What angle must two equal length lines be at, s.t. a line from the open endpoint of one, to the midpoint of the other reflects to the other midpoint?

Question

I've been thinking about this problem in my head for years, and finally sat down and drew it in CAD iterating until I could get an approximate solution, but the math is beyond me.

Here is an image with the angles marked:

Essentially, you have 2 lines of equal length. I don't think the length matters due to the problem construction. For now we can imagine it as an isosceles triangle with the point upward, giving a left and right leg.

At what angle must these lines be to each other so that if I drew a line from the end of the right leg to the mid point of the left leg, the line could then reflect off the left leg and would then hit the midpoint of the right leg. This would form a similar triangle in the top portion of the isosceles. Using what math I did remember, I labelled the drawing so that x is the angle I'm looking for, but no matter what equation I type into wolfram alpha I seem to be missing some relation since the answer always comes back in terms of t (where t is x + y)

E.g. tried: 180 = x + 2y; w+x=y; w+x+t = 180; t+y=180;

I tried including the law of cosines, but I just end up introducing a new variable for each equation, so I'm always one equation short of solving the system.

With my CAD drawing, I can measure that the angle is approximately equal to 41.4096 degrees +- 0.0002

Answer 1

In the image, I have labelled side lengths of the upper isosceles triangle as $a$, the middle horizontal line as $b$, and the horizontal base as $c$. I don't have image posting permission yet, so click the link to view my diagram.

Using the law of cosines, $$b^2 = 2a^2(1-\cos x)$$ $$a^2 = 2c^2(1-\cos x)$$

Substituting for $a^2$, we have $$b^2 = 4c^2(1-\cos x)^2$$ $$b = 2c(1 - \cos x)$$

We also know $2b = c$ because you have two proportional isosceles triangles. Substituting for $b$ we get $$\frac{c}{2} = 2c(1-\cos x)$$ and solving for $x$ yields $$x = \cos ^{-1} \frac{3}{4}$$ which lines up very well with your approximation.

Answer 2

Just to simplify some calculations, $x=2\alpha$. Then $y=90^\circ-\alpha$. Also, $$w+x+(x+y)=x+y+y$$so $$w=y-x=90^\circ-3\alpha$$ Then in the triangle from the origin, reflection point, and the intersection of the lines, you apply law of sines: $$\frac{\sin w}{L}=\frac{\sin(x+y)}{2L}$$ so $$2\sin(90^\circ-3\alpha)=\sin(90^\circ+\alpha)$$ With some simple trigonometric formula, you can the rewrite this as $$\cos\alpha(4\cos2\alpha-3)=0$$ The first term can't be zero, so $$\cos2\alpha=\frac34$$or $$2\alpha=\arccos\frac34\approx 41.409^\circ$$