Posted - 03/04/2013 : 18:15:51 Three rectangles with integer sides have identical areas. The first rectangle is 278 feet longer than it is wide. The second rectangle is 96 feet longer than it is wide. The third rectangle is 542 feet longer than it is wide. Find the dimensions and area of the rectangles.

12 L A T E S T R E P L I E S (Newest First)

someguy

Posted - 03/19/2013 : 03:55:21 Hi TchrWill, there is nothing fancy going on with those statements.

Suppose we have two squares that differ by 71137. If we choose the nonnegative integers n and k so that the smaller square is n^2 and the larger square is (n+k)^2, then the difference of these squares is (n+k)^2 - n^2 = n^2+2kn+k^2 - n^2 = 2kn+k^2.

This is where the equation 71137 = 2kn + k^2 comes from.

Solving for n in this expression gives

n = (71137 - k^2) / (2k)

Note: Given any two squares, we can always find integers n and k for which the smaller square is n^2 and the larger square is (n+k)^2. Just let n=sqrt(smaller square) and k = sqrt(larger square)-n.

TchrWill

Posted - 03/17/2013 : 16:29:53 My last post did not post for some reason or another.

Nice going guys. You sure knocked that one right out of the ballpark. A sure shorter solution than the one I received from an unknown source 15 years ago.

Someguy, I am not familiar with the expressions you used. Might there be a derivation or explanation someplace that would light the bulb in my brain. They look familiar but I just can't put their source straight.

71137 = 2kn + k^2

we can write a difference of squares as (n+k)^2-n^2=2kn+k^2)

n=(71137-k^2)/(2k)

Have posted another oldie for you to ponder if sure choose.

someguy

Posted - 03/14/2013 : 02:29:39 Hi TchrWill.

Here is the method I used last night but I have chosen a different starting point since solving for x first leads to slightly fewer cases that need to be considered.

We want z^2 + 542z - x^2 - 278x = 0 and y^2 + 96y - x^2 - 278x = 0. Applying the quadratic formula to solve for y and z leads to the observation that both x^2+278x+73441 and x^2+278x+2304 must be squares.

This means 71137 is a difference of squares. If there is a solution, then we must be able to write 71137 as

71137 = 2kn + k^2 for some positive integers k and n.

(we can write a difference of squares as (n+k)^2-n^2=2kn+k^2 )

Note that we must have 1 <= k <= 266 and k must divide 71137=(11)(29)(223). The possible values for k are k=11, k=29, and 223.

Since n=(71137-k^2)/(2k), the possible values for n are 3228, 1212, and 48. Since x^2+278x+2304=n^2, we can again use the quadratic formula to solve for x and find that only two of the possible values of n lead to integer values of x.

n=1212 leads to x=1080. n=48 leads to x=0.

Since we want rectangles in the initial problem, we reject x=0 and find there is only one solution, x=1080.

Finally, we use this to find y=1164 and z=970.

TchrWill

Posted - 03/13/2013 : 13:01:11 Nice job someguy. Quicker to say the least. Might you list the steps to arriving at the y = 1164.

The following is the clarified solution I acquired back in 1998.

While sorting out my math problem archives, I ran across this somewhat elusive problem. This gem first came across my desk back in 97-98 when I was running an online Problem of the Week Club. It was sent to me by another recreational math enthusiast. I do not know where he obtained the problem but his efforts aimed at solving it led no where at which point he sent it to me for my enjoyment. We both pursued closed form solutions with little success. He ultimately found a solution posted in the publication containing the original problem statement. The published solution left much to be desired as it was fraught with missing steps. I took it upon myself to try and fill in the steps and logic missing in the response. I make no claims for its derivation but and only filled in some of the missing logic in the response. While not a fully closed form solution, it does come close since no guess and check is required (nor is any required in someguy's answer). I will be searching my files for a clue as to where this problem originally came from and continue my efforts toward discovering a closed form solution.

Problem

Three rectangles with integer sides have identical areas. The first rectangle is 278 feet longer than it is wide. The second rectangle is 96 feet longer than it is wide. The third rectangle is 542 feet longer than it is wide. Find the dimensions and area of the rectangles.

Let x = width of rectangle 1 Let y = width of rectangle 2 Let z = width of rectangle 3

Then, x(x + 278) = x^2 + 278x = A y(y + 96) = y^2 + 96y = A z(z + 542) = z^2 + 542z = A

Completing the squares, x^2 + 278x + (139)^2 = A + 19,321 or (x + 139)^2 = A + 19,321 y^2 + 96y + (48)^2 = A + 2304 or (y + 48)^2 = A + 2304 z^2 + 542z + (271)^2 = A + 293,764 or (z + 271)^2 = A + 73,441

Letting (x + 139) = a Letting (y + 48) = b Letting (z + 271) = c

We can now write: (a^2 - b^2) = (x + 139)^2 - (y + 48)^2 = A + 19,321 - A - 2304 = 17,017 (c^2 - b^2) = (z + 271)^2 - (y + 48)^2 = A + 73,441 - A - 2304 = 71,137 (c^2 - a^2) = (z + 271)^2 - (x + 139)^2 = A + 73,441 - A - 17,017 = 54,120

Knowing that a^2 b^2 = (a + b)(a b): Any factor pairs of 17,017 can be equated to (a + b) and (a - b) allowing for solutions of a and b. For values of a, corresponding values of x are derivable and the resulting rectangle areas for each solution. Doing the same for (c^2 - b^2 = (c + b)(c - b) = 71,137 will produce solutions for c, z and A which can be compared to the values derived for a,x, and A.

Considering (a^2 - b^2) = 17,017: The prime factorization of 17,017 is 7^1(11^1)13^1(17^1). The total number of factors derives from f(N) = (a + 1)(b + 1)(c + 1).......(i + 1) where here, a, b, c, ....i are the exponents of the prime factors. Therefore, f(17,017) = (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16 namely 1, 7, 11, 13, 17, 77, 91, 119, 143, 187, 221, 1001, 1309, 1547, 2431, and 17,017.

For example: We can write (a + b) = 17,017 and (a - b) = 1. Adding, 2a = 17,018 making a = 8509 and x = 8370. The resulting area becomes A = 8370(8370 + 139) = 72,383,760. Repeating this process for the other viable factor pairs produces the following results:

Considering (c^2 - b^2) = 71,137: The prime factorization of 71,137 is 11^1(29^1) 223^1. The total number of factors is therefore 2(2)2 = 8 namely 1, 11, 29, 223, 319, 2453, 6467, and 71,137.

Equating the appropriate factor pairs to (c + b) and (c - b), we derive a set of values for c, z, and A.

Note the common area for x = 1080 and z = 970 identifying an area of 1,466,640.

Substituting the derived A back into y^2 + 96y = 1,466,640 leads to y^2 + 96y - 1,466,640 = 0.

From the quadratic formula, y = [-96+/-sqrt(9216 + 5,866,560)]/2 = 1164.

Therefore, for widths of 1080, 1358 and 1164, the corresponding lengths are 1080 + 278 = 1358, 1164 + 96 = 1260 and 970 + 542 = 1512.

The rectangles being sought are therefore 1080ft by 1358ft, 1164ft by 1260ft, and 970ft by 1512ft, all with areas of 1,466,640 sq.ft.

Any comments welcomed.

Will post another favorite soon.

someguy

Posted - 03/13/2013 : 04:22:33 Ultraglides idea is useful but note that (y^2+96y+19321) and (y^2+96y+73441) are both squares.

The difference of these two squares is 54120.

Finding two squares that differ by 54120 isn't too bad and leads to y=0 or y=1164. Since we don't want y=0, we use y=1164 and solve for x and z.

x=1080,y=1164,z=970

the_hill1962

Posted - 03/13/2013 : 00:01:14 Yes, I got tired after having my program run for a while. I plan on optimizing it so it will run faster. The first attempt at coding usually is slow and sloppy for me but that is how I do it.

quote:Originally posted by Ultraglide

I used a spreadsheet and found that y=24 is a solution (which was already found) but I got bored after 126. Note that the trivial solution x=y=z=0 works but obviously does not define a rectangle.

Ultraglide

Posted - 03/12/2013 : 21:24:03 I used a spreadsheet and found that y=24 is a solution (which was already found) but I got bored after 126. Note that the trivial solution x=y=z=0 works but obviously does not define a rectangle.

the_hill1962

Posted - 03/12/2013 : 13:28:01 Yes, looking for the "y" that makes y+96y+19321 a perfect square is the key! I do believe it is going to be an incredibly LARGE y. Finding the solution in this manner could be quite impossible. Probably almost as impossible as finding the prime factors of incredible large numbers! This is an interesting problem. I have been busy lately and have not had time to explore further in my first attempt. However, I think that this equation that Ultraglide posted is much more elegant. At least it seems to be a better way to tackle it. The code was getting messy as I tried to optimize it from my first attempt.

TchrWill

Posted - 03/11/2013 : 18:37:09 A clear and wise first attempt hill1962. However, y^2 + 96y + 19,321 is a perfect square for an integer "y". I'll let you exercise your efforts in this search and seize path. Finding a method of getting one of the three clearly leads to the remaining two.

The usual 3 equations with 3 unknowns is a cakewalk but we really have 4 unknowns here. I am attacking it myself and will hold off for a few days before I post the solution I do have. I appreciate your exploring it as I am sure that there is another, simpler solution.

Ultraglide

Posted - 03/11/2013 : 16:41:55 I took the first equation and rewrote it as a quadratic in x: x+278x-4y-96y=0 Using the quadratic formula gives x = -139(y+96y+19321). The only way this would give an integer is if the quantity under the root sign is a perfect square, which it is not.

TchrWill

Posted - 03/09/2013 : 16:11:30 Your approach would ultimately lead to solution(s)but would take quite a length of time. Indeed, a computer program could be written to make the search and seize process acceptable. I do not have a program capable of doing this, nor I have the ability to write such a program.

I have an algebraic solution sent to me some time ago but am primarily interested in finding alternative solution paths, be they simple or more lengthy. I found my 15 year old notes on this problem and am going to give it another try. I would like to see some responses from the viewers of the problem. I will ultimately post the solution sent to me 15 years ago which was not clear and required my simplifying the steps given.

I hope that my posting these problems and solutions will ultimately lead to wider interest. I feel that all of us like to tackle word problems that are not homework or class assignments. I look forward to increased interest and participation.

I welcome any and all comments and suggestions.

the_hill1962

Posted - 03/06/2013 : 15:39:57 I probably won't be of much help because what I am seeing is solving the system: x(x+278)=y(y+96) y(y+96)=z(z+542) z(z+542)=x(x+278) I don't know how to solve this system algebraically. However, since the solutions need to be integers, I looked at a table of values for x(x+278) when x=1, it is 279 x=2 gives 560, x=3 gives 843, x=4 gives 1128 and so on. Then I checked the table of values for y(y+96) to see if any of the above values were in the list since we are trying to solve x(x+278)=y(y+96) and if I found a match, that would be a solution. No luck. I went further in the list and got to x=10 in x(x+278) gives 2880. This is in the list for y(y+96) at y=24. Hoping that z(z+542)=2880 gives integer solutions, I solved it but found that it is z = -271 76321 So I kept looking over the tables and found x=370 and y=444 both gave the value 239760 but solutions to z(z+542)=238760 aren't integers! Not giving up, I saw x=522 and y=600 matched at the value 417600 but z(z+542)=417600 does not give integer solutions. I found another at x=640 and y=720. However, again, no 'good' solution for z(z+542)=239760. I was thinking of writing a computer program so that I could find matching values quickly... Do you have a programming language that you know that you could write? If not, I will see what I can do. Let me know. Of course, we could still wait for some replies here. I am sure there is an algebraic way of doing this problem. I just see there have been 17 hits on this topic but no replies.