# An Algebraic Proof of Heron’s Formula

Heron’s formula concerning $A$, the area of any triangle states:

$A = \sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad\quad\quad\quad\quad(1)$

where a,b,c are the three sides of the triangle and, $s=\frac{a+b+c}{2}$.

We are going to prove it with the aid of a CAS:

Substituting $s$ into (1), the formula becomes

$A = \frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}\quad\quad\quad\quad\quad(2)$

A triangle with three known sides is shown in Fig.1 where $x$ is part of the base of the triangle.

Fig. 1

By Pythagorean theorem,

$h^2+x^2=b^2$

$h^2+(a-x)^2=c^2$

To obtain $h^2$, we will use Omega CAS Explorer (see Fig. 2)

The function ‘eliminate’ eliminates variable $x$, returns the value of $h^2$

Fig. 2

The result is $h^2 = \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{4a^2}$, i.e.,

$h = \frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{2a}$.

Therefore, using the standard formula for triangle,

$A = \frac{1}{2}a h =\frac{ \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}$

which is (2)

This is the 1st example in my presentation at ACA 2013 titled “An Algebraic Approach to Geometric Proof Using a Computer Algebra System”.

In Fig. 3, $BA \perp CA, BD \perp CD$. Can you find the area of $\Delta ACD$?

Fig. 3

# Curvatures, Curvatures, Everywhere Curvatures

Fig. 1

In his book “What is Mathematics”, Courant wrote:

Here is my proof:

$\alpha$ is the slope-angle tells us that

$\tan{(\alpha)} = f'(x)$

which implies

$\frac{d\tan(\alpha)}{dx}=\frac{d\tan(\alpha)}{d \alpha}\frac{d\alpha}{dx}$

$=\sec^2(\alpha)\frac{d \alpha}{dx}=(1+\tan^2(\alpha))\frac{d \alpha}{dx}=(1+(f'(x))^2)\frac{d\alpha}{dx}=f''(x),$

i.e.,

$\frac{d\alpha}{dx}=\frac{1}{1+(f'(x))^2} f''(x)$.

We also know that

$s=\int^{x}_{}\sqrt{1+(f'(x))^2}\;dx$.

Therefore,

$\frac{ds}{dx} = \sqrt{1+(f'(x))^2}$.

Hence by definition, the curvature

$\kappa= h'(s)= \frac{d \alpha}{ds}= \frac{d \alpha}{dx} \frac{d x}{ds}= \frac{d\alpha}{dx}\frac{1}{\frac{ds}{dx}}=\frac{f''(x)}{1+(f'(x))^2}\frac{1}{\sqrt{1+(f'(x))^2}}$

$=\frac{f''(x)}{(1+(f'(x))^2)^{1+\frac{1}{2}}}$

$= \frac{f''(x)}{(1+(f'(x))^2)^{\frac{3}{2}}}$

I am now suggesting an exercise :

Calculate the curvature of figure in Fig. 1 at a point of your choice.

# Integration, CAS vs human

To evaluate integral

$\int x(1+x)^{19}\; dx$,

the CAS I have tried (maxima, mathematica) expands $(1+x)^{19}$ first, followed by multiply the expression by $x$. Finally, integrate term by term yields the result, a messy looking polynomial (see Fig. 1)

Fig. 1

A human being however, will more likely to recognize the fact that $(1+x)^{19}$ is the derivative of $\frac{(1+x)^{20}}{20}$ and therefore use the method of integration by parts:

$\int x (1+x)^{19}\;dx$

$=\int x (\frac{(1+x)^{20}}{20})'\;dx$

$=x \frac{(1+x)^{20}}{20}-\int x'\frac{(1+x)^{20}}{20}\;dx$

$=x\frac{(1+x)^{20}}{20}-\frac{1}{20}\int(1+x)^{20}\;dx$

$=\frac{x(1+x)^{20}}{20}-\frac{(1+x)^{21}}{420}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)$

An even better approach is:

$\int x(1+x)^{19}\;dx$

$=\int (1+x-1) (1+x)^{19}\;dx$

$=\int ((1+x)-1) (1+x)^{19}\;dx$

$=\int (1+x)^{20}-(1+x)^{19}\;dx$

$=\int (1+x)^{20}\;dx-\int (1+x)^{19}\;dx$

$= \frac{(1+x)^{21}}{21}-\frac{(1+x)^{20}}{20}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad(2)$

The difference of result from CAS and (2) is a constant, as expected (see Fig. 2)

Fig. 2