Chaitanya's Random Pages

October 29, 2011

The orthocentre of an isosceles triangle of base length 2

Filed under: mathematics — ckrao @ 9:48 pm

During one of my recent mathematical explorations, I wanted to calculate the distance of the points of an isosceles triangle to its orthocentre (points where the altitudes meet).

🙂

In other words, I firstly wanted to find x = PH in terms of h in the following figure.

Once x is found, the required distances will be \sqrt{1+x^2} and |h - x|. In this post I will show two ways of finding x.

Here is the approach I initially came up with. Since the angles at P and Q are each 90 degrees, quadrilateral QAPH is cyclic. Then by the intersecting chords theorem (or by the similarity of triangles BAQ and BHP),

\displaystyle 1.2 = \sqrt{1+x^2} \sqrt{4-y^2}

from which \displaystyle 1 + x^2 = \frac{4}{4-y^2}       …(1)

Also, since triangles BAQ and CAP are similar,

\displaystyle \frac{y}{2} = \frac{1}{\sqrt{1+h^2}}

from which \displaystyle \frac{4}{y^2} = 1+h^2       …(2)

Combining (1) and (2),

\displaystyle 1 + x^2 = \frac{4}{4-y^2} = \frac{4/y^2}{4/y^2 - 1} = \frac{1 + h^2}{h^2} = 1 + \frac{1}{h^2}.

It follows that \displaystyle x = 1/h. Who would have thought that such a simple answer would be found! Hence, we have rediscovered a way of constructing the reciprocal of a positive number h: draw an isosceles triangle with base length 2 and height h, then the height of its orthocentre above the base is 1/h.

Naturally, the question comes up of whether there is an easier way of calculating this answer and surely enough there is. If we extend CH until it meets the circumcircle of ABC at C', we find that \angle HBA = ACC' are equal (each being the complement of \angle BAC). Then we also have \angle ACC' = \angle ABC', both angles lying on the arc AC'. Both acute and obtuse angled cases are illustrated below.

It follows by Angle-Angle-Side that triangles HPB and C’PB are congruent, from which x = HP = PC'. In other words, the reflection of H in AB lies on the circumcircle! This fact is true for the reflection of H in any side, and ABC need not be isosceles.

By the intersecting chords theorem for CC' and AB intersecting at P (or equivalently, by the similarity of triangles CAP and BC'P),

\displaystyle 1.1 = x.h

or x = 1/h as required.

I later found that the second proof is also given in pp18-19 of Honsberger’s book entitled “Episodes in 19th and 20th century Euclidean Geometry”.

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