# Count the Orthocenters

It's a common knowledge that three altitudes in a triangle intersect in a point known as the orthocenter of the triangle. So let's start with a triangle ABC and draw all three of its altitudes AH_{a}, BH_{b}, and CH_{c}.

Now there is an interesting question: How many triangles with altitudes drawn are there in the diagram?

The answer is 4 and, besides ABC, there are three more triangles: AHB, BHC, AHC, H being the *orthocenter* of ABC. Indeed, since AH_{a} is orthogonal to BC so BC is orthogonal to AH. (In general terms, the relation of orthogonality is *symmetric* and instead of claiming that one line is orthogonal to another we may simply say that the two lines are orthogonal.) Again, by the symmetry of orthogonality, AC serves as another altitude in ΔAHB. Thus, in ΔAHB we have three altitudes: CH_{c}, BC, and AC that intersect at the point C, the *orthocenter* of ΔAHB.

Similar considerations apply to the triangles BHC and AHC.

## References

- D. Wells,
*You are a Mathematician*, John Wiley & Sons, 1995

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