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.
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…
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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?
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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.
(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)
Sunil Roy and I were strolling down the road – somewhere in Chelsea area with the rest of the family – when he posed an interesting puzzle for me. See if you can get it.
To give everybody a chance, refrain from posting the answer in the Comments section. Send me a personal message and I will put your name up if you solved it.
You have two dices. One has 1 thru 6 painted on its faces like a normal dice. The other has blank on every face. You can write down any integer number 1,2,3… on each of the blank faces of the second dice. You can even leave it blank – which would stand for 0. You are allowed to repeat the numbers (you can have two faces with the same number).
The question is: What numbers would you paint on the six faces of the second dice such that:
(*) When the two dice are rolled together, the sum total of the two faces up can be any integer between 1 and 12 AND
(*) The probability of any of those integers (meaning the sum of the two faces up being 1 or 2 or 3… or 12) is exactly the same.
You can assume that the two dices are unbiased. (It will be completely random which face will show up)
