30 – 3 = 3^3 ]]>

For part B, you take the two people selected from B and C and form a new group including person R. This group is shown to the psychic as the fifth seance. For the sixth seance, take the two remaining people from part A (the people who weren’t chosen) and the person selected from the fifth seance and show these people to the psychic. The person the psychic selects in the sixth seance is the second guilty person.

I’m not sure if this is correct, but I think it makes sense (hopefully).

]]>Part a) is quite simple I think. Take the ten people and divide them into three groups of 3 with one person remaining. Show each of the three groups to the psychic. The three people selected from the three groups then constitute the fourth seance that is shown to the psychic. The person selected in the fourth seance should be guilty.

I’m less certain about part b), but I’ll post my solution anyway. From our solution for part a), we have already found one guilty suspect. This person can be removed from the pool of suspects, leaving nine people. Finding the second guilty person should conform to one of the following cases

1. Both guilty persons were included in the fourth seance (i.e. two of the three groups that we split the ten people into contained guilty persons).

2. The same group that produced the first guilty person contained the second guilty person, so only one guilty person was included in the fourth seance (i.e. one of the three groups contained both of the guilty persons).

3. The remaining person is the second guilty person.

We can use these cases to figure out the second guilty person. Take the two persons remaining from the fourth seance and form a new group that includes the remaining person. This new group will constitute the fifth seance. The person selected from the fifth seance will then form a new group with the remaining two people from the group that initially contained the first guilty person (this group will constitute the sixth seance). The person selected in the sixth seance should be the second guilty person.

]]>Yes, cutting in half works too.

]]>– Cut out any shape that is symmetrical in the line that is equidistant from the given point and the centre, and turn it over.

– Cut out any shape that is 2-fold rotationally symmetrical about the point half way between the given point and the centre, and rotate it.

– Cut out an equilateral triangle with the given point and the centre as two of its corners, and rotate it. Obviously this generalises to other polygons and other orders of rotational symmetry, and david’s answer is the limiting case.

– If you only have scissors, and no pen or ruler or compass, and if you insist on simply-connected parts, then it is still possible. Fold the disk so that the given point aligns with the centre, and cut along the edge of the disk. This gives a leaf shape and a crescent. Rotate the leaf shape. ]]>