Problematics | The long and the short of growing flowers

This is the week of the Academy Awards, which will be awarded on the very morning when this instalment of Problematics comes out. However, since we already a warm-up puzzle a couple of weeks ago, we will skip movies for now and possibly go back to them later.

This week, let us talk about Ian Stewart, professor emeritus of mathematics at the University of Warwick, is one of the most prolific writers of popular mathematics and science today. Over the last several years, Professor Stewart has often provided helpful comments and insights for mathematics-based articles I have written. His books also contain a variety of puzzles, one of which I am adapting here today.

Before that, though, here’s a different puzzle that will lead us to what was discussed above. The following is adapted from a puzzle that is attributed to Heron of Alexandria, the ancient mathematician and engineer.

#Puzzle 81.1

A floriculturist marks out two rectangular flowerbeds of very different dimensions, one of them with a more disproportionate length-to-breadth ratio than the other. Because he has seeds for a variety of flowers, which he wants to grow separately, he divides both plots into as many square mini seedbeds as their respective areas allow. Each square has a side of 1m, and the number of squares in each plot is an integer. Once the seedbeds are marked out, the Flower Man sets up poles at the corners of both plots and finally fences them off with separate lengths of wire, just one circuit around each plot.

Looking at his work with satisfaction, like most of us do after a job well done, our man notices with interest that the number of 1x1 squares in one plot is exactly three times the number of such squares in the other. He then examines how much wire he has used and finds, again to his amusement, that the wire around one plot is three times as long as the wire around the other plot.

There are two things to note here, though. One, a little wire is used in wrapping it around the poles, but this small amount is not factored into the calculation. Two, this time, the plot with a lower number of squares has been fenced with a wire that is three times as long as the wire going around the plot with the higher number of squares. To put it another way, the plot with the larger area has the smaller perimeter.

What are the dimensions of the two plots?

#Puzzle 81.2

The reason I chose this week to adapt one of Professor Stewart’s puzzles is that it can follow naturally from the story you have just read. You may have seen other versions of the same puzzle elsewhere, but Professor Stewart has planted more red herrings than any other version I have seen. The puzzle in this form appears in ‘Professor Stewart’s Cabinet of Mathematical Curiosities’.

In time, pretty flowers bloom on the floriculturist’s two rectangular plots. He decides to sell them. He estimates as ₹9,000 the total value of the flowers, which includes the monetary equivalent of the time, money and effort he has spent growing them. Aiming for a round-figure profit, he advertises the sale at ₹29,000.

The first customer who arrives looks very respectable. He writes out a cheque, then exclaims: “Oh, I made a mistake; I wrote ₹30,000 instead of ₹29,000.”

“Never mind,” says the Flower Man, “I’ll get you the change.”

He comes to me and presents me the cheque, for which I give him ₹30,000 in cash. The Flower Man then returns ₹1,000 from this to the customer.

I go to the bank with the cheque, where it bounces. I confront the Flower Man. With the respectable-looking customer nowhere to be found, the Flower Man now turns to you for help. You too loan him ₹30,000, which he repays to me. In time, you demand your repayment, and the Flower Man returns your ₹30,000 too.

He now counts his losses: “A ₹20,000 profit that should have been mine, then the ₹1,000 change given to the customer, then ₹30,000 to the Problematics writer, and finally another ₹30,000 to the puzzle solver. My total losses come to ₹81,000.”

Was he right? If not, how much did he actually lose?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 80.1

Hi Kabir,

During the first leg, in 3 hours, the walkers would have travelled a distance of 3x6 = 18km while the runner would have travelled 3x9 = 27km, which includes the 18km travelled by the walkers plus a forward-and-backward distance of 9km. Thus, the runner travelled 18 + 9/2 = 22.5km running ahead and 9/2 = 4.5km running backward at a 9km/hr. Thus, time taken by the runner travelling ahead = 22.5/9 = 2.5hours (=2:30 hours), and time taken running back = 4.5/9 = 0.5 hours ( 30 minutes).

— Anil Khanna, Ghaziabad

***

Dear Kabir,

In the second leg, the runner travels 12km more to the edge of the forest. At 12km/hr, he takes 1 hour to reach the edge. By then the walkers will have travelled 6 km. On return, the distance of 12 – 6 = 6km is covered at their relative speed of 12 + 6 = 18km/hr. So, the time taken is 6/18=1/3 hours or 20 minutes, and the total duration of the run = 1 hour 20 minutes.

— Yadvendra Somra, Sonipat

#Puzzle 80.2

For this puzzle, I had promised extra points for neatness, and among the neatest illustrations have come from Professor Anshul Kumar and Group Captain (retired) RK Shrivastava. I hope no one will mind, however, that I have chosen to highlight instead the illustration sent by Inaaya, who is one of the youngest readers ever to solve a puzzle in Problematics.

Dear team,

Sharing the puzzle that is solved by my daughter —Inaaya Luthra, 8 years, student of grade 3. She has been inspired by her Nanaji, Mr Y K Munjal, who regularly sends his responses to your puzzles.

— Yashica Munjal

Solved both puzzles: Anil Khanna (Ghaziabad), Yadvendra Somra (Sonipat), Dr Sunita Gupta (Delhi), Shruti M Sethi (Ludhiana), Akshai Bakhai (Mumbai), Kanwarjit Singh (Chief Commissioner of Income Tax, retired), Sanjay S (Coimbatore), Sundarraj C (Bengaluru), Ajay Ashok (Mumbai), Amarpreet (Mumbai), Shishir Gupta (Indore), Professor Anshul Kumar (Delhi), YK Munjal (Delhi), Group Captain RK Shrivastava (retired, Delhi), Sampath Kumar V (Coimbatore)

Solved #Puzzle 80.2: Inaaya Luthra, Dr Vivek Jain (Baroda), Shri Ram Aggarwal (Delhi), Jaikumar Inder Bhatia & Disha Bhatia (Ulhasnagar, Thane)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com