Author 
Topic 

TchrWill
Advanced Member
USA
80 Posts 
Posted  03/29/2013 : 09:33:50

There are three square rugs of different sizes. The first rug has 5 more square feet than the second rug. The second rug has 5 more square feet than the third rug. What are the integer dimensions of the three rugs?
Enjoy 


someguy
Advanced Member
Canada
143 Posts 
Posted  04/12/2013 : 21:24:45

Hi TchrWill,
looks like not answering didn't help much. I have been very busy with work so did not have a chance to think about this one until today. Perhaps others would like to try it with a hint? There have been more than 50 views so I assume someone else read it.
The edges of the squares are measured in inches. 


TchrWill
Advanced Member
USA
80 Posts 
Posted  04/12/2013 : 22:01:22

Hi someguy,
quote: Originally posted by someguy
Hi TchrWill,
Looks like not answering didn't help much. I have been very busy with work so did not have a chance to think about this one until today. Perhaps others would like to try it with a hint? There have been more than 50 views so I assume someone else read it.
The edges of the squares are measured in inches.
I was praying that you would respond soon, hoping that your continued interest would nudge others in the cadre to join in the fun.
Great hint. I assume that it led you to the answer which I will hold out a few more days before posting. Something leads me to think that it was solved by others but they are reluctant to participate for some reason.
I will continue to hope for others to join in. I truly hope that those that do join in will share some problems that they have run across over the years.
Bill 


TchrWill
Advanced Member
USA
80 Posts 
Posted  04/17/2013 : 20:30:39

quote: Originally posted by TchrWill
There are three square rugs of different sizes. The first rug has 5 more square feet than the second rug. The second rug has 5 more square feet than the third rug. What are the integer dimensions of the three rugs?
Enjoy
For what its worth:
Knowing that there is only one pair of square frames that could have a difference of 5 sq.ft. in their areas, i.e., 2x2 and 3x3, or 9  4 = 5, it behooves us to convert the area difference to square inches, 5x144 = 720 sq.in., wherein we stand a better chance of finding three integral squares differing by 144.
Also, knowing that x^2  y^2 = 720 can be rewritten as (x + y)(x  y) = 720, we can equate any pair of factors of 720 to both (x + y) and (x  y) and solve for x and y. The prime factors of 720 are 2^4(3^2)5^1 meaning that 720 has (4+1)(2+1)(1+1) = 30 factors all together. Subtracting 1 from the exponent of 2 and adding 1 to the other exponents gives us 3, 3, and 2. Half the product of these 3 numbers is the number of ways that 720 can be expressed as the difference of two squares, or (3x3x2)/2 = 9 ways. To define these 9 ways we divide 720 by 2^2 giving us 180. There are 9 pairs of factors of this number; 1180, 290, 360, 445, 536, 630,, 920, 1018, and 1215. The values of x and y derive from x = (f1 + f2)^2 and y = (f1  f2)^2 as follows: (180+1)^2  (1801)^2 = 181^2  179^2 = 32,761  32,041 = 720. (90+2)^2  (902)^2 = 92^2  88^2 = 8464  7744 = 720. (60+3)^2  (603)^2 = 63^2  57^2 = 3969  3249 = 720. (45+4)^2  (454)^2 = 49^2  41^2 = 2401  1681 = 720. (36+5)^2  (355)^2 = 41^2  31^2 = 1681  961 = 720. (30+6)^2  (30  6)^2 = 36^2  24^2 = 1296  576 = 720. (20+9)^2  (209)^2 = 29^2  11^2 = 841  441 = 720. (18+10)^2  (1810)^2 = 28^2  8^2 = 784  64 = 720. (15+12)^2  (1512)^2 = 27^2  3^2 = 729  9 = 720 Upon inspection of the x and y values, it is seen that 31, 41 and 49 are the only three that satisfy the arithmetic progression requirement that the difference between their successive squares be 144. Therefore, the dimensions of the three frames were 31 in., 41 in., and 49 in.
Any shorter versions welcomed. 


someguy
Advanced Member
Canada
143 Posts 
Posted  04/21/2013 : 06:21:17

Hi TchrWill,
We started off the same, but I used that fact that a difference of squares can be written as
(n+k)^2  n^2 = 2kn+k^2
instead of factoring x^2y^2 as (x+y)(xy). We had this difference in solutions once before. It does not lead to a quicker solution, but the method doesn't require much.
Since we are looking for positive integers n and k with the property that 2kn+k^2 = 720, we know k must be an even integer between 2 and 26. (If k was odd, 2kn+k^2 would be odd. Also, since 2kn>0, k^2<720). For each of these 13 values of k, we compute n=(720k^2)/(2k) and pay attention to the cases where n ends up being an integer. 9 of the 13 cases lead to acceptable values of n and k.
k=2,n=179 > 179 and 181 k=4,n=88 > 88 and 92 k=6,n=57 > 57 and 63 k=8,n=41 > 41 and 49 k=10,n=31 > 31 and 41 **41 is the only value that occurs twice k=12,n=24 > 24 and 34 k=14,n is not an integer k=16,n is not an integer k=18,n=11 > 11 and 29 k=20,n=8 > 8 and 28 k=22,n is not an integer k=24,n=3 > 3 and 27 k=26,n is not an integer
Scanning the list of possible side lengths, we see 41 is the only value that occurs in two pairs. 31, 41, and 49 inches are the dimensions we are looking for.
Although this method has a few more cases to consider, it may be easier for others to follow (but I like your method). Perhaps you could make it easier for others to read your solution if point out that the difference between (x+y) and (xy) is even so both of these factors must be even. Dividing both sides of the equality (x+y)(xy)=720 by 4 leads to ((x+y)/2) ((xy)/2) = 180. We are looking for two numbers whose product is 180. Now, given that (f1)(f2)=180 we can say 2(f1)=(x+y) and 2(f2)=(xy) which leads to x=f1+f2 and y=f1f2.



TchrWill
Advanced Member
USA
80 Posts 
Posted  04/21/2013 : 12:58:34

Nice touch someguy. I will add that path to my solution. I find it more rewarding to uncover, or become aware of, alternate solution paths. Thanks for responding. Hope to touch bases on future problems. 


the_hill1962
Advanced Member
USA
1470 Posts 
Posted  04/30/2013 : 14:17:08

Are there any more Recreational Math problems? 



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