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Given an equilateral triangle $ABC$, we choose a point $D$ inside it, determining a new triangle $ADB$. enter image description here We draw the circles with centers in $A$ and in $B$ passing by $D$, determining the new points $F$ and $G$ on the side $AB$.

These two points define the segment $AF$ (blue), $FG$ (green), and $GB$ (red). We denote $\alpha=\overline{AF}$, $\gamma=\overline{FG}$, and $\beta=\overline{GB}$.

enter image description here

So far, I was able to prove that if $\frac{\gamma^2}{2\alpha\beta}=1$, then $D$ lies on the arc of the circle with center in the midpoint $M$ of the side $AB$, and with diameter equal to $\overline{AB}$ (in these conditions, $ABD$ is a right triangle). enter image description here

On which curves lies the point $D$ if $$ \frac{\gamma^2}{2\alpha\beta}=n, $$ where $n\in\{2,3,4\ldots\}$?

Thanks for your suggestions!

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I don't think you will find a nice description of the curve without using coordinates.

Let $A = (-1, 0)$, $B = (1, 0)$ and $D = (x, y)$. Then it is easy to prove that $$\alpha = 2 - \sqrt{(x - 1)^2 + y^2} \qquad \beta = 2 - \sqrt{(x + 1)^2 + y^2}$$ $$\gamma = \sqrt{(x + 1)^2 + y^2} + \sqrt{(x - 1)^2 + y^2} - 2$$ Therefore, the equation $\gamma^2 = 2 n \alpha \beta$ amounts to: $$\left [\sqrt{(x + 1)^2 + y^2} + \sqrt{(x - 1)^2 + y^2} - 2 \right ]^2 = 2 n \left [2 - \sqrt{(x - 1)^2 + y^2} \right ] \left [ 2 - \sqrt{(x + 1)^2 + y^2} \right ]$$

Indeed, if $n = 1$ then the equation reduces to $x^2 + y^2 = 1$, which represents the circle having diameter $AB$. If $n > 1$, the equation represents a curve having a branch shaped like an oval which becomes "pointy" as $n$ increases.

If we choose $D$ on the altitude of the triangle, it is easy to see that $\frac {\gamma^2} {2 \alpha \beta} \to \infty$ as $D$ approaches $C$. So the answer to the related question is yes.

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  • $\begingroup$ Thanks Luca! Nice answer! Do you have any idea how to draw such curves with Geogebra? $\endgroup$ – user559615 Jul 29 '18 at 9:42
  • $\begingroup$ I exploit your cleverness a bit more: Supposing to exclude the points $D$ lying on the altitude, does your answer include the case in which (also) $\alpha,\beta,\gamma$ are integers? $\endgroup$ – user559615 Jul 29 '18 at 9:46
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    $\begingroup$ On GeoGebra you can create a slider n and then simply write (sqrt((x + 1)² + y²) + sqrt((x - 1)² + y²) - 2)² = 2n (2 - sqrt((x - 1)² + y²)) (2 - sqrt((x + 1)² + y²)) to create the corresponding implicit curve. $\endgroup$ – Luca Bressan Jul 29 '18 at 9:59
  • $\begingroup$ It doesn't really make sense to ask for $\alpha, \beta, \gamma$ to be integers unless you specify the length of $AB$. $\endgroup$ – Luca Bressan Jul 29 '18 at 10:01
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    $\begingroup$ You can simply copy and paste the equation in the Input Bar. If you want to use the ImplicitCurve command, you have to input a function, not an equation: just replace = with - and give the expression as the argument. (I'm using GeoGebra Classic 5, I don't know about other versions.) $\endgroup$ – Luca Bressan Jul 29 '18 at 17:08

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