Evaluate

For (see *Exercise-1*)

*Exercise-1* Show that for .

Evaluate

For (see *Exercise-1*)

*Exercise-1* Show that for .

The proof in my post “A Computer Algebra Aided Proof in Plane Geometry” can be significantly simplified if complex numbers are used.

Consider complex numbers in FIg. 1.

Fig. 1

Since

we have,

i.e.,

Similarly,

and

give

Let denote the slope of line connecting and respectively.

From

we see that

lie on the same line.

Notice that since .

See also “Treasure Hunt with Complex Numbers“.

Given and two squares in Fig. 1. The squares are sitting on the two sides of and , respectively. Both squares are oriented away from the interior of . is an isosceles right triangle. is on the same side of . Prove that points and lie on the same line.

Fig. 1

Introducing rectangular coordinates show in Fig. 2:

Fig. 2

We observe that

Fig. 3

Solving system of equations (4), (5), (6), (7), we obtain four set of solutions:

Among them, only (10) truly represents the coordinates in Fig. 2.

Fig. 4

By Heron’s formula (see “Had Heron Known Analytic Geometryâ€¦“), the area of triangle with vortex is

.

From Fig. 4, we see that it is zero.

Therefore,

lie on the same line.

The reason we do not consider (9), (10), (11) is due to the fact that

(9) contradicts (2) since .

And,

by (1), (10) and (11) indicate which contradicts (3).

*Exercise-1* Prove “ lie on the same line” with complex numbers (hint: see “Treasure Hunt with Complex Numbers“).

* Problem:* and are squares.

** Solution**: Introducing a rectangular coordinate system and complex number

Let denote the area of triangle and respectively.

We have

Let

By Heron’s formula derived *without* *invoking trigonometric function *(see “An Algebraic Proof of Heronâ€™s Formula“),

We also have

(see “Treasure Hunt with Complex Numbers“) and,

Similarly,

Let

That is,

It is shown by Omega CAS Explorer that the expression under the square root of is the same as that of

Therefore,

*Exercise-1* The two squares with area 25 and 36 ( see figure below) are positioned so that Find the area of triangle TSC.

Besides “Wallis’ Pi“, there is another remarkable expression for the number as an infinite *product*. We derive it as follows:

From the trigonometric identity

,

we have

.

That is,

Dividing both sides by yields

or,

.

It follows that since ,

;

i.e.,

Let

(1) becomes

We know

Applying the half-angle formula

gives

Hence,

We compute the value of according to (3):

Fig. 1

*Exercise-1* Compute from (1) by letting

Solving

Even though ‘contrib_ode’, Maxima’s ODE solver choked on this equation (see “An Alternate Solver of ODEs“), it still can be solved as demonstrated below:

multiplied by

i.e.,

Integrate it, we have

By (see “Integration by Parts Done Right“),

As a result, (1-1) yields a new ODE

or,

Upon submitting (1-3) to Omega CAS Explorer in non-expert mode, the CAS asks for the range of

Fig. 1

Depending on the range provided, ‘ode2’ gives three different solutions (see Fig. 2, 3 and 4).

Fig. 2

Fig. 3

Fig. 4

Let’s also solve (1-3) manually:

If (1-3) has a constant solution

In fact, this solution can be observed from right away.

Otherwise (),

That is,

or

For let we have

Divide both numerator and denominator on the left side by ,

Write it as

Multiply both sides by

Integrate it,

we obtain

i.e.,

or,

where

For let

For

Notice when (1-2) becomes

This is a Bernoulli’s Equation with and Solving it (see “Meeting Mr. Bernoulli“),

Since we can verify (3-1) as follows:

substitute into (2-3),

Unsurprisingly, this is the same as (3-1).

*Exercise-1* Mathematica solves

But it only return one solution. Show that it is equivalent to (2-2).

*Exercise-2* Solving (1-2) using ‘contrib_ode’.

*Exercise-3* Show that (2-2), (2-3) and (2-4) are equivalent to results shown in Fig. 2, 3 and 4 respectively.