I sent this to my nephews since they have been sending me a few puzzles. See if you can beat them to it…

Orange circle inscribed inside an equilateral triangle inscribed inside a larger circle. Can you prove that the orange area is the same as the addition of the two purple areas?

Reeling under the lack of power, water supply and all that the devastating cyclone Amphan left behind, apparently my nephews hunkered down and started doing puzzles. They could not wait for power to come back – so that they could recharge their phones and send me this problem thru Whatsapp.

Good problem. About 2 years back, I had posted a similar version of the problem.

This week is another area problem. Take a rectangle ABCD (as shown here). Draw two lines from the adjacent vertices A and D to two random points E and F on the opposite side such that they intersect within the rectangle. See picture below. Nothing is drawn to scale. The area of the two triangles are 4 and 16. One of the quadrilaterals has an area of 10. What is the area of the other quadrilateral?

Again, this is geometry from elementary school with a twist to it.

Note that the original version of the problem that I had posted had the quadrilateral area as 1 instead of 10. Harmindar Matharu successfully proved that such a diagram is not possible. I should have been more careful while picking some values for those areas…

Take any square. Take any random point within the square. Draw straight lines from that point to the midpoints of every side of the square. You will get four quadrilaterals. Three of them have areas of 40 units, 60 units and 80 units. What is the size of the total square?

This is an interesting problem and requires a little visualization and some basic memory of geometry.

We just took off from the airport and as the plane started gaining height, I noticed an interesting landscape on the ground. There is a man made water body and it has been carefully dug up to leave a landmass looking like an airplane!

Which airport was I in?

This is a followup puzzle to the previous one to understand why the one-black-dot case is not possible. Some find the answer to be controversial. To me the logic is bulletproof though.

So, it is the same problem as before except that the lady skips the portion where she asks the three men to raise their hands if they see at least one white dot. Instead, after removing the blindfolds, she straightaway asks “What color dot do you have on your forehead”.

After quite some thinking, the smartest person came up with the correct answer.

What was the answer and how did he figure it out?

For completeness sake, I am restating the whole problem here…

—
Three smart men were trying to find out who is the smartest of them all. Or was at least most logical and quickest thinker. To resolve that, they went to a woman who had the reputation of being the smartest of them all. She devised a quick problem for them. She blindfolded them and painted a dot on each of their foreheads.
She then told them “I have painted a dot on each of your foreheads. The dots are either white or black in color. I am going to soon open your blindfolds. You will be able to see other’s dots but not your own. Then I am going to ask you to guess the color of your own dot”.
Saying so, she opened their blindfolds.
“Okay, now guess what color is on your forehead”.
The three men started pondering and soon one of them – who was indeed a little smarter than the others – spoke up.
“Correct”, said the lady.
Question: What was the answer and how did he figure it out?
—

Three smart men were trying to find out who is the smartest of them all. Or was at least most logical and quickest thinker. To resolve that, they went to a woman who had the reputation of being the smartest of them all. She devised a quick problem for them. She blindfolded them and painted a dot on each of their foreheads.

She then told them “I have painted a dot on each of your foreheads. The dots are either white or black in color. I am going to soon open your blindfolds. You will be able to see other’s dots but not your own. I am then going to ask you to raise your hand if you see a white dot (on either of the two friend’s foreheads). Then I am going to ask you to guess the color of your own dot”.

Saying so, she opened their blindfolds.

“You may raise your hand now if you see a white dot”

All three men raised their hands.

“Okay, now guess what color is on your forehead”.

The three men started pondering and soon one of them – who was indeed a little smarter than the others – said “I have a white dot on my forehead”.

“Correct”, said the lady.

Question: How did the man figure out what color he had?

There are three archers – Hubert, Anamika and myself. Hubert is the world renowned, highly decorated Olympics champion from Belgium who is a “sure shot”. Meaning he never misses. Then there is Anamika – who is a amateur archer but has gotten her game to a point where she scores 2 out of 3 times. And then there is Rajib – yours truly – who barely knows which end of the arrow should point towards you. And I have an average starter’s chance of success – 1 in 3.

We find ourselves in a contest where we can only survive by killing the other two. (One successful shot with the arrow is enough to kill anybody). Also, we three are highly intelligent and know how to calculate probabilities of success. (With that level of intelligence, how we got ourselves into this situation is a good question but terrible for this puzzle).

The rules are the following:
Since I am the novice, I get the first shot.
Next, Anamika – being the second best shot goes. Of course, that assumes I did not aim at her and succeeded in which case Hubert goes next.
Once Hubert has had a chance (assuming he was still alive after I and Anamika had our shots), the cycle continues.
So it is me -> Anamika -> Hubert -> repeat till there is one person standing.

The question is: What should be my strategy as the first shot to maximize my chances of still living after the whole ordeal.

Here is one answer to the puzzle… The key thing is to realize – like Srini did that the “end” coin is a problem. Only way to solve this would be to create a hole in the line first – so that the end move is secured.

(well, evening, if you are on the other side of the world).

Start by getting six identical coins. Arrange them in the pattern “A” as shown in the picture. (An equilateral triangle).

The goal is to eventually land up with the pattern “B” – again, as shown in the picture. (A straight line).

Here are the rules…
1. You cannot lift a coin – merely slide from its current position to the new position.
2. You move one coin at a time.
3. When you move a coin, no other coin changes position.
4. IMPORTANT: When you move a coin to a new position, in that new position, it MUST touch TWO other coins at least.

Finally, this is not a trick question – nor is there any sleight of the hand involved.

I am sure there are many ways of doing it but the one I got needed me 7 moves. (Not sure if that is the shortest way though)