Friday, December 22, 2017

Altitude Reciprocals

While researching some properties of altitudes of triangles, I came across the theorem, “The reciprocals of the altitudes of any triangle can themselves form a triangle” (from Wikipedia).  Since the source of this theorem was from Wikipedia, and the theorem itself was not immediately obvious, I decided that I should prove its verity before passing it on to anyone else.  I not only found out that this theorem was true, but discovered an even stronger theorem, which is, “The reciprocals of the altitudes of any triangle can themselves form a triangle that is similar to the original triangle.”

To prove this theorem, first define an altitude as a function of the triangle’s side.  Given ABC with sides a, b, and c, the law of cosines states that cos C = a^2 + b^2 – c^2/2ab


Then the altitude ha from A to a, would be ha = b sin C.  Substituting the law of cosine value of C gives ha = b sin(cos-1(a^2 + b^2 – c^2/2ab)).  Using a right angle triangle and Pythagorean Theorem, sin(cos-1(a^2 + b^2 – c^2/2ab)) = √(4a^2b^2 – (a^2 + b^2 – c^2)^2)/2ab.


The radicand 4a2b2 – (a2 + b2 – c2)2 simplifies to 4a2b2 – (a4 + b4 + c4 + 2a2b2 – 2a2c2 – 2b2c2) = 2a2b2 + 2a2c2 + 2b2c2 – a4 – b4 – c4, and so ha = b √(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4)/2ab = √(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4) / 2a.  Letting k = √(2a^2b^2 + 2a^2c^2 + 2b^2c^2 – a^4 – b^4 – c^4)/2, the altitude ha = k/a.

Using a similar argument obtains hb = k/b and hc = k/c, and so the reciprocals of the three altitudes are 1/ha = a/k, 1/hb = b/k, and 1/hc = c/k.  These three reciprocals are proportional to the original three sides of the triangle a, b, and c by a factor of 1/k, and so the reciprocals of the altitudes form a triangle that is similar to its original triangle by SSS similarity.

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