Geometry problem

Question

Two squares have integer side lengths in the ratio of 4 : 3. Their intersection is also a square with integer side lengths. If the total area shown on the surface equals 5000 square units, how long is a side of the largest square?

Answer 1

Say the squares have side lengths 4x and 3x, and so have areas 16x² and 9x². Say the square intersection has side length n, with n<3x.

If we simply add the areas of the squares, 16x²+9x², we double-count the area of the intersection. To correct the double-count, subtract off the intersection area once. So the total area is 16x²+9x²-n²=5000.

25x²-n²=5000

(5x)²-n²=5000

We need to find x and n, knowing that they're integers. One observation:

n²=5(5x²)-5(1000)

n is a multiple of 5. So if we rewrite n=5m and substitute, we get

(5x)²-(5m)²=5000

25x²-25m²=5000

Divide by 25 to get

x²-m²=200

Factor:

(x+m)(x-m)=200

Now, list all of the factor pairs of 200, and see if we can find a solution. Note that x+m must be positive, so we only care about positive factors. The possible values of x+m and x-m are:

1 and 200

2 and 100

4 and 50

5 and 40

8 and 25

10 and 20

Now to rule some of these out. If we add the factors, we get x+m+x-m=2x, an even number. That rules out (1, 200), (5, 40), and (8, 25), since those sums are odd. Let's check the other three. Note that x-m<x+m, since m is positive, so we always set x-m to be the smaller factor.

In each case, we add the two equations to get 2x=(the sum of the factors), then divide by 2 to get x. So we have

x=51, m=49

x=27, m=23

x=15, m=5

Now go back and find n in each case. Remember that n=5m, the side length of the intersection square:

x=51, n=245

x=27, n=115

x=15, n=25

But remember also that we need n<3x, which fails in the first two cases. So we rule those out, and the only remaining option is x=15.

Then the side length the largest square is 4x=60.

Answer 2

Nice problem!

If you draw the squares, you can label the large square 4x, the small square 3x and their intersection y².

The Area = 5000 = (4x)²+(3x)² -y² = 25x² - y²

x and y are integers, so let's solve x in terms of y: x = sqrt((5000+y²)/25) = sqrt(200 + y²/25) which tells us that y² = n²*25 in order to be an integer and that x is sqrt(200 + n²)

The first n that works is n=5, y²= 625, y = 25 and x=sqrt(225) = 15 (Note that 60 and 45 are greater than 25, so this makes sense physically.

The answer is 4x = 60

Please consider a tutor. take care.