A US secondary school teacher tried unsuccessfully to solve a problem from this year’s gaokao, a Chinese college entrance exam known for its difficulty. The video got very popular: “Posts about the video have been viewed 140 million times on microblogging site Weibo, and the video has been picked up by other Chinese news media.”
I was very interested in the problem, but I could only find a graphic of the problem in Chinese (so I couldn’t even copy the characters to Google Translate). Desperate, I asked if anyone on Twitter could help, and I owe a huge thanks to two kind people who provided a translation within hours:
. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Chinese Test Question
“Making a thousand decisions, even the wise will make a mistake.” Chinese Proverb
Let’s start by graphing the function in (0, 1].
(Technically the function in this intervals is not defined at x = 0, but we’ll see in a little bit that f(0) is in fact 0).
The graph of x(x – 1) is a parabola with endpoints at (0, 0), (1, 0) and its minimum value is in the middle at (0.5, -0.25). So we have:
Now we use the property that f(x + 1) = 2f(x). This means we can shift the graph to the right by 1 interval, say to (1, 2], by multiplying the height of the parabola by 2. So we get the following:
Note the graph in the second interval is 2(x – 1)(x – 2) because it is 2 times as tall, and the variable x maps to x – 1 when shifting to the right.
We can continue this pattern, and in the interval (2, 3] the graph is of 4(x – 2)(x – 3).
We can also shift the graph to the left, and each time the parabolas will be half the height, so they will get smaller and smaller.
We want to know where the function is always greater than -8/9.
The function does get smaller than -8/9 in the interval (2, -3], so we need to find the smallest such value m. We can do this by solving for the smallest x value such that the equation is true:
This is a fun viral problem that I got many requests to solve. Given the first equation, solve for the value of the second expression:
4b2 + 1/b2 = 2 8b3 + 1/b3 = ?
You are supposed to solve in 90 seconds, and no calculators are allowed. I admit I didn’t solve it that quickly. But it’s a fun one. Watch the video for a solution.
. . . . “If you are curious, you’ll find the puzzles around you. If you are determined, you will solve them.” — Erno Rubik . . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Viral Algebra Problem – Solve In 90 Seconds
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Let’s take a detour to geometry to solve this algebra problem. What is the volume of a cube with a side of x in which a smaller cube, with a side of y, has been removed? One way to express the volume is the difference of the volumes of the cubes, which is:
x3 – y3
But we can also solve for the volume by summing the volumes of 3 rectangular prisms, as seen in this diagram:
So we have:
x3 – y3 = (x – y)x2 + (x – y)xy + (x – y)y2
We can factor x – y from each term to get the difference of cubes formula:
x3 – y3 = (x – y)(x2 + xy + y2)
We can then do a neat little trick. If we substitute y = –z, we end up with the sum of cubes formula:
x3 + z3 = (x + z)(x2 – xz + z2)
It is the above formula we need to solve the original problem.
Solving the problem
Let x = 2b and z = 1/b, so that we have a sum of cubes:
Numbers that are the sum of 2 positive cubes OEIS https://oeis.org/A003325 The video length shows as 4:07 on the YouTube app, chosen since 407 is the sum of cubes of two positive numbers: 407 = 4^3 + 7^3.
This problem is adapted from a Data Scientist interview question asked at Facebook. I did a blog post in 2015, but I am re-posting as I have just posted a video about it.
You are waiting for your flight to Seattle, and to pass the time you call 3 friends in Seattle. You independently ask each one if it is raining.
All 3 of your friends say “Yes, it is raining.”
But each friend lies with probability 1/3 and tells the truth with probability 2/3.
Can you solve for probability it is actually raining in Seattle? Watch the video for a solution.
The problem can be solved using conditional probability (Bayes Rule).
There are two situations where the friends all say “Yes.” One is if it is raining and they are all telling the truth. Each tells the truth 2/3 of the time, so all three telling the truth is a probability of (2/3)3 = 8/27. The other situation is if it is not raining and all three are lying. Each lies 1/3 of the time, so all three lying is a probability of (1/3)3 = 1/27.
Here’s a probability tree of the situation.
The event “all three say yes” happens 1/3 = 8/27 + 1/27 of the time. Out of these times, there is an 88.9% = 8/9 = (8/27)/(1/3) chance that it is actually raining, and a 1/9 chance it is not raining.
Actually the above answer is incomplete. It assumes that “rain” and “not rain” are equally likely to occur. You are supposed to recognize this and ask the interviewer what the probability of rain is. Then they tell you to solve for 25% or maybe a general value p.
*Why might an interviewer do this? Most people who get an interview are fully capable of solving a textbook problem where all the information is provided. The interviewer wants to know, can you do more than solve a textbook problem? Can you identify that information is missing, and then adjust accordingly–either to ask, or to solve generally? This is a very important job skill, as clients often ask you to solve a problem with incomplete information. You might have to ask the client for more information, or provide your own general analysis to suit their needs.
Solving generally
Let’s solve it generally, supposing the probability of rain is p.
We adjust the analysis as follows. If it’s raining, which happens with a probability p, then all three tell the truth with probability 8/27. If it’s not raining, with probability 1 – p, then all three lie with probability 1/27.
Here’s an adjusted probability tree.
We want to do the ratio of the left branch probability to the sum of both branches.
The event “all three say yes” happens p(8/27) + (1 – p)(1/27) = (7p + 1)/27 of the time. We now need to calculate (8p/27) divided by (7p + 1)/27, which results in an (8p)/(7p + 1) chance that it is actually raining.
*The problem by itself is not hard–many high school students could solve it. The hard part is trying to solve this under pressure, and then having the sense to identify and ask for the missing information.
A couple of years ago I received this problem by email.
If a, b, c, d, are positive integers with a sum of 63, what is the maximum value of ab + bc + cd?
I did some research and traced it to the 2013 Australian Intermediate Olympiad. Talented students in years 7-10 (aged about 11-15) have 4 hours to solve 10 questions. There are no calculators are allowed.
Also keep in mind the age group of students. By age 15, students typically have not taken calculus. So limit your problem-solving techniques to algebra and geometry.
This problem stumped me, but I thought its solution was interesting and wanted to share. Watch the video for the answer:
. . . . “Talent is important, but how one develops and nurtures it is even more so,” Terence Tao. . . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Tricky Australian Olympiad Question
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
The key to the problem is noticing:
ab + bc + cd = ab + bc + cd + da – da = b(a + c) + d(c + a) – da = (b + d)(a + c) – da
So let’s think about this problem geometrically. Imagine a square with one side a + c and another side b + d. We wish to maximize the area of the three blue squares.
For a fixed length and width, the largest blue area occurs when a is the smallest value, which for positive integers is a = 1. Similarly, for a fixed length and width, the largest blue area occurs when d is the minimum value d = 1.
Thus, to maximize the blue area, we can maximize the area of the large rectangle with the condition a = d = 1. The problem simplifies to two variables:
max (b + d)(a + c) – da = max (b + 1)(1 + c) – 1
Now recall the sum of the variables is 63, so we can solve for b in terms of c:
a + b + c + d = 63 1 + b + c + 1 = 63 b + c = 61 b = 61 – c
So we then return to the maximization problem:
max (b + 1)(1 + c) – 1 = max (62 – c)(1 + c) – 1 = max –c2 + 61c + 61
Now we think geometrically again. The above quadratic has a graph of a parabola, and its maximum occurs at its vertex.
A parabola with equation αx2 + βx + c, where α &neq; 0, has its vertex at x = -β/(2α).
For our equation –c2 + 61c + 61, the vertex is at x = -61/(-2) = 30.5.
Since we need c to be an integer, we can test the closest values of 30 and 31. In either case, we get the same maximum value for the area:
This seemed like an impossible problem at the start. But with a little bit of geometric thinking and algebra, we would figure out the answer without too much trouble. Still, it’s not easy by any means, and I admire the 11-15 year olds who could solve this kind of problem in a timed competitive test!
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A question on the SATs in England sparked a lot of discussion, leaving some students in tears and even confusing some teachers, according to tes. The students were year 6, or about 11 years old.
The problem is to divide the grid using 2 straight line cuts to leave 3 shapes: you want 2 squares of different sizes, and 1 rectangle.
. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To The SATs question that made students cry
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
If you make 2 cuts in the grid, you will usually end up dividing it into 4 shapes.
This confused many people. How are you supposed to make 3 shapes from 2 cuts? You have to think a little creatively. Or as I like to say, you need to think inside the box.
Notice the grid is 5 squares tall, so imagine making 1 cut to make a 5×5 square. Then remove that square. The resulting rectangle is 2 squares wide, so make another cut to make a 2×2 square. This leaves a 3×2 rectangle. This is the answer:
Some people were able to see the answer right away. But I tell them to be nice to people who found it difficult! It’s not always easy to see the answer in an exam with time pressure, and this one does require a bit of creative thinking!
A few years ago this equation spread in Vines and on the internet:
9 + 10 = 21
It is not true in the standard base 10 decimal system. But what if we modify the equation a bit with other number bases? Suppose the numbers on the left are in base x and the number on the right is in base y:
(9 + 10) (base x) = 21 (base y)
For what values of x and y is this equation true? This is actually a fun little problem. Watch the video for a solution.
. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To The Viral Meme 9 + 10 = 21 Solved
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Since we have number bases, we want x and y to be positive integers. The term x/2 requires that x be a positive even number.
Also since 9 is in base x, we have x ≥ 10, as the digit 9 would not be used for a base 9 or smaller.
Thus we have the pairs of solutions:
x = 10, so y = 9 x = 12, so y = 10 x = 14, so y = 12 … x, y = 4 + x/2
So at first 9 + 10 = 21 seems like a simple false equation. But if we think about number bases, there are an infinite* number of solutions–pretty neat!
The development and spread of decimal numerals is a fascinating history. I want to share a few interesting parts from Wikipedia:
Fact 1: the base 10 system was developed by Aryabhata in India, and Brahmagupta introduced the symbol for 0.
https://en.wikipedia.org/wiki/Numeral_system (Quoting Wikipedia) The most commonly used system of numerals is the Hindu–Arabic numeral system. Two Indian mathematicians are credited with developing it. Aryabhata of Kusumapura developed the place-value notation in the 5th century and a century later Brahmagupta introduced the symbol for zero. The numeral system and the zero concept, developed by the Hindus in India, slowly spread to other surrounding countries due to their commercial and military activities with India. The Arabs adopted and modified it. Even today, the Arabs call the numerals which they use “Raqam Al-Hind” or the Hindu numeral system. The Arabs translated Hindu texts on numerology and spread them to the western world due to their trade links with them. The Western world modified them and called them the Arabic numerals, as they learned them from the Arabs. Hence the current western numeral system is the modified version of the Hindu numeral system developed in India. It also exhibits a great similarity to the Sanskrit–Devanagari notation, which is still used in India and neighbouring Nepal.
Fact 2: Fibonacci shared the method “how the Indians multiply” in 1202, but it took Europe hundreds of years to adopt the method. For all the poeple that think base 10 is natural since we have 10 fingers, I wonder: why did it take so long for Europe to adopt a “natural” system? I don’t think it’s so natural–the decimal system s a revolutionary idea, and we should give proper credit to the Indian mathematicians who developed it.
(I am intrigued at the similarities with a more recent episode. The method how the Japanese multiply is a fun way–not as revolutionary–to visualize multiplication and to learn group theory. I feel a bit like Fibonacci as others are very slow to accept the method’s value!)
https://en.wikipedia.org/wiki/Liber_Abaci#Modus_Indorum (Quoting Wikipedia) In the Liber Abaci, Fibonacci says the following introducing the Modus Indorum (the method of the Indians), today known as Hindu–Arabic numeral system or base-10 positional notation. It also introduced digits that greatly resembled the modern Arabic numerals.
(Quoting translated Liber Abaci on Wikipedia): “There from a marvelous instruction in the art of the nine Indian figures, the introduction and knowledge of the art pleased me so much above all else, and I learnt from them, whoever was learned in it, from nearby Egypt, Syria, Greece, Sicily and Provence, and their various methods, to which locations of business I travelled considerably afterwards for much study, and I learnt from the assembled disputations. But this, on the whole, the algorithm and even the Pythagorean arcs, I still reckoned almost an error compared to the Indian method.
… (Quoting Wikipedia) In other words, in his book he advocated the use of the digits 0–9, and of place value. Until this time Europe used Roman Numerals, making modern mathematics almost impossible. The book thus made an important contribution to the spread of decimal numerals. The spread of the Hindu-Arabic system, however, as Ore writes, was “long-drawn-out”, taking many more centuries to spread widely, and did not become complete until the later part of the 16th century, accelerating dramatically only in the 1500s with the advent of printing.
Recently I posted about a problem that was viral in China. Solve for the area shown in red, in the diagram below, between the semicircle and the rectangle’s diagonal.
Today’s challenge is to solve for the general case. Find a formula for the red shape in terms of r, the radius of the semicircle.
You can use the method I presented in the previous problem. But I will warn the algebra gets a little messy! It turns out there is a simpler way to solve the problem, and I credit Tony Hasler for the solution.
. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Viral Math Problem – Simple Formula For The General Solution
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Thanks to Tony Hasler for deriving this solution!
We start by setting up a coordinate system, where (0,0) is the lower left hand corner. The diagonal has the equation y = x/2 and the semicircle has the equation y = r – √(r2 – (x – r)2).
Now let’s focus on where the semicircle and line y = x/2 intersect. We need to solve the following:
r – √(r2 – (x – r)2) = x/2
r2 – (x – r)2 = (x/2 – r)2 5x2 – 12xr + 4r2 = 0 (5x – 2r)(x – 2r) = 0 x = 2r or x = 2r/5
The solution x = 2r corresponds to the upper right hand corner of the rectangle. We want the interior intersection, which will be x = 2r/5.
Since y = x/2, the y-coordinate will be r/5. So the curves intersect at (2r/5, r/5).
Now draw a line connecting the points (r, r) and (2r/5, r/5). We can then write an equation for the area of the red shape as the sum of the areas of two triangles minus the area of a circular sector.
We just need to know where the line between (r, r) and (2r/5, r/5) intersects the x-axis. As this line has a slope of 4/3, it is straight-forward to calculate its x-intercept is (r/4, 0).
The blue triangle then has an area of:
blue triangle area 0.5(r/4)(r/5) = r2/40
The green triangle, with corners (r/4, 0), (r, 0), and (r, r) has legs of sides 3r/4 and r, so its area is:
green triangle area 0.5(3r/4)(r) = 3r2/8
Now all that remains is the area of the purple circular sector. We can find its central angle using the dimensions of the green triangle. Since the opposite side is 3r/4 and its adjacent side is r, the tangent of the angle is equal to 3/4. So the angle’s measure is arctan(3/4). A circular sector’s area is 0.5r2(central angle), so its area is:
circular sector area 0.5r2 arctan(3/4)
Now remember the calculation is:
We now add the areas of the blue and green triangles and subtract the area of the circular sector to get:
r2/40 + 3r2/8 – 0.5r2 arctan(3/4)
= r2(0.4 – 0.5 arctan(3/4))
And that’s it! We can solve for the area of this kind of share for any value of the radius, whether we want r = 4, or r = 10, or r = 73 (and of course, you know 73 is the best number).
It’s great when a hard geometry problem works out to such a simple formula.
Thanks to Ata from Turkey for suggesting this! The problem is intended for the 8th grade (around 13 years old) and is adapted from a book of practice problems, Tonguc 8. Sinif Sayisal Zoru Bankasi).
Here’s the problem.
Tongul draws 2 rectangles ABCD and ABEF whose edges are integers (in cm).
The area of ABCD is 20 cm2 and ABEF is 10 cm2.
Which is the minimum possible perimeter of the rectangle CDFE in cm?
(a) 12 (b) 14 (c) 22 (d) 26
I admit I got this wrong! Can you figure it out? Watch the video for a solution.
. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Math Problem For 13 Year Olds That Stumped Me
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
I’ll first present who I approached the problem and got to the wrong answer. I drew the two rectangles as follows.
If we set AB = x, then the opposite sides EF and CD are also equal to x. Since ABCD has an area of 20, BC = AD = 20/x. Since ABEF has an area of 10, BC = AD = 10/x.
The perimeter of CDEF is then:
x + 20/x + 10/x + x + 10/x + 20/x = 2x + 60/x
Since each side is an integer, 10/x has to be an integer. Hence we can consider factors of 10 for values of x, so x could be 1, 2, 5 or 10. The perimeter values could be:
From these calculations, I thought the answer was 22. But this is not correct. Where is the mistake?
The correct answer
The trick is to think inside the box! The original diagram can be drawn with ABEF contained inside of ABCD like this:
We can now proceed similarly. Again letting AB = x, we have EF = CD = x. Then since ABCD has an area of 20, BC = 20/x = AD. Since ABEF has an area of 10, BE = AF = 10/x. We then must have EC = FD = BC – BE = 10/x.
The perimeter of CDEF is then:
x + 10/x + x + 10/x = 2x + 20/x
Again the values of x are the factors of 10, so we have:
While I didn’t solve the problem, it was a fun one. We’re always told to think outside the box. But in a problem like this, you have to think inside the box!
A rectangle contains 3 identical circles as shown in the diagram.
(the top two circles are tangent to two sides of the rectangle; the bottom circle is tangent to one side of the rectangle; each circle is tangent to the other two circles)
If each circle has a radius of r, what is the area of the rectangle in terms of r?
The problem is adapted from one of the hardest GCSE exam problems (a test given in the UK).
. . . . “Weave a circle round him thrice” Samuel Taylor Coleridge, Kubla Khan . . . . M I N D . Y O U R . D E C I S I O N S . . . . Answer To Rectangle Area From 3 Identical Circles
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
We can solve for the side lengths by drawing radii of the circles.
The horizontal distance is equal to 4 radius lengths, or 4r.
The tricky part is the vertical distance between the centers of the circles. If we connect the centers of the 3 circles, we get an equilateral triangle with a side length of 2r. So its altitude is r√3. We then add 2 radius lengths to get the entire side length, which is then r(2 + √3).