An Algebraic Proof of Heron’s Formula

May 27, 2017June 13, 2017 / Michael Xue / 5 Comments

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.

Screen Shot 2017-05-28 at 11.48.57 PM.png

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 $h^2$

Screen Shot 2017-05-28 at 9.35.40 PM.png

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

May 20, 2017October 7, 2020 / Michael Xue / 2 Comments

Fig. 1

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

Screen Shot 2017-05-18 at 11.08.51 PM.png

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

May 17, 2017July 7, 2020 / Michael Xue / Leave a comment

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)

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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: