Given a point on the circumference of a cyclic quadrilateral, prove that the product of the distances from the point to any pair of opposite sides or to the diagonals are equal.

Proof:

Let a, b, c, d be the coordinates of the vertices A, B, C, D of the quadrilateral and consider the complex plane with origin at the circumcenter of ABCD. Without loss of generality, assume that the circumradius equals 1.

The equation of line AB is

.

This is equivalent to , that is,

Let point be the foot of the perpendicular from a point M on the circumcircle to the line AB. If m is the coordinate of the point M, then

and

since .

Likewise,

,

,

and

Thus,

as claimed.

QED.

More later,

Nalin Pithwa

PS: The above example indicates how easy it is prove many fascinating theorems of pure plane geometry using the tools and techniques of complex numbers.

“Give me a place to stand, and I will move the Earth.” So, famously, said Archimedes, dramatizing his newly discovered law of the lever. Which in this case takes the form

Force exerted by Archimedes x distance from Archimedes to fulcrum equals

Mass of Earth x distance from Earth to fulcrum.

Now, I don’t think Archimedes was interested in the position of the Earth in space, but he did want the fulcrum to be fixed. (I know he said ‘a place to stand’, but if the fulcrum moves, all bets are off, so presumably that’s what he meant.) He also needed a perfectly rigid lever of zero mass, and he probably did not realize that he also needed uniform gravity, contrary to astronomical fact, to convert mass to weight. No matter, I don’t want to get into discussions about inertia or other quibbles. Let’s grant him all those things. My question is: when the Earth moves, how far does it move? And, can Archimedes achieve the same result more easily?

(Ref: Prof Ian Stewart’s Cabinet of Mathematical Curiosities).

A man started from home at 14:30 hours and drove to a village, arriving there when the clock indicated 15:15 hours. After staying for 25 min. he drove back by a different route of length times the first route at a rate twice as fast, reaching home at 16:00 hours. As compared to the clock at home, the village clock is

(a) 10 min slow

(b) 5 min slow

(c) 5 min fast

(d) 20 min fast

Problem #2.

If , then

(a)

(b) either or

(c)

(d) and .

Problem #3.

In an election, 10% of the voters on the voters list did not cast their votes and 60 voters cast their ballot papers blank. There were only two candidates. The winner was supported by 47% of all voters in the list and he got 308 votes more than his rival. The number of voters on the list was

The vertices of an -gon lie on the sides of a regular n-gon and divide its perimeter into parts of equal length. How should one construct the gon so that its area is :

(a) maximum

(b) minimum

Hint only:

[One of the golden rule of solving problems in math/physics is to draw diagrams, as had benn emphasized by the maverick American physics Nobel Laureate, Richard Feynman. He expounded this technique even in software development. So, in the present problem, first draw several diagrams.]

There exists a side of the -gon that lies entirely on a side of the n-gon. Let and . Show that . Then, for , we have and the area S of the -gon is given by

where . Thus, is a quadratic function of x. Show that is a minimal when or and is maximal when .

Let me know if you have any trouble when you attempt it,

Occasionally, people hold polls for the most beautiful mathematical formula of all time — yes, they really do, I am not making this up, honest —- and nearly, always the winner is a famous formula discovered by Euler, which uses complex numbers to link the two famous constants and . The formula is

and it is extremely influential in a branch of mathematics known as complex analysis.

(Prof. Ian Stewart’s Cabinet of Mathematical Curiosities).

This puzzle was invented by the combinatorialist P.A. MacMahon in 1921. He was thinking about a square that has been divided into four triangular regions by diagonals. He wondered how many different ways there are to colour the various regions, using three colours. He discovered that if rotations and reflections are regarded as the same colouring, there are exactly 24 possibilities. Find them all.

Now, a 6 x 4 rectangle contains 24 1 x 1 squares. Can you fit the 24 squares together to make such a rectangle, so that adjacent regions have the same colour, and the entire perimeter of the rectangle has the same colour?

(Thanks to Prof. Ian Stewart for putting this in his cabinet and I pulled it out of it !! :-))