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.

  • $\begingroup$ What's your problem in that task? Where did you get stuck? $\endgroup$ – Michael Hoppe Nov 9 '19 at 13:36
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    $\begingroup$ In which sense it slides? $\endgroup$ – user Nov 9 '19 at 13:38
  • $\begingroup$ Sliding in the sense that the triangle rotates anti clockwise. But I don't think the direction is important. $\endgroup$ – Mike Karter Nov 9 '19 at 14:45

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.

  • $\begingroup$ Thank you Michael, finally understood the approaches and solved it too. $\endgroup$ – Mike Karter Nov 10 '19 at 13:58

It is easier to examine the case where the axes are rotated 45 clockwise as shown.

enter image description here

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,


Thus, the locus is an ellipse. To get the equation for the slant ellipse, just apply the 45-degree rotation matrix.

  • $\begingroup$ How do you know that approaching the problem with rotated axes will be helpful? (Does it come from experience or there's something to observe). Also,can you explain how did you write the coordinates of vertices? I didn't get it. Sorry for the delay in response. Thank you. $\endgroup$ – Mike Karter Nov 10 '19 at 8:40
  • $\begingroup$ Got the coordinates. $\endgroup$ – Mike Karter Nov 10 '19 at 9:11
  • $\begingroup$ Thank you Quanto, finally understood the approaches and solved it too. $\endgroup$ – Mike Karter Nov 10 '19 at 13:59
  • $\begingroup$ @Mike Karter- Glad you figured out. The 45-degree rotation is based on the observation that the locus is symmetric with respect to the 45-degree line as the hypotenuse vertexes slide along the axes and it is usually convenient to derive its equation with perpendicular symmetry, especially for conics. I’d say experience helps as well. $\endgroup$ – Quanto Nov 10 '19 at 14:46

enter image description here

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.)

  • $\begingroup$ Any ideas on how to begin? $\endgroup$ – Mike Karter Nov 9 '19 at 14:50
  • $\begingroup$ @MikeKarter I guess you'd say that $A$ is at $(0,p)$ for some $p$ between $0$ and $\sqrt2$, then calculate that $E=(2-p^2,0)$ and work on from there until you have the coordinates of $K$ as a function of $p$. Then try to take that parametric equation and turn it into a function of $y$ in terms of $x$. I don't envy you. If there's a shortcut, it would take a better geometer than me to find it. $\endgroup$ – Matthew Daly Nov 9 '19 at 15:01

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