Assignment 8: Investigating Altitudes and Orthocenters

By Rebecca L. Adcock

In previous assignments, weÕve looked at triangles, orthocenters, and circumcenters. In this lesson, weÕre going to look at those objects again and see if we can find anything new. We can choose any three points and construct a triangle. WeÕll call it ABC and weÕll also construct the orthocenter of the triangle.

*Review**: We
locate the orthocenter by constructing the triangleÕs altitudes. The altitude
is a line that passes through the vertex and is perpendicular to the side of
the triangle opposite the vertex.*

Above we have the triangle ABC with the orthocenter at H. the dashed lines are the altitudes. Where they intersect is the orthocenter.

Now weÕre going to look at the three triangles created by using the orthocenter as one vertex of each. HereÕs what the three triangles look like. See below.

Now we locate the orthocenter for each of the smaller
triangles. Before we begin, wouyld you care to speculate on the locations of
the smaller trianglesÕ orthocenters? I was surprised by their locations and
even reconstructed my original triangle in case I had made a mistake. *But it
was no mistake. HereÕs what happenedÉ..*

* *

**The orthocenter for each
of the smaller triangles is one vertex of the original triangle.**

Now letÕs construct the circumcircle for the original triangle and for the three smaller triangles and see what happens.

*Review:** A
circle can be constructed given a point and a radius. To create the circumcircle,
we need to locate the circumcenter of the original triangle. The circumcenter
is the intersection of the perpendicular bisector of each side of the triangle.
*

* *

Above, weÕve located the circumcenter. So we have a point. How do we get a radius? Since the circumcenter is equidistant from the three vertices of the triangle, a line segment drawn from the circumcenter to any vertex could serve as the radius.

Here weÕve drawn the circumcircle given the center of the triangle (circumcenter) as the center of the circle and the line segment drawn from the circumcenter to a vertex of the triangle as the radius. (The green line segment visible in the yellow triangle is the radius. That just happens to be the one I chose. We could have used any one of the three.)

Now letÕs construct the circumcircle of each of the smaller
triangles using the techniques we used above. Since the yellow, green, and blue
triangles are smaller than the original, I would guess that their *circumcircles
would be smaller *than the one we just drew.
What do you think?

Here is the circumcircle created for the green triangle.

HereÕs the circumcircle for the blue triangleÉ.

And for the yellow triangleÉ

Above you see all four circumcircles for the four triangles.

And guess what? I was wrong. I had guessed that the three
circumcircles for the smaller triangles would be smaller than the circumcircle
for the original triangle. **It appears that all the circumcircles are the
same size.**

** **

Before we list our conjectures and conclusions for this exercise, letÕs look at two more pictures. I was curious what would happen if we created a triangle using the centers of the fellow, blue, and green circles as the vertices.

Since the two triangles looked similar, I tested them by taking some measurements. (Measuring does not constitute proof in geometry, it just lets you know if youÕre on the right track.) The two triangles are congruent. WeÕre on the track of something but IÕm not sure what.

Then I wondered what would happen if we connected the 6 vertices of the two triangles.

Now weÕve got a hexagon. ItÕs not a regular hexagon, which would have all sides equal and all interior angles equal. But it looks to me like the pairs of opposite sides may be parallel. LetÕs check that out.

The two black lines were constructed individually as parallel to the opposite side . I also verified the other pairs of opposite sides.

HereÕs what weÕve learned from this exercise.

Given a triangle ABC and its orthocenter H, we can construct three triangles HAC, HAB, and HBC such that:

- the circumcircles for all four triangles are congruent.
- the intersection of any two of the circumcircles belonging to the smaller triangles is a vertex of the given triangle ABC.
- a triangle formed by connecting the centers of the circumcircles for HAC, HAB, and HBC is congruent to the original triangle ABC.
- an interesting hexagon whose opposite sides are parallel and congruent is created by connecting the six vertices of the two triangles.