Construction of n-gon from the midpoints of its sides
Construct an n-gon (polygon with n sides) for which n given points serve as midpoints of its sides.
Below there is a Java applet that may help you gain insight into the problem. The applet can be in two modes: "Place points" and "Drag mouse", depending on which of the two boxes at the bottom of the applet is checked.
- In the "Place points" mode you should define n points - midpoints of the sides of a polygon. Just click the mouse anywhere inside the applet's rectangle.
- Next you can experiment with the points you thus defined. Change the mode and start dragging the mouse.
If, for whatever reason, you decide to work with a different set of points - press the Reset button and start again.
A word of advice. First of all, start with a simple case. Try just a few points, say, 3, 4, 5. Secondly, attempt to visualize a polygon and then place the points approximately at the middle of its sides. Thirdly, pay attention to the red "target" circle.
What if applet does not run? |
By now you ought to have arrived at the following conclusions:
- If n is odd it's probably possible to solve the problem.
- If n is even something goes wrong.
For odd n, the red circle indicates one of the polygon's vertices. Once you move the cursor into the red circle the problem seems to have been solved. Now, actually the applet provides an experimental device. It may help you to fathom a proof but it can't substitute for a rigorous demonstration.
Useful for the proof could be the following observation: Let there be given a segment AB and a point O_{1}. Consider a reflection A_{1}B_{1} of AB in the point O_{1}. In other words, in order to obtain A_{1}B_{1} rotate AB around O_{1} 180^{o}. Now AB || A_{1}B_{1} but they have a different orientation. If you continue reflecting A_{1}B_{1} in another point O_{2} to get A_{2}B_{2} then the segments remain parallel and equal but orientation changes again. After an odd number (1 in particular) of reflections the last segment will be parallel to AB but having a different orientation. After an even number of reflections the first and the last segment will be directly parallel (and, of course, equal). See if you can now complete the proof.
Remark 1
When n is even the segment joining the first and the last point is always the same for a given configuration of vertices. For
Remark 2
The problem could be described algebraically. Assume m_{i}, i = 1, 2, ..., n, stand for the vectors from the origin to the given midpoints. (We may also treat them as the points in an affine space.) Let a_{i}, i = 1, 2, ..., n, denoted the unknown vertices. Then a_{i}'s satisfy the following system of equations:
(1) |
a_{1} + a_{2} = 2m_{1} a_{2} + a_{3} = 2m_{2} ... a_{n} + a_{1} = 2m_{n}. |
Surprisingly this same system of equations describes an apparently unrelated problem.
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Copyright © 1996-2017 Alexander BogomolnyLet's consider the simplest case of three midpoints (n = 3). After toying with the applet for a while, it's hard not to notice that the center of the red circles does not depend on the position of the cursor but only on the three selected midpoints. Let A be an arbitrary point. Successive reflections in O_{1}, O_{2}, and O_{3} produce points A_{1}, A_{2}, and A_{3}. It seems very plausible that the center of the red circles, say M, lies in the middle of the segment AA_{3}. Put differently, the point A_{3} is obtained from A by reflection in the point M. This is something we may try to prove.
But before we do, let's create an accurate picture of what happened. Three successive reflections resulted in another reflection. And the same is probably true of any odd number of reflections. Reflections are just functions: they map points of a plane into points of the same plane. Successive application of two reflections is none other than the composition of the corresponding functions. Thus we want to show that composition of an odd number of reflections is again a reflection.
What about an even number of reflections? What about just two reflections? Composition of two reflections is a translation to the vector parallel to the line connecting the two centers of reflection and twice as long as the distance between the centers. Indeed, in ΔAA_{1}A_{2},
Translation is a transformation of a plane that moves the plane as a whole in a certain direction to a certain distance. Each point is shifted by a certain vector. Composition of two translations with vectors v_{1} and v_{2}, respectively, is again a translation corresponding to the vector sum v_{1} + v_{2}. The latter remark easily leads to a solution of our problem for an even number of points. If n is even, we can combine successive reflections in pairs to get n/2 successive translations. Obviously (e.g., by induction) the sum of any number of translations is a translation. This is true as long as we may refer to the triangular diagram above. But sometimes the sum of vectors is 0 (one side of a triangle shrinks to a point, and the triangle does not appear triangular in the least.) To make the italicized statement universally true we have to accept as a translation the identical transformation under which every point is mapped onto itself. The identical transformation has all points fixed. Since any translation moves all points by the same vector, the only translation that has a fixed point is identical.
Returning to our problem for n even. Let vectors v_{i}, i = 1,...,n/2 map, respectively, O_{1} to O_{2}
- v_{1} + v_{2} + ... + v_{n/2} = 0, and
- v_{1} + v_{2} + ... + v_{n/2} ≠ 0.
In case 1, pick any point A and obtain points A_{i} by successive reflections in points O_{i}. Then A_{2n} will coincide with A thus solving the problem. Since A is arbitrary, the problem has an infinite number of solutions.
In case 2, no solution exists.
For n odd, the problem always has a unique solution. As we already noticed, we have to show that the composition of an odd number of reflections is again a reflection. Taking the center of the latter as a starting point A which is successively reflected in points O_{i} we'll find that the point A_{n} coincides with A. To prove the statement, note that (by induction) composition of an odd number of reflections is the composition of a reflection and a translation (or of a translation and a reflection depending on how one groups our reflections.)
Choose two points A and B, shift them into A_{1} and B_{1} by vector v, and then reflect these two in a point O to obtain A_{2} and B_{2}. As we already know, the quadrilateral ABA_{2}B_{2} is a parallelogram. Let P denote the point of intersection of its diagonals. This is the midpoint of the segment AA_{2}. Now let B vary. Regardless of its position, the midpoint of the segment BB_{2} will always coincide with the midpoint of AA_{2}, that is P. Therefore, B_{2} is always obtained from B by reflection in P.
From the above discussion it follows that all planar translations form a group, and another group is formed by all translations and reflections. The group of translations is commutative (Abelian); for the sum of vectors is commutative. The combined group of translations and reflections is not commutative. It can be shown (try it) that, in general, the composition TR of a translation T and a rotation R is not the same as RT.
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