# 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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