Rotating a Triangle about Different Sides

Let the three sides of the triangle be $a, b , c$ such that $a \gt b \gt c$ . And as usual, the vertices opposite to these sides are $A , B , C$ . The volume of rotation of the triangle about side $a$ is $V_a = \dfrac{1}{3} \pi a h_A^2$ , where $h_A$ is the altitude drawn from vertex $A$ onto side $a$ . We know that

$h_A = b \sin C = c \sin B$

hence,

$V_a = 2880 \pi = \dfrac{1}{3} \pi a b^2 \sin^2 C$

$V_b = 3240 \pi = \dfrac{1}{3} \pi b a^2 \sin^2 C = \dfrac{1}{3} \pi b c^2 \sin^2 A$

$V_c = \dfrac{25920}{7} \pi = \dfrac{1}{3} \pi c b^2 \sin^2 A$

By taking the ratios of these volumes, the sine terms cancel out, and we get,

$\dfrac{V_a}{V_b} = \dfrac{2880}{3240} = \dfrac{ 8 }{ 9 } = \dfrac{b}{a}$

and

$\dfrac{V_b}{V_c} = \dfrac{ 7 (3240) }{25920 } = \dfrac{ 7 }{8} = \dfrac{c}{b}$

Thus, the sides $a, b, c$ are in the following ratio, $a : b : c = 9 : 8 : 7$ , so that, $a = 9 t , b = 8 t , c = 7 t$ for some $t \in \mathbb{R^+}$ . With this information, we can compute the sine of angle $C$ . We have,

$\cos C = \dfrac{ a^2 + b^2 - c^2 }{2 a b} = \dfrac{ (9 t)^2 + (8 t)^2 - (7 t)^2 } { 2 (9t)(8t) } = \dfrac{2}{3}$

Thus $\sin^2 C = 1 - \cos^2 C = \dfrac{5}{9}$ . Now from the first equation, we have,

$2880 = \dfrac{1}{3} (9 t)(8 t)^2 \left( \dfrac{5}{9} \right)$

From which, $t = 3$ . And we're done, because $a + b + c = 9 t + 8 t + 7 t = 24 t = 72$ .

Let the sides of the triangle be $a>b>c$ and its respective altitudes be $x,y,z$ , so that $ax=by=cz=2\Delta$ , where $\Delta$ is the area of the triangle.

Rotating the triangle about the side of length $a$ gives a pair of cones, of total height $a$ and base radii $x$ . The volume generated is thus $\frac13 \pi ax^2$ .

We can side-step some unpleasant algebra by working with ratios. From the given volumes we have

$ax^2:by^2:cz^2=2880:3240:\frac{25920}{7}=56:63:72$

Since $ax^2=\frac{4\Delta^2}{a}$ , we can write

$\frac1a:\frac1b:\frac1c=56:63:72$

Rearranging this, we get

$a:b:c=9:8:7$

so the triangle in question is similar to a $7,8,9$ triangle. Say $a=9t$ , $b=8t$ , $c=7t$ . By Heron's formula, this triangle has area $\Delta=\sqrt{720}\cdot t^2$ . Substituting in to the first given volume, we have

$2880\pi=\frac13 \pi ax^2 = \frac13 \pi \cdot \frac{4\Delta^2}{a} = \frac{4\cdot 720t^4}{3 \cdot 9t}\pi = \frac{320t^3}{3}\pi$

and we find $t=3$ , giving $a=27$ , $b=24$ and $c=21$ , for a perimeter of $\boxed{72}$ .

Let the sides be a, b & c as usual. We can now write the following equations: s=(a+b+c)/2, 4A²/(3 25920/7)=a, 4A²/(3 2880)=b, 4A²/(3*3240)=c, A²=s(s-a)(s-b)(s-c) . Wolfram returns, a=21,b= 27 & c=24 which implies, 2s=72.