Start with any triangle T1 and its inscribed circle. Draw the
triangle T2 which has its vertices at the points of contact between
the triangle T1 and its incircle. Now keep repeating this process
starting with T2 to form a sequence of nested triangles and
circles. What happens to the triangles? You may like to investigate
this interactively on the computer or by drawing with ruler and
compasses. If the angles in the first triangle are a, b and c prove
that the angles in the second triangle are given (in degrees) by
f(x) = (90 - x/2) where x takes the values a, b and c. Choose some
triangles, investigate this iteration numerically and try to give
reasons for what happens. Investigate what happens if you reverse
this process (triangle to circumcircle to triangle...)
Scheduling games is a little more challenging than one might desire. Here are some tournament formats that sport schedulers use.
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
Well done Geoffrey, University College School, London for this
This problem is actually remarkably straightforward. A little
bit of algebra is all that is needed.
Let x and y each denote 1 of the 2 numbers to be multiplied
The layout of the grids in the question then corresponds to:
The 4 ways of calculating the left hand digit are expressed
algebraically as follows:
All 4 ways are algebraically equivalent, so this is why they
give the same answer.
To explain why the Vedic Sutra works, we run through the whole
The right hand digit is ( 10 - x ) * ( 10 - y ).
The left hand digit is ( x + y - 10 ). We should remember that this
is the tens digit and multiply by 10 before summing both sides.
10 * ( x + y - 10 ) + ( 10 - x ) * ( 10 - y )
= 10x + 10y - 100 + 100 - 10x - 10y + xy
Using the Vedic Sutra, we get the result xy, which is indeed the
product of x and y.