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Problematics | How to halve a square
This week, a centuries-old geometric exercise that is easier than it looks, and a disputed division of a soft drink between a pair of twins.
A recent research paper on how AI approaches mathematical problems provides the basis for the first of this week’s puzzles. Mind you, I am not going into what AI did. The puzzle merely picks up the centuries-old problem that AI attempted to solve, with a modification that was necessary to make it look unique. The second puzzle this week is my adaptation of a version I found among Shakuntala Devi’s writings.
Puzzle #162.1
A teenager is demonstrating how to draw a square without using a ruler. The only items he uses are an eraser-tipped pencil, a protractor and a divider. With his younger brother as his audience, the teen draws a line guided by the straight edge of the protractor, then marks two points at 90 degrees from each corner, and draws two verticals. Using the divider to mark the length of the original line, the teen marks the same length on both verticals. He erases the extra lengths, then completes the square.
His brother is not impressed, saying he could have done this himself without sitting through the demonstration. The elder one throws him a challenge: “All right then, draw a square whose area is exactly half the area of my square.” The younger one is unfazed: “Sure, let me take out my ruler and get Mom’s mobile; its calculator will give me the square root that will be the side of the smaller square.”
“No ruler, no calculator. Just pencil with eraser, protractor, and divider,” the elder brother says, setting the rules. “You may use the pencil and protractor to make additional constructions on my square if you need to, and the divider to mark out any length you wish. Neither of us needs to know that length in cm or inches. Just make sure your square is half as large as mine, area-wise.”
In fact, if the smaller square is drawn by means of additional constructions on the original square, you need not even use the divider. If you can solve this one, you may want to send figures to explain your procedure. That is not compulsory, but it will help.
Puzzle #162.2
A couple going to a party leaves food for their three daughters but, kids being kids, the girls want to make something for themselves. The 10-year-old twins are called Reena and Neera, while their 12-year-old sister is Narayani. Unfortunately, the only thing they know how to make is omelettes.
On the way home from school, they stop at a grocery store. Reena has a ₹20 note and a ₹10 coin, which is just enough for 5 eggs. Neera has only a ₹20 note, which gets her 3 eggs with ₹2 to spare. Narayani has just enough money to buy a 2-litre bottle of cola.
At home, they use the 8 eggs to make a single large omelette. They split this into three equal parts, supervised by the eldest to prevent any dispute between the twins. Then they pour some cola into three glasses, and start snacking.
When the glasses are empty, a second round is filled, and the bottle is now exactly half-full. “We’ll get two more glasses each,” says Neera, but her twin objects: “You contributed 3 eggs, so you get just one more glass after this one, your second. I contributed 5, which means I should get a total of 5 glasses.”
The eldest sighs. “You two settle your dispute yourself. I will have my 4 glasses. Count the other 8 glasses as my barter exchange for my share of the omelettes.”
Let us use the same yardstick: 8 glasses is equivalent to the omelette share of one girl. Keeping the eldest sister’s 4 glasses out of the equation, what are the rightful shares of the twins from the bottle of cola?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 161.1
Hi Kabir,
In the crossword competition puzzle, 851 is the safe cumulative score that will ensure a place in the top 30 after 10 rounds. This can be determined based on the observation that the safe score must be larger than the highest cumulative score after 10 rounds that the player ranked #31 can obtain.
Consider the scenario where scores of 70 to 100 in each round are confined to the same set of 31 players. In that scenario, the sum of the scores of these 31 players in each round = 100 + 99 + … + 71 + 70 = 31 x (100 + 70)/2 = 31 x 85. After 10 rounds, the sum of their cumulative scores would be = 31 x 85 x 10 = 31 x 850.
If these players were to have the same cumulative score, this would be 31*850/31 = 850. It turns out that this is not just a theoretical possibility. A cumulative score of 850 for 31 players is indeed feasible even when their scores in each round are distinct (that is, there are no ties in any of the rounds). The details are omitted here.
Thus 850 is the highest cumulative score after 10 rounds that a player ranked #31 can obtain. Therefore, if a player obtains a cumulative score of 850 + 1 = 851, after 10 rounds, there is no way 30 players can exceed his/her score.
— Professor Anshul Kumar, New Delhi
The only other reader who got close to the correct solution is Vinod Mahajan. He rightly calculated that 855 is the highest equal score possible among the top 30 players, but then inferred that this means 856 is the safe score. There is no way that 31 players can score 855, which means that the top 30 can afford to score a little lower than 855. But since it is possible for 31 players to score 850, even 1 point more than that is safe.
#Puzzle 161.2
To determine the number(s) with the stated property, the first step is to note that a number being divisible by every integer from 2 to 18 is equivalent to its being divisible by the LCM of these numbers, which is 12,252,240. Next, searching through the multiples of 12,252,240 in the 10-digit range yields exactly four pandigital solutions: 2438195760, 3785942160, 4753869120, and 4876391520. Each of these uses every digit 0–9 once and is divisible by every integer from 2 through 18.
— Vinod Mahajan, New Delhi
Solved both puzzles: Professor Anshul Kumar (Delhi); honourable mention: Vinod Mahajan (Delhi)
Solved #Puzzle 161.2: Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Ajay Ashok (Delhi), Shishir Gupta (Indore), Sabornee Jana (Mumbai)
