No 34:
Discussion: For those of you that don’t know, \log_b a represents the quantity x, where b^x = a for any a,b where b > 0. ‘log’ is an abbreviation of ‘logarithm’. So unless you have some experience working with logarithms, it might be useful to convert the endstatement out of logarithmworld and into exponential world. As regards the statement 7^{x+7} = 8^x, it might be useful to bring everything that is raised to the power x onto one side.
Solution: This problem just involves some algebraic manipulation, made slightly more complicated by the presence of exponentials and logarithms.
Endstatement/goal: x = \log_b 7^7 \Rightarrow b^x = 7^7
7^{x+7} = 8^x \Rightarrow 7^7 = \frac{8^x}{7^x} = (\frac{8}{7})^xTherefore, b = \frac{8}{7}.
No 33:
Discussion: To start off with, the problem might seem daunting because of the sheer number of possibilities – it seems like n could be absolutely anything, so how are you supposed to maximize and minimize? The minimum is easier, because you can start at a triangle (n=3) and work your up, so start there. For the maximum value of n, note that the average internal angle will have to be quite large, so it might be worth identifying some large primes. Remember that the key to this problem is the fact that all angles must be odd primes: so they’re all integers, all odd, and all prime. Also, the sum of the internal angles of any convex ngon is 180(n2).
Solution: Minimum: n = 4. n=3 doesn’t work (in fact no odd values of n work for the same reason) because ‘odd times odd is odd’, so the angle sum cannot be divisible by 180, but it needs to be, as mentioned in the Discussion. A set of angles that work are: {97, 83, 97, 83}.
Maximum: n = 360. The maximum possible internal angle is 179 degrees (as angles are odd primes and to meet the convex condition, all angles are less than 180 degrees). Therefore, the maximum possible sum is
179n = 180(n2)
n = 360
Therefore, the answer is 360 – 4 = 356.
No 32:
Discussion: The difficulty with this problem is in converting what you know about the octagons into some property of the square that you can then use to find its area. The two properties that come to mind are sidelength and length of diagonals. The diagonals don’t seem very accessible through the current setup, but sidelength does, as the octagons do touch all four sides.
Once you’ve understood this, all that’s left is to calculate other lengths within the octagons, which can be done through trigonometry or triangle ratios (I prefer the latter for these problems because it’s faster).
Solution:
The total length of all red lines is equal to the sidelength of the square, so we compute this and then square it. We see that it is made up of two distinct distances, repeated multiple times and demarcated by the blue lines, which we have labelled x, y. Therefore, if we call the side length s, we have:
s = 5x + 4y
Of course, y = 2 is given, so we have s = 5x + 8. We can calculate x by observing the triangle outside one of the sides of an octagon. The exterior angle of any regular polygon is \frac{360}{n}, so in this case, it’s 45 degrees. Therefore, it’s a socalled ’454590′ triangle, which has ratio of sides 1:1:\sqrt{2}. Therefore, the height of this triangle is \frac{2}{\sqrt{2}} = \sqrt{2}, so x = \sqrt{2}. Therefore, s = 5\sqrt{2} + 8.
Therefore, area = s^2 = 50 + 80\sqrt{2} + 64 = 114 + 80\sqrt{2}. Therefore,
m+n = 194
No 31:
Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils! s = 5x + 8Discussion: When I first started working on this problem, I needlessly overcomplicated it: I had pencils going alternately over and under each other, when this wasn’t needed, so try and avoid this. Just go for the simplest possible configuration. Also, you realize that if you operate in the 2D plane and find something that works for only 3 pencils, you can overlay a rotated version of that 2D solution on top of itself, to get something that works for 6 pencils!
Solution:
(yes, the pencils here are “unsharpened”, but it doesn’t really matter)
No 30:
Discussion: Seeing the square roots, we know that the only way \sqrt{120\sqrt{x}} can be an integer is if 120\sqrt{x} is a square. Because of the subtraction, we know that
0 \le 120\sqrt{x} \le 120
and we want the number of times when this is a square.
Solution: As above, the question is in essence asking: for how many values of x is 120\sqrt{x} a square? The value of this expression is bounded by 0 and 120 (there are no negative square numbers), so we just count the number of square numbers in this region: there are 11 (smallest is 0^2 and the largest is 10^2).
For each of these instances, there is precisely one value of x that works, as shown below:
120\sqrt{x} = k^2
\sqrt{x} = 120k^2
x = (120k^2)^2
No 29:
Warning: Part 2 of this problem is actually far harder than anticipated. It actually turns out to be around as hard as the actual BMO2 problem, and so far beyond the level of all other POTWs.
Discussion: The original problem that inspired this one is BMO2 2021 P2. My advice for the first part of the question is just try different values of n. Clearly because we need one of each square, the minimum possible value of n \times n is 4+9=13, so n \ge 4 (as 16 is the smallest square bigger than 13). There’s still a problem here though: placing the 3×3 leaves a region of width 1 – too small for the 2×2. Therefore, n \ge 5. For the rest of the way, you can just keep reasoning in a similar way until you get to n=8, which does work.
For part 2, you might guess that the divisibility has something to do with 2 or 3, especially as n=8 is the answer for part 1.
Note: if you have checked out the problem that inspired POTW 29, you’ll see that what you are asked to prove is that if a \times a and b \times b squares can tile an n \times n, in essence, a \mid n or b \mid n. This gives you the answer to part 2 straight away (2 or 3 or both divide(s) n), and helps immensely with part 1. Therefore, you see that with 2×2 and 3×3 tiling squares, having discarded n=4, the next closest options are n=6, 8. You can eliminate n=6 by looking at what happens when you place a 3×3 and 2×2 adjacently (they leave a region of width 1 that can’t be filled), and then you try something for n=8 and it works.
Solution:
For part 1: n=8. This is minimal as shown above. An example tiling is:
For part 2:
Outline of a solution to the general case, using a \times a and b \times b squares: BMO2 official solutions.
No 28:
Discussion: When we plug in x=1, we get that P(1) = 1+1r^22020 = 2018r^2. So all we need to do is find r^2. Many of you will probably, as I did, jump to Vieta’s Formulas, which give a bunch of simultaneous equations which you can solve. However, this is not the nicest way to solve the problem – let’s look at something we haven’t used much yet: the roots. The fact that r,s,t are roots means that if we plug them into the polynomial, we get 0. We want to find r, so why not plug in x=r?
We get: P(r) = r^3 + r^2  r^3  2020 = 0
meaning: r^2 =2020 and we’re done!
Solution: Plug in x=r to get:
P(r) = r^3 + r^2  r^3  2020 = 0 \Rightarrow r^2 = 2020Then, P(1) = 1+1  r^2  2020 = 4038
No 27:
Discussion: To start with, it’s a good idea to play around with different questions and think about what the respective tribes would say. What you want is to ask a question such that the answer from both tribes is the same. One approach is to remember that “two negatives make a positive” – so if you can somehow make the liar tribe to lie twice, they will give you the right answer.
Solution: “If I asked someone in your tribe which road is correct, what would they say?” If it is a truthteller, they will give the correct path; if it is a liar, they know that someone in their tribe would say the other one, but they themselves need to lie so they say the correct path.
No 26:
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 312 and not 110? 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: 25
Discussion: In a general situation, “find the hundreds digit” would encourage me to try using modular arithmetic, e.g. taking the expression modulo 1000. However, that’s pretty highpowered, so let’s first try to find some other approach. The first thing that jumps out to me is the common factor of 15!. As there is nothing else that seems to be helpful, let’s factorize to get (20!  15!) = 15!(20*19*18*17*16  1). What would be nice at this stage is for some 0s to come up to simplify things…and they actually do (see below)!
I just want to note that this may just seem like a lucky coincidence – that in this particular problem, it just happened to be the case that (20!  15!) is divisible by 1000. Instead though, it’s more likely the problemsetter purposefully made it this way, because in a math competition setting, where you don’t have a calculator, exploiting these ‘simplicities’ is crucial.
Solution: (20!  15!) = 15!(20*19*18*17*16  1) We see that the whole number is divisible by 1000 because 15! is divisible by 1000, as it contains the product 4*5*10*15 = 3000. Therefore, the hundreds digit is 0 because the last three digits of the number will be 0.
No. 23:
Discussion: Suppose that Emilia uses R liters of Rb(OH), C liters of Cs(OH), and F liters of Fr(OH), then we have:
10%·R+8%·C+5%·F =7%and 5%·F ≤2%. R+C+F R+C+F
The equations simplify to 3R + C = 2F and 3F ≤ 2R + 2C, which gives 9R+3C ≤2R+2C ⇒5R≤C.
Solution: Therefore the concentration of rubidium is maximized when 5R = C, so F = 4R
No. 22:
Discussion: We can rewrite this equation as 3^y + 4^y  6^y = 1 where y = x(x + 2). We can then rearrange this to 3^y  6^y = 1  4^y. We can factorise this to give us a difference of two squares on the right and we can also take out a factor of 3^y on the left to give us 3^y(1  2^y) = (1 + 2^y)(1  2^y). We then bring everything to one side, 3^y(1  2^y)  (1 + 2^y)(1  2^y) = 0. We can solve this to find values of y which we can then use to find the values of x.
Solution: Since there are two values for y where the equation holds true (y = 0 and y = 1), there are therefore 4 real solutions to the initial equation.
No. 21:
Discussion: We can very easily work out the area of the triangle using Heron’s formula. We can then divide this result by 4 to get the length of the rectangle – giving us all the information we need to calculate the rectangle’s perimeter.
Solution: 32
No. 19:
Discussion: From the question, we can determine the following: N – 4 is a multiple of 7 and N – 2 is a multiple of 5. This gives us the following equations N – 4 = 7k and N – 2 = 5m. Since the maximum capacity is 40, we know k cannot be more than 5, therefore we just need to test all the values of k to get for possible values of N which we can then substitute in the second equation.
Solution: The values k = 0, 1, 2, 3 and 5 give us values of N which when substituted into the second equation would give a noninteger value for m, therefore k must equal 4. Therefore N = 32.
No. 18:
Discussion: We want to prime factorisation of our number to include the lowest powers possible in order to find the smallest number which has 2015 factors. To do this, we can use the prime factorisation of 2015 (5 * 13 * 31).
Solution: 116. We can use this prime factorisation to give us 2^{30} * 3^{12} * 5^4. Therefore, the sum of the factors is 2(30) + 3(12) + 5(4) = 116.
No. 17:
Discussion: We can draw a triangle and then note down all the angles that are equal. We can use the fact that base angles of an isosceles are equal to find three angles that are equal. From here, solving the question is very straightforward.
Solution: 84 degrees
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^2Because 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 dominocovered 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 distancetime 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:
No. 4:
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 πr^{2} 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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Zach Marinovhttps://etonstem.com/author/zach

Zach Marinovhttps://etonstem.com/author/zach

Zach Marinovhttps://etonstem.com/author/zach
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