Puzzles

Let the radius of the small circles be \(r\). The are of half of one of these circles is \(\frac{1}{2}\pi r^2\).

The side of the square is \(2r\) and so the area of the square is \(4r^2\). Therefore the area of the whole shape is \((4+2\pi)r^2\).

By Pythagoras' Theorem, the radius of the large circle is \(r\sqrt{2}\). Therefore the area of the circle is \(2\pi r^2\). This means that the shaded area is \((4+2\pi)r^2 - 2\pi r^2\) or \(4r^2\).

This is the same as the area of the square, so the ratio is 1:1.

It can be done with two colours. Let's call these red and blue.

Draw the polygon and colour it red. As each line is added to is, swap the colours on one side of the line which leaving them the same on the other side. At the end of this process you will have any polygon coloured with two colours.

How many regions will the regular shape be split into by the lines?

As You have one more coin than me, if you don't throw more heads than me, then you must throw more tails than me.

This means that the probability of you throwing more heads then me plus the probability of you throwing more tails then me is equal to one.

By symmetry, the probabilities for heads and tails must be equal and so the probability that you throw more heads is \(\frac{1}{2}\)

You toss \(n\) fair coins, and I toss \(m\) fair coins. What is the probability that you get more heads than I do?

The number of zeros at the end of a number is the same as the number of 10s in the product that makes the number. Each of these 10s is made by multiplying 5 by 2.

There will be more even numbers than multiples of 5 in \(100!\), so the number of 5s will tell us how many zeros the number ends in.

In \(100!\), there will be 20 multiples of 5 and 4 multiples of \(5^2\). This means that \(100!\) will end in 24 zeros.

How many zeros will \(n!\) end in?

Draw on the following lines parallel to those which were added in the question.

Then a grid of copies of the smaller blue triangle has been created. Now consider the three triangles which are coloured green, purple and orange in the following diagram:

Each of these traingles covers half a parallelogram made from four blue triangles. Therefore the area of each of these triangles is twice the area of the small blue triangle.

And so the blue triangle covers one seventh of the large triangle.

If the sides of the triangle were split into \(n\) pieces the the lines added, what would the area of the smaller blue triangle be as a fraction of the area of the original large triangle?

The interior angle of a regular hexagon is 120°. The interior angle of a regular dodecagon is 150°. Therefore angle CAM is 30°.

Now, consider the quadrilateral ACDE. This quadrilateral is symmetric (as the dodecagon is regular) so the angles CAE and DEA are equal. Hence:

The angles CAM and CAE are equal, so A, M and E lie on a straight line.

The vertices \(P_1\), \(P_2\), ..., \(P_n\) make up a regular \(n\)-gon and \(Q_1\), \(Q_2\), ..., \(Q_m\) make up a regular \(m\)-gon, with \(P_1=Q_1\) and \(P_2=Q_2\).

The vertices \(P_2\), \(Q_3\) and \(P_5\) lie on a straight line. What is the relationship between \(m\) and \(n\)?

The result of the multiplication can be written as:

where \(a\), \(b\), \(c\) and \(d\) are integers. Expanding the brackets gives:

Next, if \((bd-ac)^2\) and \((bc+ad)^2\) are expanded, we get:

And so:

We have written the product as the sum of two integers.

For which integers \(a\), \(b\), \(c\) and \(d\) can the result \((a^2+b^2)(c^2+d^2)\) be written as the sum of two square integers in more than one way?

By the sine rule in BCD:

By the sine rule in BAD:

\(ABD=CBD\) and \(AD=CD\), so:

Using the sine rule in ABC:

The two sines are equal and so: