I've always wanted to start a thread of interesting math puzzles we've come across over the years. Maybe it's just me, but I love mulling over random mathematic logical exercises in my head. I don't know how this would best be formatted, but I think he who solves the most recent one should get to post the next one.

The one I'm going to start out with is a problem I had on 2015 AMC8 test back in middle school. That whole test was absolutely fascinating, but this problem stayed with me most because I couldn't actually figure it out under the test's time limit. I ended up taking it home and working it out after school, and eventually had the whole family figure it out themselves. It's not the hardest question, but it's a good one to start with because it's in the vein of math problems I want to see more of. Harder questions are most welcome, but having an elegant solution is the most important factor. I blurred out the multiple choice answers because I think it makes it too easy. No hints!

Sam; I kind of like those math problems too. After thinking for about a minute, I came up with a guess of 12.5 sq in. Not sure if my reasoning is correct for that answer.

Here’s another problem. I don’t know the answer because I’m just making up the dimensions. If you have a square opening that’s 10” by 10”, how big of a cube can be passed thru that opening. I’ll have to find a similar problem on the internet and figure out the answer for those values.

Ignore the above cube problem, it’s stated wrong. See my comment below.

**dave1707 wrote:**

Sam; I kind of like those math problems too. After thinking for about a minute, I came up with a guess of 12.5 sq in. Not sure if my reasoning is correct for that answer.

Incorrect, but you seem to have at least cleared the first hurdle!

**dave1707 wrote:**

Here’s another problem. I don’t know the answer because I’m just making up the dimensions. If you have a square opening that’s 10” by 10”, how big of a cube can be passed thru that opening. I’ll have to find a similar problem on the internet and figure out the answer for those values.

I don’t see how the answer can be anything other than a 10x10x10 cube.
I believe the answer is 15 sq in, but I may be incorrect.

**MateoConLechuga wrote:**

I believe the answer is 15 sq in, but I may be incorrect.

You got it!
Sam; I got the cube through a square problem wrong. It should be how big of a cube can pass through a 10” cube area. As for your problem, I’ll have to look into it more to see why it’s 15 sq in.

Just in case someone wants to know the answer of how big of a cube can pass thru the area of a 10” cube, it’s a 10.6066” cube . See Prince Rupert's cube for more information.

I love this kind of maths problem. I find these ones quite interesting, but not everyone sees the solution to them, which is really disappointing. For example, my brother struggles with a problem for 2 hours, but I am solving it in 10 minutes. And he was feeling pretty bad, so I found this advanced

solving math for him, and since then, it comes much more accessible to math. I feel better now that he can better understand this matter. I hope I helped him, but I didn't make him any worse. Do you have any other examples of such problems? I'd like to solve a few now.

Can someone explain how you get 15 square inches?

**arusher999 wrote:**

Can someone explain how you get 15 square inches?

It's actually really easy as long as you don't overthink it, you don't even need a calculator. Just

think about how space works and what you can do with a shape to fit it into a space.

But here's how I figured it out myself:

1. You can fit a big square inside the cut-up square if you rotate it 45 degrees.

2. You can ignore the cut corners after this, they don't affect the math (other than leading to the answer of needing a rotated square).

3. Each corner of the square touches the middle of one of the edges of the square, which forms 4 equivalent right triangles with a base and height length of 2.5 inches.

4. Since they're right triangles with the same base and height, their area is also equal to their base/height.

5. Adding that up gets 10 in^2 of space that isn't in the new square.

6. Logically, taking the space that isn't from the max amount of space, 5x5 = 25 inches, would get the remaining area of our rotated square: 25 - 10 = 15 inches.

Breaking lurking status yet again to say that with some clever manipulation of the 3rd dimension, I could fit a 34 square meter square in that space

This is an interesting problem, but I still don't see how people got 15 square inches.

I'm convinced it has to be 14.5.
Edit: the only problem I see with this solution is that I am assuming the corners of the rotated square meet the outer square at the .5 marks.

**King Dub Dub wrote:**

4. Since they're right triangles with the same base and height, their area is also equal to their base/height.

5. Adding that up gets 10 in^2 of space that isn't in the new square.

Do you even know how to math? A right triangles area is equal to the length*width divided by 2. Therefore, a right triangle with both side lengths equaling 2.5 has an area of (2.5*2.5)/2=3.125. Multiply that by 4 for each corner, and the missing area you get is 12.5. Thus, the size of your 45 degree rotated square is only 25-12.5=12.5.

You can get more area by rotating the square like I did.

**Sam wrote:**

**MateoConLechuga wrote:**

I believe the answer is 15 sq in, but I may be incorrect.

You got it!

Seriously, how is that the answer?
**Michael2_3B wrote:**

**Sam wrote:**

**MateoConLechuga wrote:**

I believe the answer is 15 sq in, but I may be incorrect.

You got it!

Seriously, how is that the answer?

Simple geometry once you understand the trick. You were just looking at the wrong triangles.

that makes sense now! any more problems?

I'm sorry but how did you get 1.5 for the small triangles areas, those side lengths (the tilted ones on the corners of the red) are definitely not 1 and 3

**Michael2_3B wrote:**

I'm sorry but how did you get 1.5 for the small triangles areas, those side lengths (the tilted ones on the corners of the red) are definitely not 1 and 3

Are you dense? The height and width are 1 and 3.
**MateoConLechuga wrote:**

**Michael2_3B wrote:**

I'm sorry but how did you get 1.5 for the small triangles areas, those side lengths (the tilted ones on the corners of the red) are definitely not 1 and 3

Are you dense? The height and width are 1 and 3.

Ouch.
Let’s not harass each other here. If the height and width of the small red triangles is 1 and 3, then the hypotenuse (the sides of the 9 square area square) cannot also be 3, but it clearly is. Perhaps we are not on the same page.

**Michael2_3B wrote:**

Let’s not harass each other here. If the height and width of the small red triangles is 1 and 3, then the hypotenuse (the sides of the 9 square area square) cannot also be 3, but it clearly is. Perhaps we are not on the same page.

Wtf are you talking about, do you know basic math? The triangle's height and width have nothing to do with the length of sides. I think you are misunderstanding that the area of a triangle is (width*height)/2.
ok. guys. i have a math problem that i can't figure out myself. what is the area of the shaded portion in terms of pi?

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