Wednesday, 2 February 2011

Midpoint Madness

I have written a few times about the medians of triangles here, and here. Here is a nice problem about midpoints suitable for really bright HS students that I adapted from an old Math Central problem at the Univ of Regina.
Given the three midpoints of a triangle, find the vertices of the triangle.

Beware, the solution will follow:
S
P
O
I
L
E
R
S
P
A
C
E

I think the easiest way to solve this, in fact, the only way I can think of at the moment, is to use vectors...which are almost totally ignored in US Geometry classes...

If we label the midpoints of sides a, b and c respectively as Ma, Mb, and Mc then we know that the vector from the origin to Ma (I'll label vectors and points with the same notation, for convenience) and the vectors B and C will obey B+C=2Ma. In the same way A+B=2Mc and A+C=2Mb.(Students should draw a segment AB and its midpoint and see why this vector addition must always be true by creating a parallelogram from two copies of the vectors OA and OB)

From these three equations we may show that since
A+B=2Mc THEN 2Mc-B= A

And since B+C=2Ma we know that B=2Ma-C... substituting that into 2Mc-B= A
gives us 2Mc-(2Ma-C)= A = 2Mc-2Ma+C

Now we use c=2Mb-A to get A= 2Mc-2Ma+(2Mb-A) or A= 2Mc-2Ma+2Mb - A...

Sooooo

2A = 2Mc-2Ma+2Mb or A = Mc-Ma+Mb...
Now we can just construct the three vectors to find the vertex A.

In a similar way we could get B=Ma-Mb+Mc and C=Mb-Mc+Ma.. but once you have one of the points, you can generate the others by construcing the segments in question.


All of this can be done by conventional compass and straightedge construction. The image shows the three vectors used to construct point A (in red) and adding the two blue vectors to the vector Mc will give Point B. I have ommitted vectors for point C to avoid clutter.

I think one big idea for students is that you can place the midpoints on a blank sheet of paper and put the origin anywhere you want and it will construct exactly the same triangle. This is a nice idea for them to grasp. With interactive geometry software you can actually create the whole thing on a page without coordinates, then grab the point you called the origin and move it around and only the vectors move with it.. the midpoints, vertices and sides of the triangle stay fixed.

It is interesting that you can do a similar approach for ANY odd sided polygon, but not for even sided ones. For the even sided polygons there may not be any solution, and if there is, then there are an infinite number of them. (if you try this for four sided figures, remember that the four midpoints of a quadrilateral must form a parallelogram.... you probably proved that in HS geometry)

Using the same vector approach as above, the substitution leads to a string of vectors that equal zero..... Ma-Mb+Mc-Md+.....=0 . Thus you can pick any point on the plane to be point A, and generate the n-gons from there.

If the number of sides is odd and greater than three, it is not even necessary that the midpoints all lie on a plane. The same method will work to build a two or three space solution for any odd number of vertices.

4 comments:

Anonymous said...

Perhaps I am missing something, but for a triangle I believe there is a quicker or more direct construction.

Construct through A a line parallel to Mb-Mc. Construct through B a line parallel to Ma-Mc. Construct through C a line parallel to Ma-Mb.

The intersection points will be the vertices of the triangle.

Why? The segments joining the midpoints are parallel to the further side, and pass through the last midpoint. We are actually constructing the sides.

Jonathan

Pat B said...

JD....
Yep, that has to work too doesn't it.. In fact, we could construct a parallel to Mb-Mc that was as long as Mb-Mc and find a vertex point directly without doing any intersections

Pat B said...

Ooops, JD.

I closed my comment too early.... try both methods on a quadrilateral..... I think I originally saw the process applied to a pentagon, but tried to simplify the question...too much it seems..but I think a quadrilateral would be sufficiently complex (although I may be overlooking a simple method there too)

Anonymous said...

Yes, what I suggest works for triangles, but not polygons with more than 3 sides.

Jonathan