. . . . As oil poured on water, as a secret entrusted to the vile, as alms bestowed upon the worthy, however little, so does science infused into a wise mind spread by intrinsic force. . Bhaskara II, Lilavati . . . 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 What Fraction Of The Total Area Is Red?
(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 only care about the fraction of the area in red, label half of the small square’s length as 1, so its entire side length is 2.
The blue triangle below has legs of 1 and 2, so its hypotenuse is √(12 + 22) = √5, which is equal to the side length of the large square.
We can then use similar triangles. The small red triangle is similar to the blue triangle, so its shorter side has to be half its longer side and equal to 0.5. Similarly, the triangle on the right is similar to the blue triangle, so its longer leg a to its hypotenuse of 2 has to be in the same ratio of the blue triangle, which is 2/√5.
So we have:
a/2 = 2/√5 a = 4/√5
We then subtract this from the large square’s side length to get the small side of rectangle shaded in red:
√5 – 4/√5 = 1/√5
(image)
Using the above diagram we can solve for the relevant areas.
Red area (red triangle + rectangle) 1(0.5)/2 + √5(1/√5) = 1.25
Total area (large square + red triangle + triangle with legs 1, 2) (√5)2 + 1(0.5)/2 + 1(2)/2 = 6.25
. . . . 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 When Does The Thief Tell The Truth?
(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).
First an answer to the warm-up puzzle.
On a truthful day, the thief would say, “I will lie the tomorrow.” And on Tuesday, Thursday, or Sunday, the thief has to lie, again saying, “I will lie tomorrow.”
This leaves Saturday, which is the answer. The thief will lie on Sunday, so the thief is lying on Saturday to say, “I will lie tomorrow.”
Now the answer to the main puzzle.
Day 1: I lie on Monday and Tuesday. Day 2: It’s now Thursday, Saturday or Sunday. Day 3: I lie on Wednesday or Friday.
The thief either lies on all of these days, or the thief is telling the truth on one of these days.
Consider day 1 and day 3 statements. Can they both be lies? If day 1 is a lie, then the truth day has to be Monday or Tuesday. If day 3 is a lie, then the truth day has to be Wednesday or Friday. These are incompatible implications since there is only one truth day in the week. Hence, one of day 1 and day 3 is the truth day and the other is a lie.
Furthermore, that means day 2 is a lying day (as there is only 1 truth day, which is day 1 or day 3).
If the day 2 statement is said on Thursday, Saturday or Sunday, then it would be true. So it has to be said on another day, meaning Monday, Tuesday, Wednesday, or Friday are possible lying days–and possible days for day 2.
Could day 1 be the truth day, and day 3 be a lying day? In that case, the truth day (day 1) has to be on Wednesday or Friday. But then day 2 would be the next day, on a Thursday or Saturday, which is impossible because day 2 has to be on Monday, Tuesday, Wednesday or Friday.
So it must be that day 1 is a lying day, and day 3 is the truth day. In that case, the truth day (day 3) has to be on Monday or Tuesday. But it can’t be Monday, or else day 2 would be the day before, on a Sunday, which is impossible because day 2 has to be on Monday, Tuesday, Wednesday or Friday.
Hence day 3 has to be on a Tuesday. This means day 2 would be on a Monday, which is permissible, and day 1 would be on a Sunday–also permissible.
Thus day 3 is the truth day, and that is on a Tuesday. The thief tells the truth on Tuesday.
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!
Thanks to all patrons! Special thanks to:
Kyle Michael Anvari Richard Ohnemus Shrihari Puranik
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.