This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Equal Areas

Let

*A*be the area of the square (and the triangle).The length of a side of the square is √

*A*, so the perimeter of the square is 4√*A*Let

*l*be the length of a side the triangle. Then^{1}/_{2}*l*^{ 2}sin60 =*A*, so*l*^{ 2}=^{4A}/_{√3}. Therefore*l*=^{2√A}/_{31/4}and the perimeter of the triangle is^{6√A}/_{31/4}.Hence the ratio of the perimeters is

^{6√A}/_{31/4}:4√*A*which simplifies to**3**^{3/4}:2##### Extension

If an

*n*sided regular polygon has the area*A*, what is the length of one of its sides?#### Chessboard Squares

There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square on a chessboard.

This can be shown by counting how many positions the top left corner of the square can sit on. For example, the top left corner of a 5×5 square can be in the first four rows and columns of the board (otherwise the square will go off the board) and 4×4=16.

64+49+36+25+16+9+4+1=204.

##### Extension

How many rectangles are there on a chessboard?

#### Downing Street

The probabilities can be summarised as follows:

First person truthful | First person lying | |

Second person truthful | ^{3}/_{4}×^{4}/_{5}=^{12}/_{20} | ^{1}/_{4}×^{4}/_{5}=^{4}/_{20} |

Second person lying | ^{3}/_{4}×^{1}/_{5}=^{3}/_{20} | ^{1}/_{4}×^{1}/_{5}=^{1}/_{20} |

As they both agree, only both lying and both truthful are possible. Hence the chance of them lying is

^{1}/_{13}and the chance of them telling the truth, and it indeed being the Minister of Maths, is^{12}/_{13}##### Extension

If the first person said it was the Minister of English and the second said it was the Minister of Maths, what is the probability that it was the Minister of Maths?

#### Multiple Sums

The multiples of 3 less than 1000 are 3,6,9,...,999; the multiples of 5 are 5,10,15,...,995. Multiples of 15 (15,30,...,990) will appear in both lists so we are trying to find (3+6+9+...+999)+(5+10+15+...+995)-(15+30+...+990). This is:

$$\sum_{i=1}^{333}3i+\sum_{j=1}^{199}5j-\sum_{k=1}^{66}15k$$
$$=3\sum_{i=1}^{333}i+5\sum_{j=1}^{199}j-15\sum_{k=1}^{66}k$$
$$=3\times\frac{333\times334}{2}+5\times\frac{199\times200}{2}-15\times\frac{66\times67}{2}$$
$$=166833+99500-33165$$
$$=233168$$
Massive thanks to MathJax for allowing me to typeset maths!

##### Extension

Find the sum of all the multiples of 3 or 5 below

*n*.
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