Canadian Mathematical Olympiad

Solution:
Since similar triangles give the same value of $K P / R^{3}$, we can fix $R=1$ and maximize $K P$ over all triangles inscribed in the unit circle. Fix points $A$ and $B$ on the unit circle. The locus of points $C$ with a given perimeter $P$ is an ellipse that meets the circle in at most four points. The area $K$ is maximized (for a fixed $P$) when $C$ is chosen on the perpendicular bisector of $AB$, so we get a maximum value for $K P$ if $C$ is where the perpendicular bisector of $AB$ meets the circle. Thus the maximum value of $K P$ for a given $AB$ occurs when $ABC$ is an isosceles triangle. Repeating this argument with $BC$ fixed, we have that the maximum occurs when $ABC$ is an equilateral triangle.

Consider an equilateral triangle with side length $a$. It has $P=3a$. It has height equal to $a \sqrt{3} / 2$ giving $K = a^{2} \sqrt{3} / 4$. From the extended law of sines, $2R = a / \sin(60^{\circ})$ giving $R = a / \sqrt{3}$. Therefore the maximum value we seek is
$$K P / R^{3} = \left(\frac{a^{2} \sqrt{3}}{4}\right)(3a)\left(\frac{\sqrt{3}}{a}\right)^{3} = \frac{27}{4}.$$

From the extended law of sines, the lengths of the sides of the triangle are $2R \sin A$, $2R \sin B$ and $2R \sin C$. So
$$P = 2R(\sin A + \sin B + \sin C) \text{ and } K = \frac{1}{2}(2R \sin A)(2R \sin B)(\sin C),$$
giving
$$\frac{K P}{R^{3}} = 4 \sin A \sin B \sin C (\sin A + \sin B + \sin C)$$
We wish to find the maximum value of this expression over all $A+B+C=180^{\circ}$. Using well-known identities for sums and products of sine functions, we can write
$$\frac{K P}{R^{3}} = 4 \sin A \left(\frac{\cos(B-C)}{2} - \frac{\cos(B+C)}{2}\right) \left(\sin A + 2 \sin\left(\frac{B+C}{2}\right) \cos\left(\frac{B-C}{2}\right)\right).$$
If we first consider $A$ to be fixed, then $B+C$ is fixed also and this expression takes its maximum value when $\cos(B-C)$ and $\cos\left(\frac{B-C}{2}\right)$ equal $1$; i.e. when $B=C$. In a similar way, one can show that for any fixed value of $B$, $K P / R^{3}$ is maximized when $A=C$. Therefore the maximum value of $K P / R^{3}$ occurs when $A=B=C=60^{\circ}$, and it is now an easy task to substitute this into the above expression to obtain the maximum value of $27/4$.

As in Solution 2, we obtain
$$\frac{K P}{R^{3}} = 4 \sin A \sin B \sin C (\sin A + \sin B + \sin C)$$
From the AM-GM inequality, we have
$$\sin A \sin B \sin C \leq \left(\frac{\sin A + \sin B + \sin C}{3}\right)^{3},$$
giving
$$\frac{K P}{R^{3}} \leq \frac{4}{27}(\sin A + \sin B + \sin C)^{4}$$
with equality when $\sin A = \sin B = \sin C$. Since the sine function is concave on the interval from $0$ to $\pi$, Jensen's inequality gives
$$\frac{\sin A + \sin B + \sin C}{3} \leq \sin\left(\frac{A+B+C}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}.$$
Since equality occurs here when $\sin A = \sin B = \sin C$ also, we can conclude that the maximum value of $K P / R^{3}$ is $\frac{4}{27}\left(\frac{3 \sqrt{3}}{2}\right)^{4} = 27/4$.