A Triangle's Smaller Twin

An Exploration of the Medians of a Triangle

by

Sarah Major

This write-up is based on a prompt from Exploration 6 which states:

Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? have same area? same perimeter? ratio of areas? ratio or perimeters?) Prove whatever you find.

Prove the area of the triangle of medians is 3/4 the area of the original triangle. If the original triangle is equilateral, then the triangle of medians is equilateral. Will an isosceles original triangle generate and isosceles triangle of medians? Will a right triangle always generate a right triangle of medians? What if the medians triangle is a right triangle? Under what conditions will the original triangle and the medians triangle both be right triangles?

First, what do we know about the medians of a triangle? We know that there are three of them in every triangle because they are the segments from the midpoint of a side of a triangle to the opposite vertex. We also know that the three medians of a triangle are concurrent, intersecting at a point called the centroid.

I have discussed properties of a centroid in the past. For some of those properties, follow this link: Centroid. We know that the centroid is a sort of “balancing point” for the triangle, so it makes sense that the medians divide the triangle into six triangles of the same area, though they may be of different sizes depending on the original triangle. This is how, in the perfect conditions, the triangle would be able to balance on a pinpoint. They are equal areas going in all directions from the centroid.

Following this fact, we can figure out that the centroid is 2/3 the distance along each median. How can we prove this easily? If we had an equilateral triangle, this would be easy to see. All three of the medians would be the same length. To have six triangles of equal area, the sides of the triangles would have to correspond to 1/3 of the median’s length and 2/3 of the median's length.

The same can be said for triangles that are not equilateral.

But what we really want to know about is the triangle constructed from the medians. An interesting fact to know about constructing this triangle is that the medians only need to be shifted horizontally or vertically to create this triangle. There is no need to rotate the segments in order to fit them into their own triangle. In other words, the slope of the medians can remain the same if they are being moved to produce their own triangle. Therefore, if one is trying to construct the median triangle on Geometer’s Sketchpad, there is no need to construct circles to make the new triangle. Simply copy and paste the segments and move as needed.

When we compute the areas of the original triangle and the median triangle, we find that the area of the median triangle is ¾ that of the original triangle.

Why is this? Let's move the medial triangle so that one of the medians is overlapping its original location:

The orange segment is the median that is overlapping. We see that the portion where the two triangles overlap creates another triangle:

We can see from the above image that not only is a side of this triangle the median, but the blue segment of the triangle is also half of one of the other medians.

The area of this triangle is what the two triangles have in common. We can then form the triangles that the two triangles do not have in common:

We can actually the orange triangle around its right vertex and fit it into the red triangle:

So that's another portion of the area that the two triangles have in common. Now all we have left is the green triangle for the original triangle and the red triangle for the median triangle. If we find the areas of the red and the green and take the ratio, we get a ratio of 1/2.

How does this help us? From our earlier discussion about the centroid, we know that it is the balancing point, and the medians split the triangle into three equal smaller triangles. The green triangle is actually made up of three of these small triangles, so it is half of the area of the whole triangle. The other half is what the two triangles have in common, and this portion is a ratio of 1:1. If we add the green triangle to the original triangle's part of the ratio, we get 1:2. If the red triangle is half of the green triangle, we get the ratio 1.5:2, or 3:4.

If we found the medians of an equilateral triangle, they would, of course, all be the same length because the distances from the vertices to the midpoints of the opposite sides are all the same. Therefore, the median triangle would also be an equilateral triangle.

Can the same be said for an isosceles triangle? We know that while isosceles triangles have two equivalent sides, they also have two equivalent angles, which are the angles from the two equivalent sides and the third side. This means that the slopes of the two equivalent sides are the same absolute value because they share the same base and the same angle (this is given in a coordinate plane where the third side is either horizontal or vertical).

Therefore, the medians corresponding to these two vertices would also be equivalent because: they share the same base, they share the same angle, and the other segment involved in the angle has the same absolute value of the slope. Therefore, the distance to corresponding points on the two sides (the midpoint) from the vertices will always be the same.

However, because the other side and angle differs, the median triangle could not be an equilateral triangle.

Under this same argument, a scalene triangle will result in a scalene median triangle.

If we begin with a right triangle, will the median triangle also be a right triangle? Not necessarily.

Only one type of right triangle will result in a right median triangle: one of the segments forming the right triangle has to be half the length of the other segment of the right angle. This can be seen by constructing such a triangle:

We could form a square with vertices at the right angle, the endpoint of the short segment, and the midpoint of the longer segment. One side of the square would then pass through the midpoint of the hypotenuse.

If we construct the diagonals of the square, one would be a full median and the other would be a longer version of a median.

The lengths do not matter, however. The point is that they are perpendicular because we know that the diagonals of a square are perpendicular. Therefore, the median triangle would also be a right triangle. However, this only occurs when the specifications to make the above square are met.