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Orthic Triangle

Date: 04/09/99 at 19:13:57
From: Mohammad alazah
Subject: Geometry

Dear Sir:

Find the angles of the triangle ABC, which is similar to triangle 
A1B1C1, where AA1, BB1, and CC1 are the altitudes of triangle ABC.

I tried to guess the answer and I thought the answer must be an 
equilateral triangle.


Date: 04/10/99 at 04:19:08
From: Doctor Floor
Subject: Re: Geometry

Hi, Alazah,

Thanks for your question!

Triangle A1B1C1 is usually referred to as the "orthic triangle." The 
point where AA1, BB1, and CC1 concur is usually referred to as the 
"orthocenter," denoted by H.

When ABC is acute we get the following picture, where I have used 
A'B'C' instead of A1B1C1:

We can see that in the quadrilateral BC'HA', angle A' and angle C' are 
both 90 degrees, so BC'HA' is a cyclic quadrilateral. This means that 
angles HBA' and HC'A' must be equal, since they cut off the same arc 
of the circle circumscribing BC'HA'. In triangle BB'C we see that 
angle HBA' = angle B'BC = 90 - angle C (degrees).

So angle HC'A' = 90 - angle C.

In the same way we find angle HC'B' = 90 - angle C.

This gives in triangle A'B'C' that angle C' = 180 - 2*angle C. 

So now we have:

    angle A' = 180 - 2*angle A
    angle B' = 180 - 2*angle B
    angle C' = 180 - 2*angle C

If we have angle A = angle A', anlge B = angle B' and angle C = angle 
C', then we indeed find an equilateral triangle.

In other cases this is less clear. But, for instance, if angle A = 
angle B', angle B = angle C', and angle C = angle A', then we would 
have similar triangles too. We find

    angle C = 180 - 2*angle A
    angle A = 180 - 2*angle B
    angle B = 180 - 2*angle C

We find then 

     angle A = 180 - 2*(180 - 2*angle C)
             = 180 - 2*(180 - 2*(180 - 2*angle A))

This simplifies to 9*angle A = 540, so angle A = 60. And again we 
have an equilateral triangle.

When ABC is obtuse, let's say angle A > 90 degrees, we find other 
angle identities (as we did for acute triangles):

    angle A' = 2*angle A - 180
    angle B' = 2*angle B
    angle C' = 2*angle C

Clearly, here we cannot assume angle A' = angle A.

But again, when angle A = angle B', angle B = angle C' and angle C = 
angle A', then we would also have similar triangles. This gives

    angle C = 2*angle A - 180
    angle A = 2*angle B
    angle B = 2*angle C

We find

    angle C = 2*(2*angle B) - 180
            = 2*(2*(2*angle C) - 180

This results in 7*angle C = 180 , so angle C = 25 5/7.

This gives the triangle:

    angle C = 25 5/7
    angle B = 2*angle C = 51 3/7
    angle A = 2*angle B = 102 6/7

So here we have a second example of a triangle that is similar to its 
orthic triangle.

I hope this helped!

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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