Batch 3 - Class 104 - Minarets of Samarra
Attendance: Smiti, Muskaan, Tishyaa, Khushi, Arnav, Anishka, Palak, Ahana, Arnav, Anisha, Liza, Zorawar, Gautam, Kabir, Ishaan, Diya, Nandini, Aneyaa, Arushi, Abhiram
Class Notes:
- A man is caught on the King's property. He is brought before the King to be punished. The King says, "You must give me a statement. If it is true, you will killed by lions. If it is false, you will be killed by trampling of wild buffalo." But in the end, the King has to let the man go. What was the man's statement?
- Answer: "I will be killed by trampling of wild buffalo."
- There is one and only one DraculaThis Dracula is locked in three and only three casketsOne is solid red, one is solid yellow, one is solid blueThe caskets have inscriptionsRed says, Dracula is here (meaning inside the casket)Yellow says, Dracula is not here (not inside the casket)Blue says, Dracula is not in the red casketAt most one inscription is true.Where must Dracula be?
- Solution - Yellow
Minarets of Samarra
There is one great minaret in Samarra, but what if the Caliph decided to build twelve? Since the Caliph loves regular geometry, he decides to build those in rotational symmetry on the following grid.
However, he also wants to make sure that every minaret is visible from every other minaret. So the following doesn't work.
- Can you build the twelve minarets of Samarra?
- For the next challenge, you have to travel to the court of Suleiman the Magnificent in Istanbul in 1500s. Your job now is to build 20 minarets with mirror symmetry as shown below.
- For a simpler problem, can you find all solutions for a small city? Place 8 minarets here. (Hint: there are 4 solutions)
- Unsolved Problem: How many minarets can you cram into a big (n x n) city? This puzzle was popularized by Henry Dudeney - a puzzlemaster who lives 1857-1930.
(Worksheets are available at http://mathpickle.com/wp-content/uploads/2015/10/No-three-Minarets-in-a-line.pdf - Pages 17,19 and 34 for problems mentioned above)
Homework
Find four different fractions, in each of which the numerator is one less than the denominator, so that when all four fractions are added together, their sum will be a whole number. There is more than one solution.
- Answer:1/2, 2/3, 6/7, 41/42 - Many other solutions
- Since each number is of form (n-1)/n which is 1-1/n, and since sum of all four is integer, it means that 1/a+1/b+1/c+1/d is an integer. The largest value of this is for a,b,c,d=2,3,4,5 which is 1.28, so the integer sum must be 1, and a must be 2. Hence 1/b+1/c+1/d=1/2. Now set b = 3 to find some solutions for c and d; then set b = 4 to find other solutions, and so on.
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