I am a computer science student currently working on his master thesis. I stumbled across a geometric problem that seems obvious but I couldn't prove it for weeks.

I attached a picture with the construction and will describe it below.

  • Assume that we have a circle C with radius r = 1.
  • M is the center of C.
  • A point v is placed inside C and v!=M.
  • Assume without loss of generality that v is placed on the same horizontal axis as M and that this axis is the x-axis of the coordinate system.
  • the distance between M and v is called d (Note: $0<d<1$)
  • Two additional points are placed on the circumference of C.
  • the perpendicular bisector of the line $\overline{u_2u_1}$ goes through M
  • The angle (u_2,v,u_1) is 90 degrees and the dashed line is bisecting that angle.
  • $\phi$ is then the angle at v between the dashed line and the X-Axis.

The big question is now: Assuming $\phi$ can be changed from $0^\circ$ to $180^\circ$, which value for $\phi$ minimizes d'?

The answer is obviously $0^\circ$ but I want to prove that.

This is my current approach:

  • use the triangle M,V,u_2 to compute $\overline{vu_2}$
  • use the triangle M,V,u_1 to compute $\overline{vu_1}$
  • use Pythagoras to compute $\overline{u_1u_2}$
  • d' is minimized if $\overline{u_1u_2}^2$ is minimized

I will provide the values in much more detail below but I think a trigonometry expert can maybe give me a hint by looking at the construction only.

Here are the values (remember that r=1 and d is a constant:

Compute the values directly from $\phi$: $$ \gamma=135^\circ-\phi \\ \gamma'=135^\circ+\phi \\ $$

First triangle to compute $a = \overline{vu_2}$: $$ \beta =\arcsin{(d\sin{(\gamma)})} \\ \alpha =180^\circ-\beta-\gamma \\ a = \frac{\sin{(\alpha)}}{\sin{(\gamma)}} \\ $$

Second triangle to compute $a' = \overline{vu_1}$: $$ \beta' =\arcsin{(d\sin{(\gamma')})} \\ \alpha'=180^\circ-\beta'-\gamma' \\ a' = \frac{\sin{(\alpha')}}{\sin{(\gamma')}} \\ $$

Use $a$ and $a'$ to compute $b=\overline{u_1u_2}$: $$ b =\sqrt{a^2+a'^2} \\ $$

At this point I need to somehow show that $\phi=0$ minimizes $b$. Then it is just a small step to argue that minimizing $b$ minimizes $d'$.

I would really appreciate a solution or a useful hint on how to prove that $\phi=0^\circ$ minimizes $d'$.

Thanks in advance, Hendrik


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