# A conjecture related to Viviani's theorem

In the following picture, related to Viviani's theorem, $A=\frac{\sqrt{3}}{2}\alpha\beta$ and $A'=\frac{\sqrt{3}}{4}\gamma^2$. We suppose $\beta>\alpha$ and $\alpha,\beta,\gamma>0$.

Is there any way to infer that $\gamma^2=2\alpha\beta$ or, $\gamma^2<2\alpha\beta$, or $\gamma^2>2\alpha\beta$?

• I believe it depends. These can take on many values. Is there any relation between $A$ and $A'$. – user211599 Jun 4 '18 at 14:36
• Thanks for your comment, Jeffery. I have no idea if there is a relation between $A$ and $A'$. Since I am not an expert, and I had troubles even to find the expressions of these areas, I thought that maybe somebody more into Viviani's theorem could already know a method to evaluate this relationship. – user559615 Jun 4 '18 at 15:20

I suppose you refer to three segments $\alpha$, $\beta$ and $\gamma$ originating from a point $P$ inside an equilateral triangle and perpendicular to its three sides. Polygon $A$ is a parallelogram having $\alpha$ and $\beta$ as sides and $B$ is an equilateral triangle with side $\gamma$.

If you intend those properties to hold for any $P$ inside the triangle, then they are certainly false. To check, take for instance $P$ on the left side: then $\alpha=0$ and $\gamma>0$, whereas $\alpha>0$ and $\gamma=0$ if $P$ is on the right side.

EDIT.

Asking for $\alpha$, $\beta$, $\gamma$ to be positive doesn't change the situation: you can place $P$ so that $\gamma$ is almost vanishing while $\alpha$ and $\beta$ are not (diagram below, left), or vice versa (diagram below, right).

EDIT 2.

If you are asking, instead, for the locus of points such that $\gamma^2≷2\alpha\beta$, then it is not difficult to find that $\gamma^2=2\alpha\beta$ along an arc of ellipse, passing through two vertices of the triangle and tangent there to two sides.

To prove that, it is convenient to use an oblique system of coordinates with $x$ axis along side $AB$ of the triangle and $y$ axis along side $AC$ (see diagram below). For simplicity I also set $AB=AC=1$. Coordinates $(x,y)$ of point $P$ have a simple relation with $\alpha$ and $\beta$: $$\alpha={\sqrt3\over2}y,\quad \beta={\sqrt3\over2}x \quad\text{and}\quad \gamma={\sqrt3\over2}-\alpha-\beta\quad \text{by Viviani's theorem.}$$ Plugging these into $\gamma^2=2\alpha\beta$ gives a simple equation for $P$: $$(x-1)^2+(y-1)^2=1.$$ In orthogonal coordinates that would be the equation of a circle with center $O=(1,1)$ and unit radius. In oblique coordinates that is instead the equation of an ellipse, whose properties can be deduced from the equation:

• the ellipse is tangent to coordinate axes at $B$ and $C$ and has center $O$;
• the coordinates of its vertices $R$, $S$, $U$, $V$ can be found from the intersection of the ellipse with lines $AO$ and $UV$ having equations $y=x$ and $x+y=2$; it turns out that those vertices have coordinates ${1\pm 1/\sqrt2}$;
• from that it follows that $RO=\sqrt{3/2}$ and $UO=1/\sqrt2$.

Once you know the length of the axes of the ellipse and their direction, the ellipse is completely defined.

• Sure. Thanks, Aretino. Maybe I should have mentioned that it must be $\alpha,\beta,\gamma>0$ and, moreover, that at least $\alpha>\beta$, in the sense that we suppose that at least two segments are not equal to each other. I add it in the Question. – user559615 Jun 4 '18 at 16:05
• Those additions don't change the conclusion: you can place $P$ so that $\alpha$ and $\beta$ are both very small while $\gamma$ is not, or the other way around. – Intelligenti pauca Jun 4 '18 at 16:10
• Dear Aretino, thank you very much. Very clever explanations! I see very well your points. Very beautiful the connection with the ellipse! How can one prove this fact? – user559615 Jun 4 '18 at 17:54
• Great proof! To me it would have been really hard to get it. Thanks for your detailed and illuminating explanations! – user559615 Jun 5 '18 at 12:51