Question

An isosceles right angled triangle whose sides are 1, 1, $\sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the coordinate axes. If it slides prove that the locus of its centroid is $(3x-y)^2 + (x-3y)^2 = \frac{32}{9}$.

Answer 1

$(3x-y)^2 + (x-3y)^2 = \frac{32}{9}$

Answer 2

Let the vertices of the isosceles right-angled triangle on the coordinate axes be $A=(a,0)$ and $B=(0,b)$. Since the hypotenuse length is $\sqrt{2}$, we have $a^2 + b^2 = (\sqrt{2})^2 = 2$. As the triangle lies entirely in the first quadrant, $a > 0$ and $b > 0$.

Let $C=(x_C, y_C)$ be the vertex with the right angle. The lengths of the legs $AC$ and $BC$ are 1. $AC^2 = (x_C - a)^2 + (y_C - 0)^2 = 1^2$ $BC^2 = (x_C - 0)^2 + (y_C - b)^2 = 1^2$

The vertex $C$ can be found by considering that it lies on a circle with diameter $AB$. The midpoint of $AB$ is $M(\frac{a}{2}, \frac{b}{2})$, and the radius is $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. The vector $\vec{MC}$ is perpendicular to $\vec{AB}$ and has length $\frac{1}{\sqrt{2}}$. The vector $\vec{AB} = (-a, b)$. A vector perpendicular to $\vec{AB}$ is $(b, a)$. So, $\vec{MC} = \pm \frac{1}{\sqrt{2}} \frac{(b,a)}{\sqrt{a^2+b^2}} = \pm \frac{1}{2}(b,a)$. Thus, $(x_C, y_C) = (\frac{a}{2}, \frac{b}{2}) \pm (\frac{b}{2}, \frac{a}{2})$.

This gives two possibilities for $C$:

1. $C_1 = (\frac{a+b}{2}, \frac{a+b}{2})$
2. $C_2 = (\frac{a-b}{2}, \frac{b-a}{2})$

Since the triangle lies entirely in the first quadrant ($a>0, b>0$), $C$ must have positive coordinates. $C_1$ always satisfies this. $C_2$ only has positive coordinates if $a=b=1$, leading to $C_2=(0,0)$. For a continuous sliding triangle, $C_1$ is the appropriate vertex. So, $C = (\frac{a+b}{2}, \frac{a+b}{2})$.

The vertices of the triangle are $A=(a,0)$, $B=(0,b)$, and $C=(\frac{a+b}{2}, \frac{a+b}{2})$. The centroid $G=(x,y)$ is given by: $x = \frac{a + 0 + \frac{a+b}{2}}{3} = \frac{3a+b}{6}$ $y = \frac{0 + b + \frac{a+b}{2}}{3} = \frac{a+3b}{6}$

We have: $6x = 3a+b \quad (1)$ $6y = a+3b \quad (2)$ And the constraint $a^2+b^2=2$.

Solving for $a$ and $b$ from (1) and (2): Multiply (1) by 3: $18x = 9a+3b$. Subtract (2): $18x - 6y = 8a \implies a = \frac{18x-6y}{8} = \frac{9x-3y}{4}$. Multiply (2) by 3: $18y = 3a+9b$. Subtract (1): $18y - 6x = 8b \implies b = \frac{18y-6x}{8} = \frac{9y-3x}{4}$.

Substitute these into $a^2+b^2=2$: $\left(\frac{9x-3y}{4}\right)^2 + \left(\frac{9y-3x}{4}\right)^2 = 2$ $\frac{(9x-3y)^2}{16} + \frac{(9y-3x)^2}{16} = 2$ $(9x-3y)^2 + (9y-3x)^2 = 32$ $[3(3x-y)]^2 + [3(3y-x)]^2 = 32$ $9(3x-y)^2 + 9(3y-x)^2 = 32$ $9(3x-y)^2 + 9(x-3y)^2 = 32$ Divide by 9: $(3x-y)^2 + (x-3y)^2 = \frac{32}{9}$ This proves the locus of the centroid.