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 h_a from ∠A to a, would be h_a = b sin C.  Substituting the law of cosine value of ∠C gives h_a = 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 h_a = 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 h_a = ^k/_a.

Using a similar argument obtains h_b = ^k/_b and h_c = ^k/_c, and so the reciprocals of the three altitudes are ¹/h_a = ^a/_k, ¹/h_b = ^b/_k, and ¹/h_c = ^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.