Two squares, and it is almost Pythagorean.....
I had a student ask me this one. He thought I might want to add it to my almost Pythagorean list... I do, but could I prove it's true? He asked:
If one square is inside another with corresponding sides parallel and the vertices of the inner square are connected in order to the sides of the outer square, then the sum of the squares of the distances of the connections from opposite vertices will be equal. For the image below, GD2 + EB2 = FC2 + HA2.
We played around at the board for awhile and finally proved it with nothing fancier than the Pythagorean Theorem and simple algebra. My method was to draw a horizontal segment (labled R for right shift) to mark how far the inner square is moved from the left edge of the outer square and then a vertical one, D, for down-shift. Then with lables of L for the large squares side and S for the small squares side (not creative, but easy to remember) I could write the length of all four segments. (DG)2 for instance would be D2 + (R+S2. We expanded out the squares of all the binomials and trinomials (and he corrected me when I wrote D for R or got a negative as the product of two negatives) and then crossed off mathching parts until everything was gone.. the two expansions were the same alphabet soup of algebraic variabless. QED.
One student witnessing our work commented that we had covered the board and never written a number. I seized the moment to explain what power that was, and began to expound on the implications for possible generalizations... since L, S, R, and D could be any positive (actually two could have been negative) value, and had no relative order or size, S did not have to be smaller than L... and R did not have to be smaller than L... conclusion??? It must also be true that the squared distances are equal even if the one square is not inside the other. It could be partly inside, or totally outside and either could be larger.
AND.... pausing for breath... S could be very, very small..... approaching zero, so the degenerate case would have the smaller square as a point....giving another corollary :
For a square and a point on the same plane, the square of the distances from the point to one pair of opposite vertices will have the same total as the square of the distances to the other pair. (Why have I never seen this theorem?... Is it in textbooks?) And I propose that the condition that the point is in the same plane is unnecessary. In fact that seems almost trival since it only adds twice the square of the perpendicular distance from the point to the plane of the square... if my mind is working correctly this morning. If that is true, that also means that for any square based pyramid, the squares of the sums of the lengths of two opposite edges from the base to the vertex will also be equal.
Now that I was on the roll, I kept playing after the kids wandered away...
What about if the sides of the squares are not parallel. Playing around with Geogebra, I have determined that that is not a necessary condition. The original theorem is true for any two squares in the plane however oriented... that is, they do not have to have their sides parallel. (proof needed, submit yours and save me some time, I'll amend this part)
what about rectangles.??..... If we inscribe a square in a rectangle, then the degenerate case (the inner square is a point) is still true, but with a square and a rectangle the vertex to vertex sums are not equal... in fact, I have discovered that they will differ by a constant value (no matter where the square is or how it is rotated ) that is equal to the product of twice the side of the square times the diffence in the sides of the rectangle. This property, however, is not preserved when the square is rotated inside the rectangle. I think that there would be a different constant difference for any orientation of the corresponding sides of the square and rectangle (untested wild conjecture).
While trying to search for the history of the problem, I found a similar relation about parallelograms In the MAA reprint of Jacques Hadamard's Geometry, a similar problem appears on page
If we replace the square by a rhombus, and sum the squares of the distances from a point to a pair of opposite vertices, I believe the two sums will differer by s^2. I have not had time to prove this yet (offers accepted).
While I'm in the mood to make conjectures, If the two squares lie in parallel planes then the sum of squares will almost cerainly still be equal. I imagine a physical construction of the two squares in the plane, with elastic chords connecting the vertices. When I lift one square out of the plane and translate it along a line perpendicular to the original plane I am adding a distance of 2* h^2 to each pair of vertices sum. Since the two squares could have been in any orientation originally, it seems that no special precaution about the orientation of the squares would be needed. This would seem to have application to skew prisms, frustums of pyramids and skew frustums (I'm not sure we have such a definiton in geometry, but I mean a frustum in which one base is not similarly aligned with the other when projected on the same plane.
I suspect that if the two squares are not even in parallel planes, the sum of the squares of segments containing alternate vertices will be equal also. No time to attack that yet.
A first postscript: It seems that for any even sided regular polygon, the sums of squares of distances from a point to pairs of opposite vertices would be equal also since any two pairs of these opposite verties would form a rectangle. Likewise, if two even sided regular polygons have opposite vertices connected in sequential order, the same property is preserved as for squares.