Monday, 2 February 2015

geometry - Viviani's theorem and ellipse



With reference to this post,



A conjecture related to Viviani's theorem,



From ellipse to circle in a Viviani's theorem framework



it was suggested (thanks to user Aretino) that, in the top picture (where the triangle is equilateral and the ellipse pass through two vertex of the triangle and tangent there to two sides), P belongs to the arc of ellipse if γ2=2αβ. Here, α,β,γ are the segments whose lengths are the distances of P from each side.




My first question is:




How can I prove this?




The second question is related to the second picture:




Does P belong to the arc of circle if and only if γ2=2αβ?





Notice that, in the first picture the segments α,β,γ sum up to the altitude of the triangle, whereas, in the second one, the segments α,β,γ sum up to the side of the triangle.



The most general issue is:




Is there any relationship between the two scenarios?



Answer




For the first part please refer to my answer to the original question. I will repeat here the argument for the second case, but there must be an error in your question, because the locus of points such that γ2=2αβ is the same ellipse as in the answer to the first part. This is obvious, because
α=(2/3)α, β=(2/3)β,
γ=(2/3)γ.



A circle is indeed the locus of P when γ2=αβ.
In this case it is easier to find the answer using cartesian orthogonal coordinates, with x axis along side AB of the triangle (see diagram below).
For simplicity I also set AB=AC=1.
Coordinates (x,y) of point P have a simple relation with α and β:
y=32α,x=β+12αandγ=1αβby the given assumption.


Plugging these into γ2=αβ gives a simple equation for P:
(x1)2+(y13)2=13.

This is the equation of a circle with center O=(1,1/3) and radius 1/3.




enter image description here


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...