Choose any three by three square of dates on a calendar page.
Circle any number on the top row, put a line through the other
numbers that are in the same row and column as your circled number.
Repeat this for a number of your choice from the second row. You
should now have just one number left on the bottom row, circle it.
Find the total for the three numbers circled. Compare this total
with the number in the centre of the square. What do you find? Can
you explain why this happens?
Make a set of numbers that use all the digits from 1 to 9, once and
once only. Add them up. The result is divisible by 9. Add each of
the digits in the new number. What is their sum? Now try some other
possibilities for yourself!
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
To trace a path around a convex polygon the turn at each vertex
will be in the same direction and the total turn (the sum of the
exterior angles) is 360°. At each vertex of the polygon
where there is an acute angle the exterior angle is more than
90°. If there were four or more acute angles the sum of
the exterior angles at these vertices would be more than
360° which is impossible.
Alternatively we can use the fact that the sum of the interior
angles is (2n-4) right angles where n is the number of sides. If
there are t acute angles (all less than a right angle) and (n-t)
obtuse angles (all less than 2 right angles) and no reflex angles
then, by considering the total number of right angles:
With a six sided polygon it is possible to alternate the three
acute angles with the three obtuse angles so that none of the sides
has an obtuse angle at both ends. With seven or more sides there
must be two consecutive obtuse angles as there are at most three