A 3D graphing Calculator projected on a 2D graphing calculator

Have you played with a 3D graphing calculator? I certainly haven’t… till I found THIS!

Most of the 3D graphing calculators available online for free are really lousy, if you have tried to look. They take forever to load your equation, and you are almost always unable to change scale or perspective. Even if the calculator is really powerful, they involve so much manual coding you might as well go to Uni first before using that. Well, look no further, this 3D graphing calculator is the work of Desmos genius Thomas A. Kuczmarski!

He uses a 2D graphing calculator, something similar to geogebra which most of you would be more familiar with, to project a 3 dimensional function with (x, y, z)! Due to the awesome power of the original 2D desmos graphing calculator, this brand new 3D graphing calculator is extremely dynamic and responsive, thus making the changing of perspective extremely convenient!

Useful thing to note: The 3D calculator uses the function z = f(x,y) to graph.

Try it out here!

Genius desmos guy, Thomas A. Kuczmarski

Have fun!



You have a (solid) blue cube and an unlimited amount of (solid) red cubes, all of which are of the same size. What is the largest number of red cubes that can touch the blue cube along its sides or parts of its sides?

Touching along edges or at corner points do not count.

Problem comes from Pi Han Goh (https://brilliant.org/problems/he-was-wrong-too/?group=3BX9rjPrQ3CM&ref_id=820890)

Comment what you think is the answer, with some graphics if possible.

My 21 configuration that worked

Michael Mendrin’s 22 configuration.


Kissing Circles

So suppose you have 3 points in 2D space.

For each of these 3 points, a circle with it’s center at these points appears.

These 3 circles are not any other circles. They are tangent to each other, or they are kissing each other.

So given any 3 points, how do you find the radii of these 3 circles? Give a thought to this question. Solutions will come out tomorrow.


Lets name the points {p}_{1}{p}_{2} and {p}_{3}, and the radii of each circle be {r}_{1}, , {r}_{2} and {r}_{3}. Also, let the distance between {p}_{a} and {p}_{b} be {D}_{a,b}.

So now,

{D}_{1,2} = {r}_{1} + {r}_{2} \quad ---(1)

{D}_{3,2} = {r}_{3} + {r}_{2} \quad ---(2)

{D}_{1,3} = {r}_{3} + {r}_{1} \quad ---(3)

From (1), (2) and (3) we can find that:




Hence solved.



Cheryl’s Birthday Puzzle (Version 2: This is the Real Deal!!!)

Forget about the original Cheryl’s birthday puzzle that has supposedly “baffled the world”. Here’s a much harder puzzle modified from the original one. This is the real deal!

Before I go on, let me introduce myself. I am Damian Boh, a guest writer for this blog. The owner (and contributors to this blog), Clyde, Jackson and Julian are my ex-students in a science talent programme called CΩergy. I’m sure many of you would have read about this in the previous posts. I am currently a physics undergraduate in the UK so, greetings from London!

Cheryl’s Birthday Puzzle: The Puzzle That “Baffled the World”

p5 math qn bday

The picture above would have looked strikingly familiar to almost everyone; we have all heard about the (original) ‘Cheryl’s Birthday Puzzle’, a problem which originated in the ‘Singapore and Asian Schools Math Olympiads contest’. This problem is challenging enough to have gone viral online. It has appeared in many US and UK websites as well, with headlines like “The Math Problem That Stumped the Internet” and “Cheryl’s Birthday: Singapore’s math puzzle baffles world”.




Solutions to this puzzle can be found everywhere online.

Cheryl’s Birthday Puzzle Version 2 (A Way More Challenging Puzzle!)

If you think the above puzzle is difficult, think again. Here I shall present a Version 2 of the puzzle. It is modified to be much more challenging. Some people I know, who can solve the original puzzle in minutes, took hours to solve this puzzle, or they do not know how to solve it at all! I did not create this puzzle and I got it from my Facebook feed, the original source is (sadly) unknown.

Here goes:

Cheryl bday 2

Now try solving it. If you need some motivation, I must point out that I thoroughly enjoyed this puzzle and it is definitely worth your time!

The solution:

This is going to be a long explanation as the question is a difficult one. Forgive me if I get a little long winded at times. I just need to ensure that everyone can follow the logic and not lose track of any detail in this puzzle.

Bear these 3 things in mind:

1. Albert knew only the month of her birthday.

2. Bernard knew only the day of her birthday.

3. David was given one particular date from the list and all 3 of them knew this date has a different day and month from her birthday.

Let us analyse in order each statement that was made by Albert, Bernard and David. We will see how each sentence allows us to eliminate certain dates off the list until we are left with the correct one.

“Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.”

The first part of the sentence is of course unnecessary. We know that all Albert knows is the month of her birthday and there is more than one possible date for each month.

Now let’s look at the second part “…but I know that Bernard does not know too.” First we ask ourselves how Bernard can know the exact date of Cheryl’s birthday from just the day of her birthday. This is only possible if the day of Cheryl’s birthday has only one corresponding possible date from the list.


It is easy to see that there are two dates in the list which corresponds to this: June 18 and May 19. If Cheryl’s birthday is one of these dates, Bernard would have gotten 18 or 19 as the day and he would know Cheryl’s birthday immediately because there is only one date in the list with 18 (or 19) as the day.

Albert can only be sure that “Bernard does not know Cheryl’s birthday too” if he was sure that the date is definitely not June 18 or May 19. How could he be so sure? He knows her birthday must definitely not be in May or June. In other words, Albert definitely must not have gotten the months May or June. So we eliminate all the dates corresponding to these two months.

Now we are left with:


“Bernard: I did not know when Cheryl’s birthday is, and now I still don’t.”

Bernard would have deduced all the above and eliminated May and June off his list as well. However, despite having this new list, he still does not know Cheryl’s birthday. Following the same reasoning given earlier, there is only one possibility in the list below corresponding to each of the days 15, 17, and 22.


If Cheryl’s birthday falls on any of the dates Jul 15, Aug 22 or Sep 17, Bernard would have gotten the days 15, 22 or 17 and would know for sure what her birthday was. Since he still didn’t know her birthday, we eliminate these dates off the list.

Now we are left with:


“Albert: I still don’t know when Cheryl’s birthday is. Having said that, I am sure David still does not know too.”

Albert updated his list accordingly.

From the first part of the sentence, he still does not know when Cheryl’s birthday is. If Cheryl’s birthday falls on Jul 16. Albert would have gotten July as the month and because there is only one possible date corresponding to the month July in this updated list (see below) he would have known her birthday immediately.


Since he still doesn’t know her birthday, we proceed to eliminate Jul 16 off the list as well.

Now we are left with:


Now let’s look at the second part of Albert’s sentence, things get a little tricky here.

“Albert: …….. Having said that, I am sure David still does not know too.”

Let’s look at the updated list above, remember David updates the list to this one as well. Let us consider all the scenarios in which David would be able to know Cheryl’s birthday for sure:

1. David gets the date Sep 14 and since the only date in the list which has both a different month and day from Sep 14 is Aug 20, he knows Cheryl’s birthday is Aug 20.

2. David gets the date Sep 20 and since the only date in the list which has both a different month and day from Sep 20 is Aug 14, he knows Cheryl’s birthday is Aug 14.

If David has gotten any other date he would not have known what Cheryl’s birthday was. For example if he has gotten the date Aug 14, Cheryl’s birthday could have been on Sep 16 or Sep 20 and he could not have known for sure which one.

For Albert to be so sure that David still does not know Cheryl’s birthday. Albert must have known that David could not have gotten the date Sep 14 or Sep 20. How can he be so sure? Because Albert got the month September, this is Cheryl’s birth month and since the date given to David must have a different month from the actual birthday, the date that David was given could not have been in September.

Now we know that Albert must have gotten the month September and the birth month is definitely in September.

Now we are left with:


“David: I knew neither the day nor the month right before Albert said his last sentence1, but after he did, now I know what month it is.”

Now let us go back to the list before Albert said his last sentence:


David said that he did not know the month right before Albert said his last sentence, i.e. when his updated list looks like this one above. If David had gotten the month August in the date that Cheryl has given to him, he would be able to eliminate this month from the above list (since his date must have a different month from Cheryl’s birthday) and concluded that Cheryl’s birthday is in September. (even before listening to David’s last sentence)

Hence the date given to David must not be in August. Keep this in mind for now.

The second part of his sentence “…but after he did, now I know what month it is.” is redundant because as explained earlier, everyone was able to deduce after Albert’s last sentence that her birth month must be in September.

“Bernard: I did not know when Cheryl’s birthday is right before Albert said his last sentence, but after he did, now I know when Cheryl’s birthday is.”

Again we go back to the list before Albert said his last sentence:


Following the same logic as explained in the first part of the solution, in the list above there is only one possible date corresponding to the day 16. If Cheryl’s birthday were in Sep 16, Bernard would have gotten the day 16 and would have known Cheryl’s birthday. However, he said that he still doesn’t know. So we eliminate this date off the list.

Together with the fact that we know Cheryl’s birthday must be in September. We are now only left with two possible dates for her birthday, as shown below.

Now we are left with:


Let’s look at the second part of the above sentence “…but after he did, now I know when Cheryl’s birthday is.” This sentence is redundant. Since Bernard can now narrow the birth month to September, he could of course use the day that Cheryl has given him to deduce the exact date of her birthday.

“David: Then I also know when Cheryl’s birthday is.”

David updates his list and now he is also only left with the two dates: Sep 14 and Sep 20. From these two dates he was able to eliminate one of them. This means that the day of the date Cheryl has given him must have been either a 14 (so he could eliminate Sep 20) or 20 (so he could eliminate Sep 14).

Recall that earlier I told you to keep in mind that the date given to David must not be in August.

Of course the date given to David must not be in September as well, since this is Cheryl’s birth month.

We conclude that the date given to David must have been in either May, June or July, and it must lie on either the day 14 or the day 20.

From the list (below), the only possible date that David must have gotten is Jun 20, after which he proceeds to eliminate Sep 20 from the list and concludes that Cheryl’s birthday is on Sep 14!


The answer:

The date that Cheryl told David is Jun 20.

Cheryl’s birthday is on Sept 14.

Hooray, we have now solved the puzzle! =)

“Albert: Now I know too.”

Yes, Albert. We all know you are as smart as us and managed to deduce everything and solved the puzzle too!

I hope you guys enjoyed solving this puzzle as much as I do. And I hope this solution was helpful especially in areas where you got stuck.


~signing off at 1.30am (UK time) (writing the solution to this puzzle took way longer than I thought)


Mathematical model of the Rolling Shutter Effect on propellers (2 propellers)

First tings first, I am going to use LaTeX code to express important mathematical expressions along with pictures. One of the reasons is because this post may be better understandable with a graphing calculator, preferably https://www.desmos.com/calculator because it uses LaTeX coding, so feel free to click on the simulators given in this post and typing your own equations for a better understanding of this post. Here is an example of the function in picture and the function in LaTeX:


\(\left(\frac{x}{\left|0.3f\left(g\left(a\right)\right)-2\right|}+0.1f\left(g\left(a+\mod \left(a,2\right)\right)\right)\right)^2+\left(\frac{y}{\left|0.3f\left(g\left(a+2\right)\right)-2\right|}+0.1f\left(g\left(a+\mod \left(a,2.5\right)\right)\right)\right)^2<2\)

This function is taken from another simulator I built a longtime ago https://www.desmos.com/calculator/clr9waooqe.

Occasionally, I will be using only LaTeX code and when I use it, it will be enclosed in \(…\) or \[…\]. If you can’t read basic LaTeX code, you can enter https://www.desmos.com/calculator and paste the code into the text box.

Back to the topic!

You might have seen something like this in a photo:

Rolling Shutter Effect

Notice how distorted the propellers are. This is because the propeller is moving very fast, and the shutter takes the picture from left to right/right to left. I would not explain much on the rolling shutter effect because information about it can already be found online.

To get you started, here is the Rolling Shutter in Wikipedia: http://en.wikipedia.org/wiki/Rolling_shutter

And here is a simulator I built using Desmos Graphing Calculator (The second most powerful online graphing calculator I know) to give you a visual sense of this effect: https://www.desmos.com/calculator/kxiqxtbaj3

I will be explaining on the derivation of the functions in the simulator.

Derivation of the equation of the shape of the distorted propeller (which is the curved black line in the pic below):


So, lets first assume the picture as a square with corners (1,1), (1,-1), (-1,1), (-1,-1) on the xy coordinate plane and the shutter moves from left to right.


Ok lets define

\({v}_{s}\) as the velocity of shutter

\(w\) as the angular momentum of propeller (Radians per unit time)

\({ \theta  }_{ i }\) as the initial displacement of propeller before the picture is taken (in radians)

These definitions are the same as I used in the simulator, so you can check it out above to understand what I mean. Note that I used time as \(a\) in the simulator instead of \(t\) which I would be doing so in this post.

Essential: Radians is just another way of expressing angle. E.g:

360 degrees corresponds to 2pi in radians.

180 degrees corresponds to pi in radians.

And so on.

Now, we can eliminate \({v}_{s}\) as what matters is the ratio of \({v}_{s}\) and \(w\). So for now, we can assume \({v}_{s} = 1\)

deleteSince \({v}_{s} = 1\), after time \(t\), the shutter would have moved by \(t\) towards the right from \(x=-1\), so the equation of the shutter would be \(x = t – 1\).

Now for the propeller. So now, after time t, the propeller would be displaced by \(w \times t + { \theta  }_{ i }\).

deleteSo the equation of the propeller would be \(\tan  \left( -wt+\theta _{ i } \right) x\)


Now… here is the interesting part. Every point on the distorted shape of the propeller is where the propeller and the shutter intersect at a certain point in time. Since the shutter displacement is equal to the value of \(t\), we can find the point at which it intersects at time \(t\):

\(y=\tan  \left( -wt+\theta _{ i } \right) x\\ x=t-1\\ \boxed { x=t-1\\ y=\tan  \left( -wt+\theta _{ i } \right) (t-1) } \)


So now, it is clear that the point at which the propeller and the shutter would intersect at time \(t\) is given as:

\(\left( t-1,\tan  \left( -wt+\theta _{ i } \right) (t-1) \right) \)


For those who recognise it, yes it is a parametric equation.

So now for solving the parametric equation:

And so the boxed equation above is the equation of the shape of the distorted propeller with \({v}_{s} = 1\).

However, we are not done yet.

To take \({v}_{s}\) into account, all we have to do is to divide every \(w\) in the equation with \({v}_{s}\), so the full equation would be:

\(y=x\tan { \left( -\frac { w }{ { v }_{ s } } (x+1)+{ \theta  }_{ i } \right)  } \)


Done. If you have any questions you can email me at nailuj.noop@gmail.com

Thanks! (I think you can guess who I am)

Update: I found a LaTeX to WordPress converter thanks to Mr Damien Boh. I’ll be using that in the future.