Here's the next question in the Math-E-Magic series of puzzles and problems. This question is taken from the harder section towards the back of the book (I was breaking us all in gently with my previous posts). I'll paraphrase the problem text, which is taken from "A Square Deal," on page 190.
A man has a piece of land he wants to leave to his three children. To his first child, he leaves an L-shaped piece of land, comprising the whole plot except a square that has been cut out of the north-east (top-right) corner. To his second child, he leaves this middle section, except for a smaller square that is taken from the corner of the plot - this smaller square goes to his third child.
The first child gets the area shown in yellow; the second child gets the area shown in pink, and the third child gets the area shown in blue. The squares have sides a, b and c, as shown. The diagram is not to scale.
The one constraint is that the first two children must receive land which has the same total area. There are many solutions to this problem, but the question is: can you identify the simplest solution with the smallest numbers for the sides a, b and c?
Firstly, let's express the question mathematically:
The first child receives the area a2 - b2. The second child receives the area b2-c2. These two areas are the same, hence a2 - b2 = b2-c2 and we must exclude the obvious b=0 and a = -c.
Trial and error will work here, but there are a number of tips that you can use to find one solution. Notice that Pythagorean triples (a2 + b2 = c2) are no help; we're solving a different puzzle here.
Tip: a and b are going to be closer together than b and c. Rearranging our puzzle expression a2 - b2 = b2-c2 gives us a2 = 2b2-c2 and if a2 > 2b2 then there is no solution; b must be greater than a/2.
Starting with the simplest numbers (as suggested in the question), I found a solution: a = 7, b = 5 and c =1, which has the convenient feature of probably being the simplest since c=1.
7*7 - 5*5 = 49 - 25 = 24
5*5 - 1*1 = 25 - 1 = 24
QED
With these values, the two children get an area equal to 24 square units,while the third child gets the remaining one square unit. Total = 24 + 24 + 1 = 49.
Having found my own solution, I checked the answers in the book. Interestingly (and controversially) the Math-E-Magic answers section has this to say:
"The proper dimensions are as shown [a = 14, b= 10, c=2]. There is no solution if the smallest square is only one mile on each side. There is however, a solution where the smallest square has a side of two. THe plots of the second and third child each measure 96 square miles, and the total are of the father's plot was 196 square miles."
Interestingly, the authors overlooked the simplest solution I found, in the same way as I overlooked theirs. I'll get in touch with them and let them know :-) Also, it's worth noting that my solution (a=7,b=5,c=1) is obtained by halving their solution (a=14,b=10, c=2) - this applies to all integer multiples (a=21, b=15, c=3 is another solution, as is a=28, b=20, c=4, and so on).
Here's a list of the solutions that I identified - it's not exhaustive, but it's what I could find in about 30 minutes of trial and improvement:
a=7, b=5, c=1 (the simplest), and all other multiples of these (14,10,2; 21,15,3; 28,20,4, etc.).
a=17,b=13,c=7
a=23,b=17,c=7
Two final notes on my solutions:
- In their simplest form, all of the variables are prime numbers. This could be just coincidence, but I've not found a solution which contains non-prime numbers (in the simplest form).
- I was very surprised (although I probably shouldn't be) to find two solutions with c=7.
A man has a piece of land he wants to leave to his three children. To his first child, he leaves an L-shaped piece of land, comprising the whole plot except a square that has been cut out of the north-east (top-right) corner. To his second child, he leaves this middle section, except for a smaller square that is taken from the corner of the plot - this smaller square goes to his third child.
The first child gets the area shown in yellow; the second child gets the area shown in pink, and the third child gets the area shown in blue. The squares have sides a, b and c, as shown. The diagram is not to scale.
Firstly, let's express the question mathematically:
The first child receives the area a2 - b2. The second child receives the area b2-c2. These two areas are the same, hence a2 - b2 = b2-c2 and we must exclude the obvious b=0 and a = -c.
Trial and error will work here, but there are a number of tips that you can use to find one solution. Notice that Pythagorean triples (a2 + b2 = c2) are no help; we're solving a different puzzle here.
Tip: a and b are going to be closer together than b and c. Rearranging our puzzle expression a2 - b2 = b2-c2 gives us a2 = 2b2-c2 and if a2 > 2b2 then there is no solution; b must be greater than a/2.
Starting with the simplest numbers (as suggested in the question), I found a solution: a = 7, b = 5 and c =1, which has the convenient feature of probably being the simplest since c=1.
7*7 - 5*5 = 49 - 25 = 24
5*5 - 1*1 = 25 - 1 = 24
QED
With these values, the two children get an area equal to 24 square units,while the third child gets the remaining one square unit. Total = 24 + 24 + 1 = 49.
Having found my own solution, I checked the answers in the book. Interestingly (and controversially) the Math-E-Magic answers section has this to say:
"The proper dimensions are as shown [a = 14, b= 10, c=2]. There is no solution if the smallest square is only one mile on each side. There is however, a solution where the smallest square has a side of two. THe plots of the second and third child each measure 96 square miles, and the total are of the father's plot was 196 square miles."
Interestingly, the authors overlooked the simplest solution I found, in the same way as I overlooked theirs. I'll get in touch with them and let them know :-) Also, it's worth noting that my solution (a=7,b=5,c=1) is obtained by halving their solution (a=14,b=10, c=2) - this applies to all integer multiples (a=21, b=15, c=3 is another solution, as is a=28, b=20, c=4, and so on).
Here's a list of the solutions that I identified - it's not exhaustive, but it's what I could find in about 30 minutes of trial and improvement:
a=7, b=5, c=1 (the simplest), and all other multiples of these (14,10,2; 21,15,3; 28,20,4, etc.).
a=17,b=13,c=7
a=23,b=17,c=7
Two final notes on my solutions:
- In their simplest form, all of the variables are prime numbers. This could be just coincidence, but I've not found a solution which contains non-prime numbers (in the simplest form).
- I was very surprised (although I probably shouldn't be) to find two solutions with c=7.
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