Create my own exam - Math Problem Example

A line passing through the centre of a circle is the diameter of the circle (Mosteller 109). If two or more diameters are in the same circle then all of them are equal in length irrespective of their position. Therefore, if the two sides from both triangles that pass through the middle of the circle were taken as o for Triangle A and x for triangle B, then o will be equal to x. The side will also be the longest on either triangle. In the equation o + p + q = x + y + z we can eliminate o and x since they cancel each other keeping in mind that o = x, hence we will be left to prove that p + q = y + z.

To prove this we will first have to identify the angles formed by the two triangles. If triangle A has angles O, P, Q where angle P and Q join sides p and q to side o respectively, then angle O is opposite to side o. On the other hand, triangle B has angles X, Y, Z where by angle X is opposite to side x and angles Y and Z join sides y and z to side x respectively. If we start with both triangles as isosceles triangles then sides p + q = y + z because for both triangles the longer side is equal (Kac and Ulam 167).

In the event that the shape of either triangle changes then the following, changes will also take place. Let us start with triangle A, a change in the shape of the triangle from an isosceles triangle an irregular triangle this will cause angle O to increase. The change will also be associated with change in length of side p and q, where with every increase in side p side q will be subsequently decreasing and vice versa. The total length of the two sides will be maintained that is p + q for all the changes will remain the same.

In triangle B the same principle will also apply such that a change in the shape of the triangle from an isosceles triangle to an irregular triangle then angle X will be increasing and an increase in side y will lead to a decrease in side z and vice versa. The total length in this case will be maintained that is to say that y + z will remain the same in whichever shape the triangle changes. Since p + q = y + z for the isosceles triangle then the same principle will apply for a change in shape of the triangles.

Theorem 2 If a rectangle is drawn inside a chessboard with its sides parallel to the sides of the chessboard, then the number of complete dark squares will NOT be equal to the number of light squares covered by the rectangle. That is if the rectangle covers m complete dark squares and n complete light squares then m ? n. Taking note that not necessarily that all the squares covered by the rectangle will be complete. Proof: Suppose we take a chessboard (which is square in shape) with sides C and D where C = D and inside the chessboard there are ‘n’ equal squares, which are made of one unit of each side, therefore n = C ?

D (Garder 510). The dark squares are denoted by b and the light squares denoted as w. A rectangle with length J and width K is drawn inside the chessboard where the side J is parallel to C and K is parallel to D. Further, J should not be equal to C and K should not be equal to D, meaning the rectangle is smaller than the chessboard and there should be no three sides of the rectangle touching the sides of the chessboard. If the rectangle is drawn to fit exactly two units of C and one unit of D then the rectangle will