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. Yours, Alazah
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 http://mathforum.org/dr.math/
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