No 12:

Discussion: When I initially saw this problem, I ignored the fact that it was an ordering of objects as opposed to a plain old sequence of numbers. This was a mistake. If you do go down this path though, you find that nothing really works – you try differences of terms, quadratic stuff, big degree polynomials, and nothing really works. So then you think hang on, why is it 3-12 and not 1-10? This is a good question to ask. Also, you think, what are “mathematical objects”? What does it mean that they “can be labelled”? The first thing that comes to mind for me is regular polygons, and I guess the labelling refers to the number of sides? Clearly then the ordering can’t be dependent on properties like angle size or anything to do with mathematical properties of the shape itself, as these increase uniformly with the number of sides. So you zoom in on the word “ordering.” What types of “ordering” are there? Numerical and alphabetical are the main ones.

Solution: The “mathematical objects” are regular polygons, labelled by the number of vertices/sides. The ordering is alphabetical based on their names!

Decagon, dodecagon, hendecagon, heptagon, hexagon, nonagon, octagon, pentagon, square, triangle

So the answer is 98543.

No. 11:

Discussion: So we have an infinite tower of exponents. And there all of unknown value. Ouch. Well let’s think about what we would do if it were just x^x=2… Unfortunately, it’s not as easy as it seems because the base and exponent are variable. So is the problem impossible unless you know some highly advanced math? Thankfully not. The trick is to use/abuse the infinite tower of exponents. If x^{x^{x^{...}}} = 2, then we can raise x to the power on each side, to get: x^{x^{x^{...}}} = x^2.

Solution: We take x to the power of each side of the equation to get:

x^{x^{x^{...}}} = x^2

Because of the nature of infinity, the left side is still equal to 2, so we have x^2 = 2, meaning x=\pm \sqrt{2}. What does it mean to raise something to an irrational power? That’s a different story.

No. 10:

Discussion: The condition on how many times we can use the scale means that we can only weigh one bag. So somehow we need to collate information from 10 bags into one bag to determine which bag has the fakes (we can’t not collate as this means we weigh only one bag which will rarely reveal the fakes). We could try putting all of the coins from all of the bags together, but this turns out to be useless as this mass is always going to be fixed (at 101g). Then what about if we include different numbers of coins from each bag? This is the key idea.

Solution: Order the bags in a line (it doesn’t matter how you do this). Take the first bag, which has 10 coins. Then add 9 coins from the next bag, 8 from the one after that, 7, 6, 5, 4, 3, 2, 1. Now, weigh this bag. If the xth bag has the fakes, then the mass of those coins will be 1.1x, meaning that the weight of the bag will be 0.1x grams more than the weight if there were no fake coins. Therefore, you can work out which bag has the fakes.

Note: the weight if there were no fake coins would be 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 by a standard formula. Therefore, the bag’s weight will be of the form 55 + 0.1x.

No. 9:

Discussion: part 1 of the problem is pretty straightforward. Part 2 is where it gets trickier. Lots of experimentation, especially with small cases as they’re easier to deal with, is important to get the message across. General advice for “is it possible” problems is to assume it can be done and try lots and lots of times. Only if you have been unsuccessful at providing what the problem wants should you strive to prove impossibility.

Solution: For a normal chessboard, yes we can. Just place 4 dominoes in line in each row, end to end, and repeat for all 8 rows. For the mutilated chessboard, it’s not possible. Here’s why. Every time we place a domino, we cover two adjacent squares that are always differently colored. Therefore, every domino-covered board must have an equal number of each of the two colors (assuming we color as for the chessboard). As a result, because the mutilated chessboard has more black squares than red, it cannot be covered.

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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