Inscribed Equilateral Triangle Trig - Expression for P/A Problem: The vertices of an equilateral triangle, with perimeter P and area A, lie on a circle with radius r. Find an expression for $\frac{P}{A}$ in the form $\frac{r}{k}$, where k ∈ Z+.
Hi, I'm having a bit of trouble solving this problem. What I've tried is using the formula for the area of an equilateral triangle (a = $\frac{3\sqrt{3}}{4}$$r^2$). Since one side is equal to $\frac{P}{3}$, I inserted that into the formula a = $2rsin(60)$ to get $P=3(2rsin(60))$. That meant that $\frac{P}{A}$ -> $\frac{3(2rsin(60))}{\frac{3\sqrt{3}}{4}r^2}$ = $\frac{2\sqrt{3}}{3\sqrt{3}r}$ = $\frac{2}{3r}$. But I don't think this is the correct answer because when I put them back into the formulas I get different values for the radius! If anyone can help and explain what I did wrong it would help a lot. Thanks!

I get that $A = \frac{P^2\sqrt{3}}{36}$. Also, $r = \frac{P}{3\sqrt{3}} \implies P = 3\sqrt{3}r$. This effectively matches your results. Thus,
$$\frac{P}{A} = \frac{36}{\sqrt{3}P} = \frac{36}{9r} = \frac{4}{r} \tag{1}\label{eq1A}$$
In your calculations, since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, then $P = 3(2r\sin(60^{\circ})) = 3\sqrt{3}r$. You multiplied by $4$ from the denominator, which should give $12\sqrt{3}$, so perhaps you just dropped the initial $1$ digit to get $2\sqrt{3}$. Multiplying your result by $6$ to compensate gives $\frac{12}{3r} = \frac{4}{r}$, i.e., my result in \eqref{eq1A}.
Since you confirmed in your comment below that, instead of $\frac{r}{k}$, the source question actually asks for the form of $\frac{k}{r}$, then \eqref{eq1A} shows that $k = 4$.
