Locus of centroid for a right triangle with hypotenuse vertexes moving alone $x$ and $y$ axes An isosceles right triangle, with unit length for the equal sides, lies entirely in the first quadrant with the ends of hypotenuse on the coordinate axes. Find the equation of locus of centroid of the triangle if it slides.
Any hints to begin or the full solution will be appreciated. 
 A: 
Very simple geometric solution:

We slide the triangle on the coordinate system's bisectors instead of its axis.   Let $AB$ the hypotenuse with $A(u,u)$ and $B(v,-v)$. From $|AB|^2=2$ we get $u^2+v^2=1$.  Verify (via inner product or Pythagoras) that $C(u+v,0)$ is the triangle's third point. Now the centroid is given by $(2u+2v,u-v)/3$, which gives $x^2+4y^2=8/9$ as the equation of the desired curve.

Former (algebraic) solution:

Let $A(u,0)$ and $B(0,v)$ be the endpoints of the triangles hypotenuse, hence $u^2+v^2=2$.  The centroid is the average if the three vertices.  To compute the third vertex $C$ one might find the intersection of the circle around $(u/2,v/2)$ with radius $\sqrt2/2$ and the perpendicular line of $AB$ through that point.
For the first quadrant you'll get under heavy use of $u^2+v^2=2$ that its first coordinate is $(u+v)/2$, its second coordinate will be $(u+v)/2$ as well.  (Fill in the gaps, please.)  As $C$ lies on the line $y=x$ I'm sure that there's a simpler geometric way to achieve this result.
Now calculate the centroid.  Using polar coordinates it's quite easy to find that the curve is indeed an ellipse as Matthew suggested.
A: It is easier to examine the case where the axes are rotated 45 clockwise as shown.

Let the angle $\alpha$ be the variable for parametrization. Then, the coordinates of the vertexes are 
$$A(\cos\alpha, \cos\alpha),\>\>\>\>\>
B(\sin\alpha, -\sin\alpha),\>\>\>\>\>
C(\cos\alpha+\sin\alpha,0)$$
The coordinates for the centroid of the triangle ABC are
$$x=\frac13 (x_a+x_b+x_c)=\frac23 (\cos\alpha+\sin\alpha)$$
$$y=\frac13 (y_a+y_b+y_c)=\frac13 (\cos\alpha-\sin\alpha)$$
which yields the following equation,
$$\frac{x^2}{\frac89}+\frac{y^2}{\frac29}=1$$
Thus, the locus is an ellipse. To get the equation for the slant ellipse, just apply the 45-degree rotation matrix.
A: 
Hint: I suspect you're in for a wall of algebra, but it looks like it's going to be the arc of an ellipse.  (This trace was done for all $E$ on the right half plane even when $A$ was below the x-axis, so your arc will be smaller than this.  I drew the complete one for the sake of highlighting the elliptical nature of the locus.)
