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I have everything pretty much figured out everything but I need help proving the unique point formed by the three perpendiculars in the picture

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  • $\begingroup$ are you and Elizabeth Hill in the same class??????? math.stackexchange.com/questions/1275863/… $\endgroup$ – Will Jagy May 11 '15 at 20:29
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    $\begingroup$ @WillJagy I was wondering the same thing. and neither wants to give much more info $\endgroup$ – Willemien May 12 '15 at 5:14
  • $\begingroup$ @Willemien, except for the time of year, this could be a Summer class or workshop in the U.S. The exercise is quite easy in the upper half plane model with Mobius transformations from $SL_2 \mathbb R$, but those techniques amount to major theorems, far beyond what these students know. So, the person or organization teaching this has something very particular in mind; since we have no way of knowing what that is, i decided to just let it go. They will be told, eventually, how it was to be done. $\endgroup$ – Will Jagy May 12 '15 at 16:58
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    $\begingroup$ @WillJagy thanks , but i did think for this question the beltrami klein model is easier (but that is a detail) but I do guess they haven't used that either $\endgroup$ – Willemien May 12 '15 at 20:30
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  1. In the hyperbolic plane all the "great triangles or ideal triangles" (whose angles are all zeros) are congruent. So if you can prove something in the case of one of them will be true for all of them.

  2. Consider the Klein model and take a special great triangle: one of the great triangles that look equilateral in the Euclidean eye. The perpendicular (Euclidean) bisectors dropped from the vertices to the opposite sides will meet in the center of the Klein circle.

  3. If a hyperbolic line goes through the Klein center and is perpendicular to another hyperbolic line in the Euclidean sense, will be perpendicular to that line in the hyperbolic sense as well.

  4. Every great triangle will have this property. (Not that they meet at "the centre" : ) but the perpendiculats mentioned above have to meet.)
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Another way to prove that there is a point where all the perpendiculars meet:

The triangle is equilateral so a reflection over one of the perpendiculars will exchange two sides, and will also exchange the 2 other perpendiculars.

you can proof that 2 perpendiculars intersect

and because that points on the mirror line stay at the same point the 3 perpendiculars meet at one point.

The length of $|AB|$ is $\frac{1}{2} \ln(\frac{7 +3 \sqrt{5}}{2}) = 0.97.... $ but the proof is rather cumbersome. but the proof is rather cumbersome. (yep the Beltrami-Klein model)

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