Solutions to Brain-Twisters I

The solutions too will often involve the use of figures and tables. As in the puzzles page, the figures are in gif format and should be downloadable by text-based browsers like Lynx.

Solution to Puzzle #1

We are given that one of the two signs is true and the other false. Could it be that the first is true and the second false? Certainly not, because if the first sign is true, then the second sign must also be true - that is, if there is a lady in Room I and a tiger in Room II, then it is certainly the case that one of the rooms contains a lady and the other a tiger. Since it is not the case that the first sign is true and the second one false, then it must be that the second sign is true and the first one false. Since the second sign is true, then there really is a lady in one room and a tiger in the other. Since the first sign is false, then it must be that the tiger is in Room I and the lady in Room II. So you should choose Room II.

Solution to Puzzle #2

If Sign II is false, then Room I contains a lady; hence at least one room contains a lady, which makes Sign I true. Therefore, it is impossible that both signs are false. This means that both signs are true (since we are given that they are either both true or both false). Therefore, a tiger is in Room I and a lady in Room II, so again you should choose Room II.

Solution to Puzzle #3

Both rooms contain ladies. This can be proved as follows:

Sign I says in effect that at least one of the following alternatives holds: there is a tiger in Room I; there is a lady in Room II (The sign does not preclude the possibility that both alternatives hold).

Now, if Sign II is false, then a tiger is in Room I, which makes the first sign true (because the first alternative is then true). But we are given that it is noth case that one of the signs is true and the other one false. Therefore, since Sign II is true, both signs must be true. Since Sign II is true, there is a lady in Room I. This means that the first alternative of Sign I is false, but since at least one of the alternatives is true, then it must be the second one. So there is a lady in Room II also.

Solution to Puzzle #4

Angle BDE = 30 degrees. Angle-chasing by itself is not sufficient to find this angle. A complicated general formula can be found by repeated use of the sine rule, or, in this case, a simple geometrical construction can be exploited.

Download this figure for the solution.

Mark E' on AC such that angle E'BC = 20 degrees. The three triangles EBC, BE'C and DE'B are all isosceles. Therefore, BEE' is equilateral, and triangle EE'D is isosceles. But DE'E = 40 degrees, and since BDE + 40 degrees = 70 degrees, angle BDE must be 30 degrees.

Solution to Puzzle #5

The Supreme Ruler's plan will not work.

Consider all first-born children. One-third will be male, one-third female, one-third hermaphrodite. Mothers who give birth to hermaphrodites will be sterilized.

The remaining mothers may have second children. One-third of the second-born will be male, one-third female, one-third hermaphrodite. Again, mothers of the hermaphrodites will be made sterile.

The remaining mothers may have third children, and so on. This obviously generalizes to families of any size. The proportions of sexes will always be 1:1:1.

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Revised: July 29, 1996