Complex Numbers versus Projective Geometry - One problem, Two solutions

The Problem

Suppose ABC is any triangle. D, E, F are points on BC, CA, AB respectively such that \(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB}\). Prove that the centroids of triangles ABC and DEF coincide.

A little Complex Number

Let A, B, C be points on the Complex plane with complex coordinates a, b, c.

The origin of the plane is made to coincide with the centroid of triangle ABC (the complex coordinates of the vertices adjusted accordingly).

Hence the complex coordinate of the centroid G is 0.

\(G = \frac{a+b+c}{3} = 0\) implying a+b+c= 0.

Next we find the complex coordinates of D, E, F in terms of a, b, c and the given ratio \(\frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = k\) (say)

Using the section formula we find that

\(d = \frac{ck+b}{k+1}\)
\(e = \frac{ak+c}{k+1}\)
\(f = \frac{bk+a}{k+1}\)

where d, e, f are the complex coordinates of D, E, F respectively.

Hence the centroid of triangle DEF is given by \(\frac{d+e+f}{3} = \frac{\frac{ck+b}{k+1} + \frac{ak+c}{k+1} + \frac{bk+a}{k+1} } {3} = 0\) using (a+b+c) = 0.

Thus the centroid of triangle ABC and DEF are same.

A little Projective Geometry

We project triangle ABC into a plane T such that the projection is an equilateral triangle.

(to be continued)


 

Is it a prime number?

353 is a prime number. So is 7919 (in fact it is the 1000th prime). There are 25 primes between 1 and 100. From 1 to 1000 there are 168 of them.

It is difficult to check whether a number is prime or not. One simple method is to try and divide the number with smaller numbers. For example to find out whether 351 is a prime or not, you start checking by dividing the number by small numbers such as 2. Of course, it is an odd number so we immediately know that it is not divisible by 2. Next we try by 3 and it works! Indeed 117 times 3 is 351.

But suppose we working with 899. To check if it is a prime or not we start dividing it small numbers. 2 will not divide as the number is odd. But 3 will not divide either. If we keep on checking, we will see that up till 28 no number divides it (899 is 29 times 31).

How far should we check to be sure whether a number is prime or not? Suppose n is the number. Then we should try and divide 'n' with all prime numbers up till \( \sqrt n \). Indeed if n is not a prime number then it must have a divisor smaller than or equal to \( \sqrt n \) apart from 1. This fact has a simple proof.

Suppose n is not prime. Then it is composite (that is it has a divisor apart from 1 and it self).
Let p be a divisor of n. then \( q = \frac {n}{p}\) is also a divisor of n.
That is n = pq
Now we claim that one of p or q is smaller than or equal to \( \sqrt n \).
If our claim is false then both of them are greater than \( \sqrt n \).
p > \( \sqrt n \) ; q > \( \sqrt n \)
This implies \(p \times q > \sqrt n \times \sqrt n \ = n \).
But p*q is not greater than n.
Hence our claim is correct.

Therefore when we have a number to check whether it is a prime or not, it is sufficient to try and divide it with all the prime numbers smaller than or equal to \( \sqrt n \) (since if it is composite then it will have a divisor which smaller than or equal to \( \sqrt n \) and this divisor is either itself prime or is divisible by a still smaller prime number).

Hence to check whether 641 is prime or not it is sufficient to check with all primes smaller than \( \sqrt {641} \) ~ 25. That is we try to divide it with 2, 3, 5, 7, 11, 13, 17, 19, 23. None of them divides 641. Thus we conclude that it must be a prime.

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