With reference to this post,
A conjecture related to Viviani's theorem,
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?
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 $\gamma'^2=2\alpha'\beta'$ 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 $1/\sqrt3$.
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