# Viviani's theorem and ellipse

With reference to this post,

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 $\gamma^2=2\alpha\beta$. Here, $\alpha,\beta,\gamma$ 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 $\gamma'^2=2\alpha'\beta'$?

Notice that, in the first picture the segments $\alpha,\beta,\gamma$ sum up to the altitude of the triangle, whereas, in the second one, the segments $\alpha',\beta',\gamma'$ sum up to the side of the triangle.

The most general issue is:

Is there any relationship between the two scenarios?

• The link and picture are nice. Though, it would be nice if you explained the picture more (e.g. it is assumed that the lines are tangent to the ellipse). Triangle is equilateral and $\alpha,\beta, \gamma$ are actually lengths (even in the original post, it was not so clear). – quantum Jun 5 '18 at 8:15
• True, I edit these info immediately. Thanks for the remarks! – user559615 Jun 5 '18 at 8:16
• I gave a proof editing my answer to the original question. A slight change to that proof will also answer your second question. – Aretino Jun 5 '18 at 10:42
• Wonderful, Aretino! Thanks again! But then the second assert is false, right? – user559615 Jun 5 '18 at 12:44

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 $\gamma'^2=2\alpha'\beta'$ is the same ellipse as in the answer to the first part. This is obvious, because $\alpha'=(2/\sqrt3)\alpha$, $\beta'=(2/\sqrt3)\beta$, $\gamma'=(2/\sqrt3)\gamma$.
A circle is indeed the locus of $P$ when $\gamma'^2=\alpha'\beta'$. 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 $\alpha'$ and $\beta'$: $$y={\sqrt3\over2}\alpha',\quad x=\beta'+{1\over2}\alpha' \quad\text{and}\quad \gamma'=1-\alpha'-\beta'\quad \text{by the given assumption.}$$ Plugging these into $\gamma'^2=\alpha'\beta'$ gives a simple equation for $P$: $$(x-1)^2+\bigg(y-{1\over\sqrt3}\bigg)^2={1\over3}.$$ This is the equation of a circle with center $O=(1,1/\sqrt3)$ and radius $1/\sqrt3$.
• Bright as always, Aretino. To me, it seems then that the answer to the second question is "yes": but, what do you mean with "there must be an error in your question because in this case $\gamma^2<2\alpha\beta$?", I don't get this observation. – user559615 Jun 5 '18 at 15:11
• My proof was wrong, but the condition $\gamma^2=2\alpha\beta$ leads to an ellipse, not to a circle: I edited my answer to make that clear. The quantities $\alpha$, $\beta$, $\gamma$ in your subsequent question are not the same as the ones used here. – Aretino Jun 5 '18 at 16:16
• I see, and it is very intriguing! But my question is related to the case in which the side of the triangle coincides with the diameter of the circle (as in the original picture). In this case, what is the relation between $\alpha', \beta',\gamma'$? Is it still $\gamma'^2=2\alpha'\beta'$? In any case, thank you a lot for your nice reasoning! – user559615 Jun 5 '18 at 17:09