\cos(36)-1)^3$

Since $k=3$

Plugging in our values (note that x=2015 is irrelevant to the problem, the information is extraneous) $\cos(36)=\frac{1}{4}(1+\sqrt{5})$ we get

$\left(\dfrac{-1+\sqrt5}{2}\right)^3$

Thus $\left(\dfrac{-x+\sqrt{y}}{z}\right)^k$ , x=1, y=5, z=2, k=3

$\therefore xyzk=(1)(5)(2)(3)=30$

1/(golden ratio)^3. Notice that as you zoom in on the right corner, all the figures will be similar.

using trigonometry $p_{n+1}$ $p_{n}$ = $p_{n}$ $p_{n-1}$ sin36 sec18.
sec18 and sin36 can be found in terms of cos 36.
sec 18 sin 36 comes out to be ( -1 + $\sqrt{5}$ )/2.
hence ratio is $( -1 +\sqrt{5}$ )/2)^3.