Inscribed Equilateral Triangle Trig - Expression for P/A

Question

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!

Answer 1

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 $k = 4$.

Answer 2

Notice that if you draw radial lines from the center of the circle to the three vertices of the inscribed triangle, the inscribed triangle is subdivided into three identical triangles. Each smaller triangle has a base that is one-third of the equilateral triangle's perimeter and has height $h = \frac12 r.$

So the total area of the equilateral triangle is the sum of the areas of the three smaller triangles, that is, $$A = \frac12 h P = \frac14 r P.$$

From that equation it immediately follows that $$\frac PA = \frac 4r.$$

This way involves minimal calculation, in particular no trig functions or square roots. You can get it even faster if you recall that the area of any regular polygon is $\frac12 hP$ where $P$ is the perimeter and $h$ is the radius of the inscribed circle.