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.

Euler Line

Most Geometry classes go over the different properties of triangles, including the orthocenter, centroid, and circumcenter.  The orthocenter is the point of concurrency of the three altitudes (a line segment through a vertex and perpendicular to a line containing the opposite side).

Orthocenter O of ∆ABC

The centroid is the point of concurrency of the three medians (a line segment through a vertex and through the midpoint of the opposite side).  One property of a centroid is that it cuts each median in a 2:1 ratio (see here for the proof).

Centroid P of ∆ABC

The circumcenter is the point of concurrency of the three perpendicular bisectors (a line segment through a vertex and through the perpendicular bisector of the opposite side). 

Circumcenter Q of ∆ABC

A little known fact about these three points of concurrency is that the orthocenter, centroid, and circumcenter always lie on the same line (with the exception of an equilateral triangle, in which the orthocenter, centroid, and circumcenter are actually the same point).  This was proved in 1765 by the Swiss mathematician Leonhard Euler and is consequently named after him.  Here are some examples of the Euler Line in different triangles:

Orthocenter O, Centroid P, and

Circumcenter Q of Acute ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Right ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Obtuse ∆ABC

A clever proof for the Euler Line can be found here and involves proving the similarity of two triangles containing the orthocenter O, centroid P, and circumcenter Q, or more specifically, the similarity of ∆OAP and ∆QDP in the following diagram:

First, we must recognize that in the next diagram, Q is both the circumcenter of ∆ABC and the orthocenter of ∆DEF.  Also, because D, E, and F are midpoints of the sides of ∆ABC , ∆ABC is similar to ∆DEF at a 2:1 ratio.

Therefore, since ∆ABC and ∆DEF are at 2:1 ratio, the vertex to orthocenter of ∆ABC (segment AO) and the vertex to orthocenter of ∆DEF (segment DQ) are also at a 2:1 ratio (side: AO ~ DQ).  Second, as mentioned previously, the centroid cuts the median at a 2:1 ratio, which means segments AP and DP are also at a 2:1 ratio (side: AP ~ DP).  Finally, ÐOAP is congruent to ÐQDP because they are alternate interior angles to the parallel lines formed by AO and DQ (angle: ÐOAP @ ÐQDP).  This is sufficient to prove that ∆OAP is similar to ∆QDP by SAS similarity.  Since the triangles are similar, ÐAOP is congruent to ÐDQP, which makes OP and QP segments of the same transversal OQ.  In other words, O, P, and Q (the orthocenter, centroid, and circumcenter) lie on the same line.