2
$\begingroup$

Exercise

Find the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.


Attempt

Given

  • Point $P$
  • Plane $M$

Construction

Mark an arbitrary point $Q$ on $M$.

Draw line segment $l$; $l$=$PQ$.

Bisect $l$ at some resultant point $R$.

Draw plane $N$; $N || M$, $R$ lies on $N$.

Answer

$N$ is the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.


Question

I can give you the answer, as well as the construction, however I don't know how to prove that my solution is legit.


Postscript

  • This exercise is found in Kiselev's Geometry; Book II: Stereometry (English adaptation).
    • It is Exercise 17, in Chapter 1: Lines and planes, Section 2: Parallel lines and planes.
$\endgroup$
  • $\begingroup$ "I'm having a hard time proving it!" Proving what? The question asks you to find the locus. Do you have a guess as to what the locus should be? $\endgroup$ – Joey Zou Sep 2 '16 at 5:00
  • $\begingroup$ @JoeyZou : Yes, I do know it, and even how to construct it. I've updated my question. However, how can I prove that my answer is indeed correct? $\endgroup$ – Fine Man Sep 2 '16 at 5:02
  • $\begingroup$ There is a basic flaw in your proof: It is true you have obtained the right plane, by using a particular point. How can you be sure that by taking another point, and applying the same process, you will find the same plane ? $\endgroup$ – Jean Marie Sep 2 '16 at 8:45
  • $\begingroup$ @JeanMarie : That's exactly my dilemma. Intuitively I know I'm right, but how can I prove that regardless of what point $Q$ I take, I'll get the same result? $\endgroup$ – Fine Man Sep 2 '16 at 17:21
2
$\begingroup$

Your transformation is an homothetic transformation $h$ whose center is the given point $S$ and ratio $\lambda=\frac{1}{2}$.

The image of the plane by $h$ is another plane. More precisely the image is the plane passing through the images of three independent points lying on the initial plane.

$\endgroup$
0
$\begingroup$

I complement the (very accurate) solution of @ mathcounterexamples.net by an analytical proof:

If the plane $(P)$ has equation

$$\tag{1}ux+vy+wz=h$$

and the point, say $P_0$, has coordinates $(x_0,y_0,z_0)$, the locus of midpoints is the set of points having coordinates $$\begin{cases}X&=&\frac{x+x_0}{2}\\Y&=&\frac{y+y_0}{2}\\Z&=&\frac{z+z_0}{2}\end{cases} \ \ \ \Leftrightarrow \ \ \ \begin{cases}x&=&2X-x_0\\y&=&2Y-y_0\\z&=&2Z-z_0\end{cases}$$

Plugging these expressions into (1), we get:

$$u(2X-x_0)+v(2Y-y_0)+w(2Z-z_0)=h$$

which induces the following constraint on the set of points $(X,Y,Z)$:

$$\tag{2} uX+vY+wZ=\dfrac{1}{2}(ux_0+vy_0+wz_0)$$

The geometrical interpretation of (2) is that the set of points $(X,Y,Z)$ belongs to a certain plane $(P')$ parallel to the initial plane $(P)$ (they share the same normal vector $(u,v,w)$).

Moreover, as we have proceeded by equivalent conditions, the solution is the whole plane.

$\endgroup$
  • $\begingroup$ This isn't what I'm looking for, but I won't vote you down. I want the answer to be just Euclidean-geometric, not algebraic-geometric. The book I go by isn't algebraic-geometry. Hence, I didn't add "algebraic-geometry" tag. $\endgroup$ – Fine Man Sep 2 '16 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.