No. 8:

Discussion: If you think about it, you try some places, and quickly realize that nothing ‘normal’ works, and that there must be some specific location. So because of the hinting in the problem, you try the South Pole, and see what’s happening there, and this turns out to be the key. We see that in order to end up back where we started, we need to be 1 mile away from the circle around the South Pole which has circumference 1 mile, so that we end up exactly where we started, and can come straight back to our previous destination. The radius of this circle is just 1/2\pi, using the formula for the circumference of a circle. Therefore, we think that the actual answer is 1 + 1/2\pi miles from the South Pole.

Proof: No new ideas here, just think through the discussion, and see that if we start at any point 1 + 1/2\pi miles from the South Pole, when we travel south 1 mile, we hit the circle with radius 1/2\pi miles. This means that the circumference is 1 mile, so when we travel east 1 mile, we end up exactly where we were, and so travelling north 1 mile brings us back to our starting point.

No. 7:

Discussion: This problem feels quite intuitive, but it’s a bit more difficult to solidify your thoughts into a rigorous proof. Think about the monk going up and down, and eventually you have the idea to picture a second monk going up at the same pace as the monk did on the first day, along the same route. As the two monks are passing along the same path, they must meet each other at some point, and at this point, they will be at the same place at the same time.

Proof: Draw a distance-time graph of the journey of the monk going up and the monk going down. The lines must cross somewhere, as they are not parallel and they must end up at each other’s starting points. Where they cross is where the monk is at the same time and place on both days.

No. 6:

Flip two of the three switches to on, and leave them on for a long time (let’s say 30 minutes). Then, turn one of them off, run upstairs, and observe/feel the bulbs.

Why this works:

If, when you go upstairs, a light is on, you know which switch controls it because only one switch is still on.

If, when you go upstairs, none of the lights are on but one of the bulbs is hot/glowing orange a bit, you know that this lamp was on but was just turned off, so the switch is the one you turned off right before going upstairs.

If, when you go upstairs, none of the lights are on and none of the bulbs are hot/glowing orange a bit, you know that the bulb was never turned on, meaning that the switch which controls it is the one which was never turned on.

No. 5:

4 congruent L-shaped pieces (similar to the bigger shape) (Credits to IMOmath)

No. 4:

4 congruent trapezia (Credits to IMOmath)

No. 2: π : √2+1

Solution: Assume the diameter of the large circle is 2, this must mean the diagonal of the square is 2. Using Pythagoras’ Theorem, we know that the side length of the square is therefore √2. The height of the top segment must be half of the result we get from subtracting the height of the square from the diameter of the circle and dividing by 2 (as there is another equal segment at the bottom). Therefore the height of the top segment is (2 – √2)/2. From this we can determine the height of the triangle by adding the height of the square to the height of the top segment which is (2 + √2)/2. As the base of the triangle lies on a side length of the square, we know the base must be √2. Therefore, we use the formula bh/2 to determine that the area of the triangle is (√2 +1)/2. Now, we know the radius of the smaller circle must be half the length of the square which is √2/2. We then use the formula πr2 to determine that the area of the smaller circle is π/2. Therefore the ratio of the areas is π/2 : (√2 +1)/2 which we simplify to π : √2+1.

No. 1: 90√6 (Click here to see the full solution)

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