enter image description here

The task is to find the centroid of the given triangle (see the image above). We also should use the fact that the volume of a cone of radius $r$ and height $h$ is $V = \frac{1}{3}\Pi r^2h$. My solution:

$1)~$ Denote the dots as following: $~O(0,0),~~ R(a, 0),~~ Q(a, b),~~ P(a, c)~$;

$2)~$ Find the volume of the cone with radius OR and height PR:

$$V = \frac{1}{3}\Pi r^2h$ = $\frac{1}{3}\Pi a^2c$$

$3)~$ Find the volume of the cone with radius OR and height RQ:

$$V = \frac{1}{3}\Pi r^2h$ = $\frac{1}{3}\Pi a^2b$$

$4)~$ If we rotate triangle OQP about the line $~x = a~$, then according to Pappus theorem it's volume is equal to:

$V = 2\Pi pA$, where

$p -~$ distance from centroid of OQP to axis of revolution (x = a);
$A -~$ area of triangle OQP;

$5)~$ The next step is to find area of OQP. We will do this with formula for right triangle $\frac{1}{2}ab$ $(a$ and $b$ are edges$)$. The area of triangle ORP is equal to $\frac{1}{2} ac$, and the area of OQR is equal to $\frac{1}{2} ab$. Thus, the area of OQP is equal:

$$A = \frac{1}{2} ac - \frac{1}{2} ab = \frac{1}{2} a(c-b)$$

$6)~$ We can now find the volume of solid, which will be created by revolving triangle OQP about line $x = a$ using the following approaches:

$$V = \frac{1}{3}\Pi a^2c - \frac{1}{3}\Pi a^2b$$ (difference of cone volumes)

$$V = 2\Pi pA = 2\Pi p\frac{1}{2} a(c-b)$$ (Pappus Theorem)

This to expressions are equal. We simplify the equality and receive that $p = \frac{a}{3}$ (the distance from centroid to axis of revolution $(x = a)$.

$7)~$ As this distance is equal to $p = \frac{a}{3}$, then $x$ coordinate of centroid is equal to $p = \frac{2a}{3}$. Also, we know that centroid of triangle is on median. So we draw the line between dots $(0, 0)$ and $(a, \frac{c+b}{2})$. This line has an equation $y = \frac{c+b}{2a}x$. If we replace $x$ with $\frac{2a}{3}$, the $y$ is equal $\frac{c+b}{3}$. So, the coordinates of centroid is $\left(\frac{2a}{3} , \frac{c+b}{3}\right)~$.

However, the textbook I use give the following answer $\left(\frac{2a(a-b}{3(c-b)} , \frac{c+b}{3} \right)~$.

I will be very grateful if anybody can explain why the answers is different. Thank you in advance! (Please, note that y coordinate is correct.)

  • 1
    $\begingroup$ The centroid of a triangle has barycentric coordinates $1:1:1$, which here corresponds to $\frac13(O+P+Q)$. Your solution is correct. $\endgroup$ – amd Sep 11 at 9:59

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