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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!

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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 what your results. Thus,

$$\frac{P}{A} = \frac{36}{\sqrt{3}P} = \frac{36}{9r} = \frac{4}{r} \tag{1}\label{eq1}$$

In your calculations, $\sin(60^{0}) = \frac{\sqrt{3}}{2}$, so $3(2r\sin(60)) = 3\sqrt{3}r$. 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{eq1}.

Another issue is that the $r$ is in the denominator, not the numerator as your question asks. Perhaps one of the $2$ fractions were meant to be their reciprocal, i.e., $\frac{A}{B}$ or $\frac{k}{r}$, so then $k = 4$?

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  • $\begingroup$ Ah I'm sorry for the typo, the question does ask for $\frac{k}{r}$. I don't know why that wasn't an obvious mistake. Thank you so much for the clarification! $\endgroup$ – James R Aug 9 at 2:27
  • $\begingroup$ @JamesR You're welcome for the clarification. Also, I hope my answer resolved your main question. Although I expressed $A$ and $P$ somewhat differently than you did (I just calculated them using symmetry and various $30-60-90$ triangles), as I stated and you can verify if you wish, they are basically the same results as you obtained. $\endgroup$ – John Omielan Aug 9 at 2:31
  • $\begingroup$ @JamesR If there's anything else you would like me to ask me about regarding my answer, please let me know. Otherwise, if you found it useful, please consider up voting and/or accepting my answer. Thanks. $\endgroup$ – John Omielan Aug 10 at 5:56

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