# 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 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 $$k = 4$$?
• 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! – James R Aug 9 at 2:27
• @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. – John Omielan Aug 9 at 2:31