Can we find the area of a pedal triangle?

Can we find the area of a pedal triangle?

Question

A triangle $ABC$ has sides $BC$, $CA$, and $AB$ of lengths $a$, $b$ and $c$ respectively, and angles at $A$, $B$ and $C$ are $\alpha$, $\beta$ and $\gamma$ where $0 \leq \alpha ,\beta ,\gamma \leq \frac{1}{2}\pi$.

1. Show that the area of $ABC$ equals $\frac{1}{2}bc\sin\alpha$. Deduce the sine rule $\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}.$

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1. The points $P$, $Q$ and $R$ are respectively the feet of the perpendiculars from $A$ to $BC$, $B$ to $CA$, and $C$ to $AB$ as shown. Prove that $\text{Area of }PQR\text{} = (1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma) \times (\text{Area of }ABC\text{}).$

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1. For what triangles $ABC$, with angles $\alpha$, $\beta$, $\gamma$ as above, does the equation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$ hold?

Suggestion

A triangle $ABC$ has sides $BC$, $CA$, and $AB$ of lengths $a$, $b$ and $c$ respectively, and angles at $A$, $B$ and $C$ are $\alpha$, $\beta$ and $\gamma$ where $0 \leq \alpha ,\beta ,\gamma \leq \frac{1}{2}\pi$.

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1. The points $P$, $Q$ and $R$ are respectively the feet of the perpendiculars from $A$ to $BC$, $B$ to $CA$, and $C$ to $AB$ as shown. Prove that $\text{Area of }PQR\text{} = (1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma) \times (\text{Area of }ABC\text{}).$

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1. For what triangles $ABC$, with angles $\alpha$, $\beta$, $\gamma$ as above, does the equation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$ hold?

Use this applet to explore the behaviour for different triangles $ABC$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/BcAFHM9w/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

The blue triangle $PQR$ is called a “pedal triangle” because it is formed from the feet of perpendiculars.

Remember that in the question, the angles at $A$, $B$ and $C$ cannot be obtuse.

When does the blue triangle disappear?

Solution

A triangle $ABC$ has sides $BC$, $CA$, and $AB$ of sides $a$, $b$ and $c$ respectively, and angles at $A$, $B$ and $C$ are $\alpha$, $\beta$ and $\gamma$ where $0 \leq \alpha ,\beta ,\gamma \leq \frac{1}{2}\pi$.

1. Show that the area of $ABC$ equals $\frac{1}{2}bc\sin\alpha$.

We take $AB$ as the base, which has length $c$, and $CR$ as the height, which has length $AC \sin(\angle CAR) = b \sin\alpha$. Then the triangle has area $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}cb \sin\alpha.$

By choosing a different side as the base of the triangle and using the same method, we see that the area is also given by $\text{Area}(ABC) = \frac{1}{2}bc\sin\alpha = \frac{1}{2}ac\sin\beta = \frac{1}{2}ab\sin\gamma.$ If we divide through by $\frac{1}{2}abc$, and then take the reciprocals of the equations, we get the sine rule $\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}.$

It’s going to become very tedious writing the word “area” repeatedly in our equations. So we’ll use the notation $[ABC]$ to mean the area of triangle $ABC$.

We have $[PQR]=[ABC]-[AQR]-[BRP]-[CPQ].$

Consider the triangle $AQR$. Using our formula from the first part, we have $[AQR] = \frac{1}{2}AQ \times AR\times\sin\alpha.$ But we know $[ABC]=\dfrac{1}{2}bc\sin\alpha$, and so $\sin \alpha = 2\dfrac{[ABC]}{bc}.$ Substituting, we get $[AQR] = \frac{AQ \times AR}{b \times c} \times [ABC].$

But if we look at the right-angled triangles $RAC$ and $BAQ$, we see that $\cos\alpha = \dfrac{AR}{b}=\dfrac{AQ}{c},$

and substituting gives $[AQR] = \cos^2\alpha \times [ABC].$

Similarly, $[BPR] = \cos^2\beta \times [ABC],$ and $[CQP] = \cos^2\gamma \times [ABC].$

So finally

$$\begin{align*} {}[PQR] &{}= [ABC] - [AQR] - [BPR] - [CQP]\\ &{}= (1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma) \times [ABC]. \end{align*}$$

1. For what triangles $ABC$, with angles $\alpha , \beta , \gamma$ as above, does the equation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$ hold?

If $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$, then the area of the triangle $PQR$ is zero.

This can only happen when two of $P$, $Q$ and $R$ coincide - at the right angle of the triangle.

So this equation holds only if the triangle $ABC$ is right-angled.

In fact, it is easy to see that we have an if and only if relationship here: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma =1$ if and only if the triangle $ABC$ is right-angled.