Interesting Math Problems

[QillerDaemon]

and you are wrong if you think there is only one way to cut up a square into 5 rectangles and have them all be of equal area.

Yes, but there is one additional restriction: one of the rectangles has one side length being 4 units and the other length unknown. There is only one way to divide up a square following those two restrictions.

[Animal]

and you are wrong if you think there is only one way to cut up a square into 5 rectangles and have them all be of equal area.

Yes, but there is one additional restriction: one of the rectangles has one side length being 4 units and the other length unknown. There is only one way to divide up a square following those two restrictions.

no. i gave you two examples that do it and both divide the square up entirely differently.

[Animal]

i will put my answer a different way.

If you have a square with a side of 10. then i can cut that into five equal area rectangles with one of the rectangles having dimensions of 5x4.

If you have a square with a side of 5. then i can cut that into give equal area rectangles with one of the rectangles having dimensions of 4x1.

In both of those examples, I cut the square up entirely differently.

[Reservoir Dog]

10 units.

It seems like this answer would work as well. You could have 2 rectangles that are 5x4 and then 3 that are 10x2. The area of each of the rectangles in this example is 20 square units. The Square it all started from was 10 units x 10 units. I suspect there are infinite solutions to this problem.

To be honest, I'm rethinking it. The 10.2 answer could be correct.

[QillerDaemon]

and you are wrong if you think there is only one way to cut up a square into 5 rectangles and have them all be of equal area.

Yes, but there is one additional restriction: one of the rectangles has one side length being 4 units and the other length unknown. There is only one way to divide up a square following those two restrictions.

no. i gave you two examples that do it and both divide the square up entirely differently.

I'll try to post the diagram that came with the problem. When I posted the original question, I thought the diagram was "obvious" so didn't include it. But I got to thinking about that diagram, and why it made the solution to the problem so easy. I got in contact with the fellow who posted the question originally, and he said that the diagram was correct, and the *only* correct diagram for the stated problem. The issue is that you can look at the diagram, and think that due to the couple of integer side measurements, all the side measurements are integer. But if you divide 256 units squared by five rectangles, you don't get an integer area per rectangle. So even if one side of a rectangle is integer, the other side will not be integer. In fact, only three of the rectangles have an integer side length on one side. Rounding to one decimal digit produces two more "integer" side lengths, but that's some hard rounding.

Yes, you can cut up the square into shapes of equal area, but the restriction of one rectangle (possibly at least one) having a side length of four units puts the number of slicing solutions down to exactly one. Well, really two, only because two of the rectangles have the same dimensions, so combined into one new rectangle, could be sliced perpendicular to the original cut and still be a valid solution. But that's a triviality that makes no difference to the original diagram or to the solution.

[Animal]

and you are wrong if you think there is only one way to cut up a square into 5 rectangles and have them all be of equal area.

Yes, but there is one additional restriction: one of the rectangles has one side length being 4 units and the other length unknown. There is only one way to divide up a square following those two restrictions.

no. i gave you two examples that do it and both divide the square up entirely differently.

I'll try to post the diagram that came with the problem. When I posted the original question, I thought the diagram was "obvious" so didn't include it. But I got to thinking about that diagram, and why it made the solution to the problem so easy. I got in contact with the fellow who posted the question originally, and he said that the diagram was correct, and the *only* correct diagram for the stated problem. The issue is that you can look at the diagram, and think that due to the couple of integer side measurements, all the side measurements are integer. But if you divide 256 units squared by five rectangles, you don't get an integer area per rectangle. So even if one side of a rectangle is integer, the other side will not be integer. In fact, only three of the rectangles have an integer side length on one side. Rounding to one decimal digit produces two more "integer" side lengths, but that's some hard rounding.

Yes, you can cut up the square into shapes of equal area, but the restriction of one rectangle (possibly at least one) having a side length of four units puts the number of slicing solutions down to exactly one. Well, really two, only because two of the rectangles have the same dimensions, so combined into one new rectangle, could be sliced perpendicular to the original cut and still be a valid solution. But that's a triviality that makes no difference to the original diagram or to the solution.

I still disagree. I will post a picture if I can

math_01.jpg

[QillerDaemon]

Like many of these types of problems, the creator starts from the solution and leads back to the final problem statement. You start from a square of 256 square units (so 16 units to each side), then how can you cut up the square into five equal area rectangles so that one (possibly only one, or maybe more than one) rectangle has a one side length of four units. The real idea is to show how that the diagram produced by these restrictions is the only initial condition that leads back to the solution. That's kind of what I'm working on, the proof of initial concept.

Your solution is likely correct for some problem, just not this one. Like I said, I thought that the diagram was "obvious" so didn't include it, that was a mistake. The great Martin Gardner made this mistake a few times in his old SciAm column, so I can only feel so bad about it.

[Animal]

New problem:

A square is cut up into five rectangles of equal area. The rectangles will not necessarily have the same dimensions. The side length of one rectangle is four units. How many units is the original square? The answer is a number, not a formula.

I am not going to read down, in case anyone has posted the answer. Let me think about this.

My first guess would be that the Original Square is 5 units x 5 units.

Four of the rectangles would be 4x1 and the fifth rectangle would be 2x2. All having an area of 4 units.

I kinda regret asking this problem now, only because you have to know there is only one way to cut up a square into five rectangles and have them all have the same area, *and* have one of the rectangles having a side length of 4 units at the same time. In cutting up the square, it turns out that there's a smaller square made up of three of the equal-area rectangles, and from there it's simple math. The side length of the original square turns out to be 16 units, with total area 256 units.

You can cut up a square in all sorts of ways that involve the component rectangles having equal areas and with each rectangle having its own dimensions. But only one way to do it such that one rectangle has a side length of four. I thought that was obvious, but it really isn't.

Here is what you posted originally:

New problem:

A square is cut up into five rectangles of equal area. The rectangles will not necessarily have the same dimensions. The side length of one rectangle is four units. How many units is the original square? The answer is a number, not a formula.

That was how the problem was stated. There is no mention of 256 units. Simply a square of any size. I chose a 10 x 10 square. And I cut it into 5 rectangles, shown in my diagram. One of them has a side equal to 4. Actually, two have a side equal to 4, but that doesn't matter in the way the question is worded. All 5 of the rectangles (5x4, 5x4, 10x2, 10x2, and 10x2 have equal area of 20 square units). If you add up the area of those 5 rectangles you get 100 square units which is the area of the square to begin with.

Now, rather than talk about what the author did, tell me what it is in my solution that doesn't meet the context of the question.

[Animal]

If you take a square that MUST have 256 units, then that square is 16 x 16 units.

Now, if you cut that square up into 5 rectangles then each one of them would have an area of 51.2 square units. Which is an odd choice, but considering the size of the original size of the square, you are stuck with it.

Now, if you take 51.2 and divide by 4 you get 12.8. So one of the rectangles has to have a length of 12.8 by 4. So, that square (16x16) could be cut into 4 rectangles 12.8 x 4 and one other rectangle that is 3.2 x 16. Which would work, but its just one more example of a solution that works.

[QillerDaemon]

Since we're dredging up old threads...

Three rectangles lay in a row on a line. The rectangle on the left (first one) has an area of 12 units, the rectangle on the right (third one) has an area of 20 units. The middle rectangle (second one) sits between the the left and right rectangles with a width of 3 units. The total base length of the three rectangles is 12 units. The middle rectangle is higher than the rectangle on the left by five units, and is also higher than the rectangle on the right by four units. What is the area of the middle rectangle? As there are multiple solutions, find only the smallest whole number solution.

[Deathproof]

Since we're dredging up old threads...

Three rectangles lay in a row on a line. The rectangle on the left (first one) has an area of 12 units, the rectangle on the right (third one) has an area of 20 units. The middle rectangle (second one) sits between the the left and right rectangles with a width of 3 units. The total base length of the three rectangles is 12 units. The middle rectangle is higher than the rectangle on the left by five units, and is also higher than the rectangle on the right by four units. What is the area of the middle rectangle? As there are multiple solutions, find only the smallest whole number solution.

16 units

[Animal]

24

[rule34]

Yeah. 220... 221, whatever it takes.

[Antknot]

Pie are found, cornbread are square

[Animal]

the first rectangle can't be 1 unit tall with an area of 12 since the entire base of the 3 rectangles is only 12. So that rules out 1 as its height.

If the height of rectangle 1 is 2, then the base is 6 (since area = 12). That would make the height of Rectangle 2 to be 7 (5+2). With an area of 21 (7 x 3). And that would leave the base for the final rectangle to be 12-6-3 = 3. Since its area is known to be 20 and that isn't divisible by 3, then we can rule out the height of Rectangle 1 being 2.

Now, if the height of retangle 1 is 3. The base is 4 (since area =12). That would make height Rectangle 2 to be 8 (5+3). With an area of 24 (8 x 3). And that would leave the base for the final rectangle to be 12-4-3 = 5. Since its area is 20 and that is divisible by 5, then this answer works and it would be the smallest area for rectangle 2.

So, the answer is 24.

Rectangle 1 is 3 x 4 (height x base)
Rectangle 2 is 8 x 3
Rectangle 3 is 4 x 5

Total of the base = 12

Extra credit for showing my work.

[Animal]

A gun and a bullet cost $110. The gun costs $100 more than the bullet. What does the bullet cost?

[rule34]

$5

[Animal]

$5

excellent.

[QillerDaemon]

Extra credit for showing my work.

Interesting! You solved the problem essentially "backwards" compared to a more traditional solution path. But that's alright, we were only interested in the answer. A more traditional solution path would be:

Let a be the horizontal side length and b be the vertical side length of the left rectangle. So A(rec L) = 12 = ab
Let x be the horizontal side length and y be the vertical side length of the right rectangle. So A(rec R) = 20 = xy

"The total base length of the three rectangles is 12 units" -> 12 = a + 3 + x -> x = 9 - a
Length y is one unit larger than b -> y = b + 1
So A(rec R) = 20 = xy = (9 - a)(b + 1) = 9b + 9 - ab - a. We know that quantity ab = 12.
So 20 = 9b + 9 - 12 - a -> 20 = 9b - 3 - a -> 23 = 9b - a. This is a linear equation in (a,b).

So try some values of a to get a whole number value for b.
So a = 1 means b = 24/9, a = 3 means b = 26/9. If a = 4, then b = 27/9 = 3.

So a = 4 and b = 3, then x = (9 - a) = (9 - 4) = 5, and y = (b + 1) = (3 + 1) = 4.
A(rec R) = 20 = xy = 5x4, check.

So horizontal side length of rec C = 3, and vertical side length = b + 5 = 3 + 5 = 8 (could also be (b + 1) + 4, same answer)
Therefore A(rec C) = 3x8 = 24.

[Animal]

Extra credit for showing my work.

Interesting! You solved the problem essentially "backwards" compared to a more traditional solution path. But that's alright, we were only interested in the answer. A more traditional solution path would be:

Let a be the horizontal side length and b be the vertical side length of the left rectangle. So A(rec L) = 12 = ab
Let x be the horizontal side length and y be the vertical side length of the right rectangle. So A(rec R) = 20 = xy

"The total base length of the three rectangles is 12 units" -> 12 = a + 3 + x -> x = 9 - a
Length y is one unit larger than b -> y = b + 1
So A(rec R) = 20 = xy = (9 - a)(b + 1) = 9b + 9 - ab - a. We know that quantity ab = 12.
So 20 = 9b + 9 - 12 - a -> 20 = 9b - 3 - a -> 23 = 9b - a. This is a linear equation in (a,b).

So try some values of a to get a whole number value for b.
So a = 1 means b = 24/9, a = 3 means b = 26/9. If a = 4, then b = 27/9 = 3.

So a = 4 and b = 3, then x = (9 - a) = (9 - 4) = 5, and y = (b + 1) = (3 + 1) = 4.
A(rec R) = 20 = xy = 5x4, check.

So horizontal side length of rec C = 3, and vertical side length = b + 5 = 3 + 5 = 8 (could also be (b + 1) + 4, same answer)
Therefore A(rec C) = 3x8 = 24.

i thought about setting up functional equations for each rectangle, but based on the numbers, it looked to me that there could only be a very few solutions that would work. so i figured that starting with 1 for the height of the first rectangle and working up would be faster. since the answers had to involve whole numbers, that really reduced the possible solutions.

[Animal]

Three crackheads live under a bridge downtown. They agreed to buy new rocks of crack. Alan and Burt would go get the crack and Carl would stay to protect their shopping carts and boxes. Burt bought 75 rocks of crack behind the bus station and Alan bought 45 rocks. Once they got back to the bridge, they split the rocks equally. Carl paid $140 for the crack. How much dollars did
Burt and Alan get from that sum, considering they equally split the crack?

[QillerDaemon]

Three circles are stacked so they are touching each other. What is the area of the space between them if the radius of each circle is 4. Bonus credit, if the radius of each circle is x.

[Deathproof]

It's kinda scary to me how smart you guys are. For real, that's not my stroking your collective egos. Math, talk about a torture device. I took enough in HS to pass (barely)!!

We're not really all that smart. We just look like it compared to CTC.

[stonedmegman]

Three circles are stacked so they are touching each other. What is the area of the space between them if the radius of each circle is 4. Bonus credit, if the radius of each circle is x.

4

[Animal]

Three circles are stacked so they are touching each other. What is the area of the space between them if the radius of each circle is 4. Bonus credit, if the radius of each circle is x.

i'm not sure i understand the question. I am pictures 3 balls (with radius 4) all stacked on top of each other like balls of ice cream on a cone. So, I don't know what you mean by the space between them?