Puzzles

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so it is not possible

If the second hand moves continuously instead of moving every second, will there be a time when the hands of the clock are all 120° apart?

Pick two points (A and B) of the same colour (say red). Consider the points:

If at least one of C, D or E is red, then three equally spaced red points have been found. If C, D and E are blue, then they are three equally spaced blue points.

Will there be four equally spaced points of the same colour?

Yes. To show this, let's suppose there is an arrangement of pins with no monochrome triangle.

Let the middle pin be blue (if it is not blue, then swapping all the colours will make it blue without adding/removing monochrome triangles).

Pins 8, 2 and 6 make a triangle. Therefore one of these pins must be blue. By symmetry, any of these situations is equivalent to 8 being blue.

Pins 9 and 4 must be red to prevent a blue triangle.

Pin 3 must be blue to prevent a red triangle.

Pin 6 must be red to prevent a blue triangle.

But now whichever colour 10 is, it will make a monochrome triangle. Therefore there must be three pins of the same colour which lie on the vertices of an equilateral triangle.

What is the smallest \(n\) such that any mixture of \(n^2\) red and blue pins arranged in a square grid must contain four pins of the same colour which lie on the vertices of a square?

Many more puzzles like this one can be found in my post on the Chalkdust blog.

The maximum you can get is £50, with the taxman getting £28. Here is how to get it:

Can you always beat the taxman if £1 up to £\(n\) are available?

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?