Assignment #8

by

Jim Meneguzzo

The following discussion with start with an investigation of the orthocenter and circumcenter of triangles. Let us look at the following graph:

The following graph contains four triangles, one large triangle(ABC), and then three other triangles constructed from the orhtocenter(H) of the first triangle. Our first investigation dealt with finding the orthocenters of the three inner triangles and it was found that they are the non common vertice of the outer triangle, that is, HAB orthocenter is C and HBC orthocenter is A, and so forth.

The next investigation dealt with finding the circumcenters of the four triangles and these are points K,N,Q, and P above. Listed below is the circumcenter with its triangle:

ABC-K

HBC-N

HAB-P

HAC-Q

At first glance, it seems that the circumcenter for the outer triangle is the only one inside it's triangle, this seems to hold true if you adjust the graph to make one of the inner triangles become an outer triangle as such:

Moving H around creates some neat inquiries, such as, moving H so that it is concurrent with one of it's triangles vertices, such as A, when this happens it seems to make K and N become concurrent and they seem to be the midpoint of the side opposite H and A, let us view a graph to illustrate this:

This also holds true when H is made concurrent with the other two vertices of it's triangle(B and C).

In summary, it has been shown that in constructing an acute triangle with three inner obtuse triangles created by the orthocenter of the acute triangle, you get some interesting results about the orthocenters and the circumcenters. First, the only circumcenter that lies within it's triangle will be that of the acute triangle. Second, when the orthocenter of the acute triangle becomes a vertex of the acute triangle, the cirumcenter of the acute triangle and the circumcenter of one of the obtuse triangles become concurrent and they are the midpoint of the side opposite the vertex created by the orthocenter of the acute triangle.