Blog Post #79: For Elizabeta!

Dear Readers,

It’s quite late now in Belgium, but I still really wanted to write this wee article tonight as a nice, extra boost for Elizabeta, a truly outstanding Maths student (at Musica Mundi School in Waterloo, Belgium) who always produces terrific work!

Let’s have a fun word puzzle, right now!

Leave out just the very last letter of I DREAM CHESS, and use all of the other ten letters to make the name of a famous Greek mathematician from long ago…

I’m going to directly congratulate you for thinking of ARCHIMEDES, whom Elizabeta and I were talking about just yesterday!

Now for some bonus Maths…!

Imagine a single vertical column of identical spheres stacked together in a cylindrical tube such that the spheres touch the wall of the tube. The bottom sphere also touches the base of the cylinder, and the uppermost sphere touches at the top.

What fraction of the cylinder’s volume is not occupied by the spheres?

Enjoy figuring out the answer like Elizabeta will certainly do, before checking the solution given down below…

We also have a fun chess puzzle, as follows…

It’s White to move, but she also has an invisible piece sitting on a particular square! Which type of piece is it? Well, every legal type, except for the one she has, would allow White to deliver checkmate right now. However, given what she has, she can only force checkmate in 3 moves. It’s yours to find!

Pictured: A ‘Fantastic Four’ group of students!

Keep smiling like they do!

With very best wishes as always,

Paul Motwani.

P.S.=Puzzle Solutions!

In the chess puzzle, if White were to have either a pawn, bishop or extra queen on f6, then Qxg7 would be checkmate immediately.

Having a knight on f6 wouldn’t work because Black’s king would be in check and so it would then have to be Black to move instead of White.

What about having a white rook on f6…?

Then 1 Qe6+ Kh7 2 Rxh6+! gxh6 3 Qf7# makes a very neat checkmate!

In the spheres/cylinder Maths puzzle, suppose that there are n identical spheres each of radius r. Their total volume is 4/3*Pi*r cubed*n.

The volume of the cylinder will be Pi*r squared*2r*n because the total height of the cylinder, or of n stacked spheres, is 2r*n.

The cylinder’s volume simplifies to 2*Pi*r cubed*n.

When we work out the spheres’ volume as a fraction of the cylinder’s volume, nearly everything cancels to leave us just with (4/3)÷2, which equals 2/3.

So, the spheres together occupy two thirds of the cylinder, and therefore one third of the cylinder’s volume is not occupied by the spheres.

A wee joke to finish:

Why did the student not know formulas about spheres?

He didn’t get around to learning them!

 

Author: Paul A. Motwani

My name is Paul Motwani, but my colleagues, my students and their parents mostly call me "Mr. Mo"! My middle initial, A, stands for Anthony, because I was born on the official feast day of St. Anthony of Padua, the patron saint of miracles and of lost souls. I love teaching Mathematics and Chess, and giving fun-packed talks and shows in schools and clubs. The popular ingredients of Math, Chess, Mystery and Magic are my "Fantastic Four", and I give prizes too! I am an International Chess Grandmaster, and (loooooong ago!) I was the World Under-17 Champion. I am the author of five published chess books and hundreds of newspaper articles. I live with my wonderful wife and son in Belgium. I also love music, movies and puzzles. I blog at paulmotwani.com. My e-mail address is pmotwani141@gmail.com. You can find me on Facebook, too.

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