Week of May 16, 2022
Question 4 from Team Round, HMMT November, 2013
Week of May 9, 2022
Question 19 from AMC 10A, 2021
Week of May 2, 2022
Question 11 from Ritangle 2021, Stage 1 (page 22 here)
Easter break puzzles
Q1: Problem 22 from the Senior Maths Challenge 2016
Q2: Problem 5 from the HarvardMIT Math Tournament, February 2022, Guts Round
Q3:
Discussion:
The obvious issue with approaching this question is that you have a variable in the base and exponent. How can we rectify this? How can we bring the power down so that it’s on the same level as the base? Take a logarithm of both sides. Which one does it make sense to use? We already have a logarithm base 2 in the exponent, so let’s keep it base 2. Then, we end up with a quadratic which can be solved.
Solution:
We take the logarithm base 2 of both sides:
\log_2{x^{\log_2{x^3}n}} = \log_2{13}
Therefore,
\log_2{13} = (\log_2{x^3}n)(\log_2{x}) = (3\log_2{x}n)(\log_2{x})
Expanding:
3(\log_2{x})^2  n\log_2{x}  \log_2{13} = 0
If we let the solutions to this equation be x_1, x_2, we know that x_1x_2 = 32 from the initial condition in the problem (there are only two roots at most because this is a quadratic equation. In the case where there is a repeated root, we would just have x_1^2 = 32. Clearly we must pick n such that there is at least one real root though, as if both roots are complex, the product of the real roots is nonexistent.). The solutions to the quadratic above will be values of \log_2{x}, rather than values of x, so we take the logarithm base 2 of the product of the roots to get that \log_2{x_1} + \log_2{x_2} = 5.
This represents the sum of the roots of the quadratic above, which can also be expressed as \frac{n}{3} by Vieta’s formula, meaning that 5 = \frac{n}{3}, so n = \textbf{15}.
Week of Mar. 21, 2022:
Discussion: If we let t, b represent the number of tenors and basses selected respectively, we see that the requirement that they differ by a multiple of four can be expressed as t  b \equiv 0 \, \textrm{mod} \, 4. One way to approach this question is to simply list out all of the possible options, count them up, add them, and mod by 100 (or mod by 100 as you add them to keep the numbers smaller and easier to deal with). But this seems grossly inefficient, and there is a better way.
What’s annoying is that the difference needs to be a multiple of four. It would be much easier if the sum had to be a multiple of four, as then we could just pick singers without worrying about whether they are tenors or basses. We can make this conversion by letting b' = 8  b, meaning that we get that t  8 + b' \equiv 0 \, \textrm{mod} \, 4 \Rightarrow t + b' \equiv 0 \, \textrm{mod} \, 4. Now, we can just pick groups of 0, 4, 8, etc. and we’re good.
Solution:
We let t, b represent the number of tenors and basses selected respectively, and let b' = 8  b represent the number of basses not selected. Therefore, we see that the requirement that t  b \equiv 0 \, \textrm{mod} \, 4 morphs into t + b' \equiv 0 \, \textrm{mod} \, 4.
Therefore, we can consider a group of 14 objects, where 6 of those are tags which if selected mean that that tenor is selected, and the other 8 of which are tags which if selected mean that that bass has not been selected. The only restriction this gives is that, as at least one singer must be selected, we cannot have t = 0, b' = 8 at the same time, and the only possible valid values of t + b' are 0, 4, 8, 12.
Therefore, N = \binom{14}{0} + \binom{14}{4} + (\binom{14}{8}  1) + \binom{14}{12}
(We subtract one for the case where b' = 8, t=0)
Simplifying: N = 1 + 1001 + 3003  1 + 91
Taking N modulo 100, we have that N \, \textrm{mod} \, 100 \equiv 1 + 1 + 3  1 + 91 = \textbf{95}.
Week of Mar. 14, 2022:
Discussion:
Work backwards. It seems like f(n) needs to be minimal for n to be minimal, so see what the smallest possible output of the function f can be at every stage, and what numbers work as inputs.
Solution:
We want to find the smallest n such that f(f(f(n))) is not a onedigit number, i.e. f(f(f(n))) \ge 10.
We note that if n is as small as possible, then f(n), f(f(n)), f(f(f(n))) will all be as small as possible; otherwise, n could be made even smaller by reducing the value of one of its digits.
The minimum value of f(f(f(n))) = 10, as it needs to not be a onedigit number, and 10 is the smallest twodigit number.
The smallest value of f(f(n)) for which f(f(f(n))) = 10 is 19. We strive to minimize digits, and see that with one digit, the largest value we can get is 9. Therefore, we need two digits, and we want to minimize the tens digit, so we set that to 1 and have the ones digit be 9.
The smallest value of f(n) for which f(f(n)) = 19 is 199 by similar reasoning. We strive to minimize digits, but with two digits, the largest sum we can get is 9 + 9 = 18. Therefore, we need three digits. We strive to minimize the hundreds digit, so we set it to 1 and have the other digits be 9s.
The smallest value of n for which f(n) = 199 is 19999999999999999999999 (there are 22 9s). We strive to minimize digits, and as the largest possible value of a digit is 9, we see that we need at least 199/9 = 22.\overline{1} digits, i.e. 23 digits. We want the first digit to be as small as possible, and it turns out we can have 1 at the start, followed by 22 9s.
Therefore, \textbf{n = 19999999999999999999999} (there are 22 9s) is the smallest possible value of n.
Week of Mar. 7, 2022:
Problem 7 from the HarvardMIT Math Tournament, November 2021, General Round
Week of Feb. 28, 2022:
Discussion:
The trick with this problem is to express all of the conditions specified in the question algebraically, and then to stay focused on the goal of finding abcd. Note that abcd can be obtained from squaring a sum of pairs of a,b,c,d, and that you may need to square and combine different expressions in order to eliminate certain terms so that all that’s left is an expression with abcd and some numbers. Also remember that you might not need to find an equation with abcd as a whole – you might only need to find part of the product if you already have the other parts of the product.
Solution:
Looking into the conditions provided, we have that the nth term of each of the sequences is:
u_n = a + b(n1)
v_n = c + d(n1)
u_n + v_n = 5 + 8(n1)
u_nv_n = w_n + k(n1)^2 = 6 + 7(n1) + k(n1)^2
Plugging in the expressions for u_n and v_n into u_n + v_n and u_nv_n, we get that:
(a+c) + (b+d)(n1) = 5 + 8(n1)
[a+b(n1)][c+d(n1)] = 6 + 7(n1) + k(n1)^2
Equating coefficients of (n1)^0, (n1)^1, (n1)^2 (which we can do because we know that these equations must be true for all values of n; see this post from Brilliant if you’re still unsure), we get that:
a+c = 5, b+d = 8, ac = 6, ad + bc = 7, bd = k
We see that when we square ad + bc, we will get abcd within that expansion. Let’s see if we can also get the value of ab + cd, as it’s something else that when squared will also give some terms of abcd.
40 = (a+c)(b+d) = ab + ad + bc + cd = ab + cd + 7 \Rightarrow ab + cd = 33
Let’s now square ab + cd, ad + bc:
33^2 = (ab + cd)^2 = a^2b^2 + c^2d^2 + 2abcd, 7^2 = (ad + bc)^2 = a^2d^2 + b^2c^2 + 2abcd
We now see that if we are going to work out the value of abcd in this way, we need to find the value of the sum of the pairs of squares above. We square a+c, b+d and manipulate:
25 = (a+c)^2 = a^2 + c^2 + 2ac = a^2 + c^2 + 12 \Rightarrow a^2 + c^2 = 13
64 = (b+d)^2 = b^2 + d^2 + 2bd = b^2 + d^2 + 2k \Rightarrow b^2 + d^2 = 64 2k
We now multiply the two expressions above together:
13 \times (642k) = (a^2 + c^2)(b^2 + d^2) = a^2b^2 + a^2d^2 + b^2c^2 + c^2d^2
Therefore,
7^2 + 33^2 = (ab+cd)^2 + (ad+bc)^2 = (a^2 + c^2)(b^2 + d^2) + 4abcd = 13(642k) + 4abcd
We can also express abcd as 6k, meaning we get:
1138 = 13(64)  26k + 24k \Rightarrow k = 153
Therefore, abcd = 6 \times 153 = 918 = \textbf{918}
Week of Feb. 21, 2022:
Discussion:
Consider how the new polynomial relates to p(x), and how the roots are related as well. When counting the roots, don’t forget the Fundamental Theorem of Algebra.
Solution:
x^{12} + ax^8 + bx^4 + c = p(x^4)
Therefore, we can factorize this degree 12 polynomial as
(x^4  9002)(x^4  2009)(x^4  (2009 + 9002i\pi))
We now identify the nonreal roots of each of the three separate quartics. The equation y = x^4  9002 is a quartic curve, and is just y = x^4 (which is like y= x^2 but flatter near the origin) shifted down, meaning it has only two real roots. The same is true for y = x^4  2009. Because of the Fundamental Theorem of Algebra, each of these two quartics must have two nonreal roots so that they each have four roots in total. As for
(x^4  (2009 + 9002i\pi)) = 0 \Rightarrow x^4 = (2009 + 9002i\pi)
it is not possible to have a real number that when multiplied by itself gives a complex number, so the quartic has four nonreal roots by the Fundamental Theorem of Algebra.
Therefore, the total number of nonreal roots is 2 + 2 + 4 = 8 nonreal roots.
Week of Feb. 14, 2022:
Discussion: Because only one prime can be in each subset, which prime is actually being considered becomes irrelevant. Just pick a prime and count how many combinations of nonprimes can be added to it.
Solution:
Within 1,2,3,..., 12, there are 5 primes (2,3,5,7,11) and 7 composite numbers (nonprimes). For each prime that we pick, none of the other primes can be in the same subsets. We have 7 composite numbers to potentially add to each subset, and each of them is either in a subset or not in a subset, meaning we have two options as to what we do with each composite number, giving 2^7 = 128 subsets containing a particular single prime number. As the primes are indistinguishable in this case (the value of the prime is irrelevant), we can just multiply by 5 to get that the total number of subsets that contain exactly one prime is 128 \times 5 = \textbf{640}.
Week of Feb. 7, 2022:
Discussion: Think about what it means for none of the logarithms to be integers. Then, try some small cases, e.g. b = 2, 3, 4, 5 and see what happens.
Solution:
For none of the logarithms from \log_b{m} to \log_b{(m+2017)} to be integers, it means that none of m, m+1, m+2, ..., m+2017 can be integer powers of b, so for each b, we need to find the smallest n such that b^{n+1}  b^{n} > 2018. Ensuring that the gap between these consecutive integral powers of b is larger than 2018 would mean that none of m, m+1, m+2, ... m+2017 are integer powers of b. Then, we can just select m_b = b^n + 1.
Trying some small cases, we see that m_2 = 2^{11} + 1 = 2049, m_3 = 3^7 + 1 = 2188, m_4 = 4^5+1 = 1025, m_5 = 5^4 + 1 = 626. Therefore, it seems like 2188 may be the largest value we get. Let’s consider what happens for different fixed values of n.
The first inequality in each set comes from the fact that b^{n+1}  b^{n} > 2018 is required for there to be no integer powers within the range of the 2018 numbers m, m+1, m+2, ... m+2017. The second inequality is because b^{n}  b^{n1} < 2018, as n needs to be minimal. The values obtained from the inequalities can be obtained through using the quadratic formula, and then a bit of trial and error for the larger powers.
If n=0, b  1 > 2018 \Rightarrow b > 2019 \Rightarrow m_b = 1+1 = 2 < 2188
If n=1, b^2  b > 2018, b1 < 2018 \Rightarrow 45 < b < 2019 \Rightarrow m_b < 2019 + 1 = 2020 < 2188
If n=2, b^3  b^2 > 2018, b^2  b < 2018 \Rightarrow 12 < b < 45 \Rightarrow m_b < 45^2 + 1 = 2026 < 2188
If n=3, b^4  b^3 > 2018, b^3  b^2 < 2018 \Rightarrow 6 < b < 12 \Rightarrow m_b < 12^3 + 1 = 1729 < 2188
If n=4, b^5  b^4 > 2018, b^4  b^3 < 2018 \Rightarrow 4 < b < 6 \Rightarrow m_b < 6^4 + 1 = 1297 < 2188.
What we’ve shown is that for all values of b where 5 \le b, m_b < 2188. As we’ve also examined b=2, 4 and found m_b to be smaller than 2188, we can conclude that the largest number in Mary’s sequence is m_3 = \textbf{2188}.
Week of Jan. 31, 2022:
Discussion:
Of course draw a diagram to start. Then consider how you can use the fact that \triangle AED and \triangle BEC have the same area. Something you may want to consider using is an expression for triangle area that doesn’t involve a perpendicular height. Click here for a link with different area expressions that may be useful.
Solution:
We use the area expression A = \frac{1}{2}ab \sin{C} for a generic triangle \triangle ABC, where C is the angle between the sides with lengths a, b.
Applying the formula to this problem, we get that:
\frac{1}{2}BE \times EC \sin{BEC} = A = \frac{1}{2}AE \times ED\sin{AED}
As \angle BEC = \angle AED, we have that
BE \times EC = AE \times ED \Rightarrow \frac{BE}{ED} = \frac{AE}{EC}
Therefore, and because \angle AEB = \angle DEC, by SAS similarity, \triangle AEB \sim \triangle DEC.
Using this newfound information we have that, \frac{9}{12} = \frac{AE}{EC} = \frac{x}{14x} \Rightarrow x = 6, where we let x to represent the length AE and use the fact that AC = 14.
Therefore, \textbf{AE = 6}.
Week of Jan. 24, 2022:
Discussion: We thought this problem could be solved in a nice geometric way using Cavalieri’s Principle (which it turns out only works for volumes), instead of using integration. Unfortunately, we need to use integration, in particular surface areas of revolution. Sorry about that. So if you haven’t done any calculus yet, in particular integration, feel free to come back to this solution when you’ve learned a bit more math.
Solution:
In order to obtain a single slice of a sphere, we will revolve a slice (between x = \frac{2rk}{n} and x=\frac{2r(k+1)}{n}) of the circle x^2 + y^2 = r^2 around the xaxis.
Therefore, the surface area of one slice is:
2\pi {\displaystyle \int_{\frac{2rk}{n}}^{\frac{2r(k+1)}{n}} y \sqrt{1+(\frac{dy}{dx})^2} \, dx }
We rearrange the equation of the circle to get y = \sqrt{r^2  x^2}.
Also, \frac{dy}{dx} = \frac{x}{\sqrt{r^2  x^2}}, so 1+(\frac{dy}{dx})^2 = \frac{r^2  x^2 + r^2}{r^2  x^2} = \frac{r^2}{r^2  x^2}.
Therefore, the surface area is:
2\pi {\displaystyle \int_{\frac{2rk}{n}}^{\frac{2r(k+1)}{n}} \sqrt{r^2  x^2} \frac{r}{\sqrt{r^2  x^2}} \, dx } = 2\pi {\displaystyle \int_{\frac{2rk}{n}}^{\frac{2r(k+1)}{n}} r \, dx } = 2\pi {\displaystyle [rx]_{\frac{2rk}{n}}^{\frac{2r(k+1)}{n}}}.
This results in \frac{4\pi r^2}{n}. As the surface area of an entire sphere with radius r is 4\pi r^2, we have shown that all slices have equal surface area, namely \frac{1}{n} of the total surface area of the bread!
Week of Jan. 17, 2022:
Discussion:
This problem initially might seem daunting, as a,b,c could contain any prime powers. Look to the first line just below the equation though. It might helpfully restrict what a,b,c can be.
Solution:
Because we know that \ln{a} + \ln{b} + \ln{c} = p\ln{2} + q\ln{3}, we know that each of a,b,c can only be made up of powers of two and three, as if we combine the logarithms, we get that:
\ln{abc} = \ln{2^p3^q} \Rightarrow abc = 2^p3^q
Therefore, let’s let a=2^x3^y, b=2^z3^w, c=2^m3^n. Substituting this into the equation given in the question, we get:
2^{5x}3^{5y} = 2^{4z}3^{4w+1} = 2^{3m+1}3^{3n}
Comparing exponents of 2 and 3 separately (as no positive integer power of 2 can ever be divisible by 3 and vice versa, due to uniqueness of prime factorization), we get that :
5x = 4z = 3m+1, 5y=4w+1=3n
As we need to find the smallest a,b,c, we need to minimize x,y,z,w,m,n. We can begin to solve these equations using least common multiples (LCM).
LCM(5,4) = 20, so for the first equation on the left, we need to find the smallest multiple of 20 that is one more than a power of three. 20 = 2 + 18. 40 = 1 + 39, so we’re done. For the equation on the left, x=8, z=10, m=13.
LCM(5,3) = 15. 15 = 3 + 12. 30 = 2 + 28. 45 = 1 + 44, so this is the smallest multiple of 15 which satisfies the second equation. Therefore, y=9, w=11,n=15.
Therefore, we have that a=2^83^9, b=2^{10}3^{11}, c=2^{13}3^{15}.
So \ln{a} + \ln{b} + \ln{c} = \ln{abc} = \ln{2^{31}3^{35}} = 31\ln{2} + 35\ln{3}, so \textbf{p+q = 66}.
Week of Jan. 10, 2022:
Discussion: Think about what it means for the frog to reach lily pad 8, in terms of each of the jumps between lily pads.
Solution: In order for the frog to reach lily pad 8, it needs to not fall during each of the 7 jumps between lily pads. The probability of this happening is just the product of all of the probabilities of it not falling between lily pads:
(1\frac{1}{2})(1\frac{1}{3})(1\frac{1}{4})(1\frac{1}{5})(1\frac{1}{6})(1\frac{1}{7})(1\frac{1}{8})
which simplifies to
(\frac{1}{2})(\frac{2}{3})(\frac{3}{4})(\frac{4}{5})(\frac{5}{6})(\frac{6}{7})(\frac{7}{8}) = \frac{1}{8}
Therefore, 100m+n = 100(1) + 8 = \textbf{108}.
Week of Dec. 13, 2021:
Discussion: The 12% might seem daunting at first, because it’s not obvious how this will change anything. Start by considering what it means for a cuboid to be positioned within the sphere. What relationship must there be between 8 points on the surface for them to be considered vertices of a cuboid? What would it mean if every single possible set of these points contained at least one black point?
Solution:
If we begin to construct a cuboid within a sphere, we only need to select a single point to define the whole cuboid. The other 7 points are just reflections of the initial point (and further reflections of those points) through the xy, yz, and xz planes (drawn above on the right). This means that each of the sets of points that form a cuboid is unique – you can’t have overlap between these sets, because given a single point, you can determine one cuboid uniquely.
Now let’s see what happens if we assume every single one of these sets of points have at least one black point. Let’s consider all of the points in 1/8 of the sphere, in particular in one of the 8 equal regions created by the intersections of the xy, yz, and xz planes mentioned above. The number of these points corresponds to the number of possible cuboids because all cuboids contain one of these points, because of how cuboids are created. Therefore, if each cuboid has one black point, the total number of black points must be at least the number of points within this eighth of the sphere (as there is no overlap between sets of points that form a cuboid). Therefore, 12.5% of the sphere must be black to ensure no completely white cuboids can be formed.
As only 12% of the cuboid is colored black, we know that we can indeed find a cuboid with no black vertices.
Week of Dec. 6, 2021:
Discussion: In order to solve this problem, you need to take advantage of the fact that you have an as many points as you like. Looking at a small grouping of points is unlikely to help. What you need to do is consider a sufficiently large set of points so that you can draw conclusions from that set. The Pigeonhole Principle is likely to be helpful.
Solution: We prove that this is true for k colors. We take a column of k+1 points. By the Pigeonhole Principle, there must be two points with the same color in this column. We now take k^{k+1} + 1 of these columns. The number of different colorings of a single column of k+1 points is k^{k+1}, as each point could be one of k different colors. Therefore, if we take k^{k+1} + 1 of these columns, there must be at least two columns with the exact same coloring (i.e. colors in same position). Because each of these columns is guaranteed to have at least two points of the same color, we know that there are two columns with two points of the same color that are located in the same rows, so we can join them to form a monochromatic rectangle.
Week of Nov. 29, 2021:
Discussion: You can’t just look at the fact that N‘s leading digit is 7 and deduce things from that; the number of digits is important. Try and express N in a way such that taking an rth root becomes easier.
Solution: We observe that N can be written as N = \frac{7}{9}(10^{313}1). When we are taking roots, because 10^{313} is so large compared to the 1, we can kind of forget about the 1.
Therefore, f(r) = (\frac{7}{9}10^{313})^{1/r} = (\frac{7}{9}10^{313 \text{mod} r})^{1/r}
The mod is there because if 313 = rk + b, (10^{rk+b})^{1/r} = 10^k \times 10^{b/r} i.e. the value of k doesn’t matter, as it just produces a 10^k that doesn’t affect the value of the leading digit.
Therefore:
f(2) = (\frac{70}{9})^{1/2} \approx 2
f(3) = (\frac{70}{9})^{1/3} \approx 1
f(4) = (\frac{70}{9})^{1/4} \approx 1
f(5) = (\frac{7000}{9})^{1/5} \approx 3
f(6) = (\frac{70}{9})^{1/6} \approx 1
Therefore, f(2) + f(3) + f(4) + f(5) + f(6) = 2 + 1 + 1 + 3 + 1 = \textbf{8}.
Week of Nov. 22, 2021:
Prompt: What does the fact that everyone shook different numbers of hands mean? Now consider who shook 0 hands and who shook 8 hands? What do you notice?
Solution: 9 different answers means all of the numbers 08 showed up, as it’s not possible for anyone to shake 9 hands (as there are 10 people there in total and everyone knows their spouse).
Let’s consider the person, A, who shook 8 hands. That means that A’s spouse must have shaken 0 hands, as everyone else has at least one handshake from A. Now consider the person, B, who shook 7 hands. B’s spouse must be the one with 1 handshake (from A), as everyone else has shaken hands with A and B. We can consider everyone in this way, pairing people off, to get that the people who shook 6 and 2 hands are a couple, 5 and 3 are a couple, and so the person who shook 4 hands must be paired with someone else who shook 4 hands. There is no duplicate in the set of people questioned, meaning it must be John and his wife who shook 4 hands each.
Week of Nov. 15, 2021:
Discussion: If you’re not clear on how to start this problem, try setting up some simultaneous equations based on the information given. Variables would definitely help.
Solution: Let the rate at which Arthur, James, and Zain solve problems be a, j, z respectively. By rate, we mean ‘number of problems per minute.’
Therefore, we have (by the equation distance/speed = time):
\frac{1}{a + j + z} = 10 \Rightarrow a + j +z = \frac{1}{10} \frac{2}{a+j} = 30 \Rightarrow a + j = \frac{1}{15} \frac{5}{j+z} = 100 \Rightarrow j+z = \frac{1}{20}Therefore, we have a + j + j + z = \frac{1}{15} + \frac{1}{20} = \frac{7}{60}
Subtracting the first equation from this, we get: j = \frac{7}{60}  \frac{1}{10} = \frac{1}{60}.
Therefore, it takes James 60 minutes to solve one problem by himself.
Week of Nov. 8, 2021:
Solution: Let Paul’s sequence be represented by 1 + 3k, and Penny’s by 20175n. As we want the number that Paul and Penny count at the same time, we need k = n. Therefore, we want to solve
1+3k = 2017  5k \Rightarrow 8k = 2016 \Rightarrow k = 252Therefore, the number is 1 + 3(252) = \textbf{757} = 20175(252).
Holiday puzzles:
Q1: Question four from PUMAC Number Theory Round A 2020
Q2: Problem 9 from AMC 2019 12B
Q3: Puzzle 2 from MathsBombe 2021 (solution added below from official solutions)
The answer is 53.
To work out whether on day n Graham will have an even or an odd number of grandchildren visiting, we look at the prime factorization of n and count the number of different factors greater than one. For example, 12=2×2×3 and so 12 has factors 2,3,4,6,12 greater than one. Graham will have one grandchild visiting as a result of the `every second day’ rule, one grandchild visiting from the `every third day’, one from every fourth day, one from every sixth day, and one from every twelfth day. So 5 children will visit on day 5. It is then quite quick to go through and compute that an odd number of children visit on 53 of the 60 days, the only days on which an even number of children visit are those indexed by a square (1,4,9,16,25,36,49).
The puzzle has a link with Euler’s totient function, where \phi(n) counts the number of numbers m \leq n which are coprime to n. In this case, we want to ask whether n  \phi(n) is odd or even. There is a nice Wikipedia page on Euler’s totient function where you can learn more.
Q4: Problem 11 from Purple Comet High School Round, 2009
Q5: Problem 10 from General Round, HMMT November 2018
Q6: Problem 10 from Ritangle 2019
Q7: Question 12 from SMC 2017
Q8: Question 17 from Senior Kangaroo 2016
Q9: Question 7 from MathCounts 2021 State Target Round (solution added below from official solutions)
The points of intersection of y = x^2  x  2 and y = x+1 occur where the two expressions for y are equal. So,
x^2  x  2 = x + 1 \Rightarrow x^2  2x  3 = 0 \Rightarrow (x+1)(x3) = 0
So, the points of intersection occur when x = 1, 3. Substituting for x in the equation y = x+1, we see that when x=1, y = 1 + 1= 0, and when x=3, y = 3+1 = 4. Thus, the points of intersection are (−1, 0) and (3, 4), and the distance between these two points is
\sqrt{(31)^2 + (40)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
Q10 (bonus): Problem 9 from General Round, HMMT November 2018
No. 41:
Notation pointer: [x,y] means the interval between x,y inclusive. (x,y] means the interval between x,y excluding x and including y. (x,y) means the interval between x,y exclusive.
Discussion: You have to remember that this is a uniform distribution over the real numbers. Some things worth noting about this are that the size of a continuous region is equivalent to the width of the region. Also, the probability of any single real number being picked is 0, which makes sense, as there are infinitely many real numbers in any continuous region. From here, break the problem down. What happens if Laurent picks a number greater than 2017? What happens if not?
Solution:
If Laurent picks from the interval (2017,4034], she is guaranteed to select a larger number than Chloé. The width of this region is 40342017= 2017. The width of the original region is 40340= 4034 (note that it doesn’t matter that the interval excludes 2017 because the probability of exactly 2017 being chosen is 0). Therefore, the probability that her number is in this region is 2017/4023 = 0.5.
If Laurent picks from the interval [0,2017], then we can rephrase the situation from “Laurent and Chloé pick two numbers from [0,2017]” into “two numbers are randomly chosen from [0,2017] and then one is randomly assigned to Laurent and the other to Chloé.” We can assume that the numbers are distinct because we’re picking from a continuous interval (we can think of one number being chosen first. Then the probability of the second number being equal to the first number is 0 as discussed above). Therefore, there is a 50% chance that Laurent gets the bigger number. Therefore, the probability that Laurent’s number is bigger if she picks from the interval [0,2017] is 0.5, and the probability that she picks from this interval is 0.5, so the combined probability is 0.5 \times 0.5 = 0.25.
Therefore, the overall probability that Laurent’s number is bigger is 0.5 + 0.25 = 0.75.
Note: You can also use integration to solve this problem when looking at the second case, when the numbers are both chosen from the interval [0,2017]. We can let Laurent pick x, so then Chloé’s number must be chosen from the interval [0,x). The probability of Chloé’s number being in that region is \frac{x0}{20170} = \frac{x}{2017}. Now we need to calculate \sum_x \frac{x}{2017} for all x between 0 and 2017. This is not straightforward because we are summing over an infinite number of numbers. This is where integration comes in. The value we need to calculate is:
\int_{0}^{2017} \frac{x}{2017} dx = \frac{1}{2017}\int_{0}^{2017} x dx = \frac{1}{2017}[\frac{x^2}{2}]_0^{2017} = \frac{2017}{2}
This means that the sum of the probabilities that Chloé’s number behind Laurent’s number is \frac{2017}{2} (note that we don’t need to divide this by 2 even though we let Laurent pick first because we’re not also counting cases where Chloé picks first). Therefore, the probability that Laurent’s number is greater is \frac{2017}{2} / 2017 = \frac{1}{2}, and then we follow the same reasoning as before (we divide by 2017 because we need to divide by \int_{0}^{2017} 1 dx = [x]_0^{2017} = 2017, as the probability that Chloé picks anywhere at all in the interval is 1, for each number that Laurent picks).
No. 40:
Discussion:
Hint 1 (for solution 1): Each constant product is either 1 or 1. Consider them separately and try adding them up / finding out how many of them there are.
Hint 2 (for solution 2): What do a_ia_j remind you of? Seems like squares to me…
Solution 1: There are \binom{100}{2} = 4950 products of constant in total. Each product is either 1 or 1.
1s only come occur when a 1 is paired with a +1. This can be done in 39 \times 61 = 2379 ways. Therefore, the overall sum is “no. of constant – 2(no. of negative constant)”, because the total number of positive constant equals their sum, so we are subtracting the negatives to get the positives and then subtracting the actual value of the negatives the second time.
Therefore, the answer is 49502(2379) = 192.
Solution 2:
We square the sum of all constant, then subtract the sum of the squares, and divide by 2.
a_1 + a_2 + ... + a_{100} = 61  39 = 22.
a_1^2 + a_2^2 +... + a_{100}^2 + 2a_1a_2 + 2a_1a_3 + ... + 2a_{99}a_{100} = (a_1 + a_2 + ... + a_{100})^2 = 22^2 = 484.
a_1^2 + a_2^2 + ... + a_{100}^2 = 100 (as the square of each constant is just 1).
Therefore, a_1a_2 + ... + a_{99}a_{100} = \frac{484100}{2} = 192.
No 39:
Discussion: ‘Sum’ could be intimidating if it implies that there are many values of n that work, and it certainly could seem like there are lots of values to find. The best way to overcome this sense of fear is to dive right in. Think about what conditions must be true for the expression to be integral. n+7 and n1 are integers, so square rooting them either gives an integer or an irrational (you don’t need to worry about the denominator being a fraction. You can prove that the square root of an integer cannot be rational but nonintegral by straightforward proof by contradiction). Then, think about divisibility, which will provide some extra conditions to narrow the possible values of n down to a manageable number.
Solution: We see that for the entire expression to be integral, n1 has to be a square, meaning n = k^2 + 1 for some integer k. Therefore, for the expression to be an integer, we must have k \mid k^2 + 8 (if we substitute in for n in the numerator). Therefore, k \mid 8, so k = 1, 2, 4, 8 (k could technically be negative, but it doesn’t matter as the square is unchanged). The values of n that correspond to these values of k are n = 2, 5, 17, 65. We leave it as an exercise for the reader to check that these values of n are valid.
Therefore, the sum of all positive n for which the expression is integral is 2 + 5 + 17 + 65 = 89.
No 38:
Discussion: In its current form, its not very clear what the minimum is. In order to obtain a minimum, you generally need to use an inequality. Some examples include the trivial inequality (squares are always nonnegative) and AMGM (arithmetic mean >= geometric mean). It does seem like the minimum is just 0, because we have the sum of two squares. However, it’s important to remember that the minimum is an attainable lower bound, and if you check, you’ll see that it’s not possible for both brackets to be simultaneously zero. Therefore, the sensible thing to do is to expand and then try to rearrange to get something better.
Solution: We expand to get
x^2y^2  2xy + 1 + x^2 + 2xy + y^2 = x^2y^2 + x^2 + y^2 + 1
We then factorize to get
(x^2+1)(y^2+1) \ge 1
because squares are always nonnegative, so each bracket is at least 1. Therefore, the answer is 1.
No 37:
Discussion: It might be initially daunting to think of the entire grid at once, already laid out, so think about building it up from a single number, and what conditions are imposed. Exploring certainly helps – see how big you can get the sum to be, and what numbers you can get where. You soon see that the colors of the numbers relative to each other are highly restricted. You can prove that this is the case by thinking about how colors and parity (even/odd) vary between adjacent squares.
Solution: We can think of the numbers as being laid out in order from 1 to 25. After each number is placed, the next one must be placed in an adjacent square, and so the color must change. Therefore, every time the parity changes, the color changes – all the evens are on one color and all of the odds are on the other color. As there are 13 squares with color identical to the center color (call this red), and there are only 12 evens, the odds must be laid out on the red squares.
Therefore, the sum is \sum_{k=1}^{13} (2k1) = 2\sum_{k=1}^{13} k  13 = 2 \times \frac{13 \times 14}{2}  13 = 169.
Note: \sum_{k=a}^b f(k) is just a fancy way of saying ‘sum the expression f(k) for all integer values of k from a to b.’ f(k) does not need to contain k; it could just equal 1 or x, for example. If you don’t know how I expanded the final summation of k, check this article out.
No 36:
Discussion: The problem statement is far longer than it needs to be – make sure you distill it down. After parsing, it just turns into the sum of an arithmetic series, with the question being: how many terms of the arithmetic series with first term 5 and common difference 0.25 do you need so that the sum is greater than or equal to 1000?
Solution: The sum of the first n terms of a generic arithmetic series is \frac{n[2a+(n1)d]}{2}, where a is the first term and d is the common difference. The footnote of this article (also linked above) explains why this formula works. So, we have the inequality:
1000 < \frac{n[10+0.25(n1)]}{2} \\ \Rightarrow 2000 < 10n + 0.25n^2  0.25n \\ \Rightarrow 0.25n^2 + 9.75n  2000 > 0
To solve this quadratic inequality, we find the roots of the quadratic equation, and then note that because of this being a ushaped parabola (the coefficient of the n^2 is positive), the solutions are for n larger than the larger root or smaller than the smaller root. This will become clearer in a second.
We use the quadratic formula to find that n \approx 72.0437, 111.044 . We discard the negative value as n represents a number of roads, and so must be a whole number.
Based on the graph (for which all we need is the general shape, and so in reality a sketch will suffice), we see that 0.25n^2 + 9.75n > 2000 for n > 72.0347, i.e. n \ge 73. Therefore, a man in the tribe of Zimmer must walk 73 roads before he officially becomes a man.
No 35:
Discussion: Drawing a diagram is the first step to overcoming the initial strangeness of this problem. After that, there is a surprising variety of paths you can take. It is possible to use calculus (volumes of revolution) to determine the volume of the hole by calculating the volume of the spherical ‘caps’ on the end, Cavalieri’s principle to do the same but without calculus, or use a bit of cheeky reasoning. We’ll opt for the the cheeky approach here.
Solution: The crux of this approach is to assume that the solution must be a single numerical value, as otherwise there is ‘insufficient data’ to reach a conclusion. Therefore, consider the case when the radius of the hole is 0. In this situation, the diameter of the sphere would be 6, so the volume left after ‘drilling’ would be \frac{4\pi r^3}{3} = 36\pi.
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,
\textbf{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!
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

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