Buddyboy3000

With only four folds, it seems impossible to get back to four sides. However...

These are just some questions for clarification. 1. Can your opponent put a point on a spot that is not one of the original locations? 2. Does the square have to be straight, or can it be tilted?

The line is the area that all the centers can be at. No center of a circle that goes through the two points can be off that line. 1. All points off a circle have to be the same distance to the center, which is the radius. This goes also with points (0,3) and (2,0). Imagine a line connecting them to the center of the circle. These lines would be marked as the variable X. So, you form a isosceles triangle. Two sides are X, and the other is the sqrt(13), which is the line connecting the two points to each other. In an isosceles triangle the top would be directly above the center of the base. In this case, the base would be the sqrt(13). The top would be the center of the circle. This is where you would get the line, because it is perpendicular with the line connecting the two points, and intersects through the middle of it. 2. To answer the first thing you asked me, the variable A will affect the area. It depends on which arc you want to measure, the minor arc or the major arc. As A reaches towards positive infinity, the area with the minor arc will get bigger, and the same thing with the major arc. When A goes the opposite way, the minor arc will get smaller in area, and the major arc will forever be at 0. Bigger and smaller area is based on when A is 0, and the circle is at its smallest. When I say A gets larger in absolute value, this means that the circle will just get bigger. This is why I do not believe that there can be a center anywhere off this line. If so, then your question has me stumped.

The line that I got was the same as CaptainEd got. However, this is what I got for finding the circles... I have not yet figured out the second part of the problem with the area, but this is what I have for now.

Using the strategy I used in the first connect the dots, this is my answer.

I finally found a way to make the road system less than 7.5 miles.

Oops... I did the math wrong. My idea is actually about 7.88 miles long. Instead ls 2*sqrt(2.6)+sqrt(8)+sqrt(1.6), it is 2*sqrt(2.5)+sqrt(8)+sqrt(1.6)+2*sqrt(0.1) I made the yellow sides a tiny bit longer, and forget the short roads to towns 2 and 8. I am going to continue and try to find a shorter path.

This will work to make the road system shorter, but you will no longer have connection to town #5. To find the shortest road system, I found it useful to use Desmos Calculator, if anyone else likes the idea. Anyways, this is my proposal...

I agree with Logophobic. Finding the additional paths of KAYAK, you have 6 different patterns that can happen. Two are on the sides, and four are at the corners. In total, there are 102 different ones, doubled to get 204 paths. The patterns are different shapes that KAYAK can make on the edges. It is not really the mathematical way, but it works the same way.

Maybe when it says," ...too large for the bus," it means that it will exceed the weight limit of the bus. If that is the case, then it will be impossible to put the gold bar inside, because the bus would not handle it.

How about this...

This is the only possible answer I could think of.

This is the most reasonable answer.

So you are saying that in perspective, all of the sides that are common, can't be the same length. Does that mean that a line horizontal and a different line vertical can still be the same number? If so...

If I understood your question correctly...

To-Do List for the Morning Go to sleep Wake up at 7:02:39:752 EXACTLY Look for alarm clock Look for sledge hammer Bring sledge hammer to bed Set alarm clock to 7:59:59:999 Go back to sleep Wake up due to alarm clock Yell, "I woke up before 8:00!" Pick up sledge hammer Destroy alarm clock Fix alarm clock (again). Walk down the hallway sleepily Fall down the stairs to wake yourself up Make sure to hit the ground with your left knee (don't forget it). Look for the Bandaids to cover your bleeding knee Walk back upstairs Walk back down the hallway Yell, "FIRE!" and wake everyone else up. Let everyone gather around you to help. Say, "Oops! I thought today was somebody's birthday Repeat steps 13-16 Go to the store Pick up some more Bandaids Go to the cash register Ask the person behind you in line, "What did you have for breakfast?" When the person replies, yell at the top of your lungs, "I FORGOT TO EAT BREAKFAST YESTERDAY!" Sprint towards the breakfast food aisle Realize you are not hungry Go back home

These might work for two of them. Number 2 Number 4

I showed this to my sister, and we both got an idea.

when you realize you are on the New Logic/Math Puzzles section of the forum doing complicated math in your free time We are all math nerds!

I took me a little while to figure this out. Now that I did... ...I still can't stop laughing!

When I do not look at the +, I see a blank spot moving around in the pink dots. When I do look at the +, the blank spot looks like a green dot. This is because your mind can not take the rapid change of colors quickly.

What do you mean by gender clues?

First, I need to see if it is theoretically possible. Counting all of the triangles, I got 96. Divided by 2 for all of the diamonds is 48. 48 divided by 3 is 16. So, 16 of each diamond needs to be used. I think it would be best to start placing them at the edges. I am able to do the outer edge evenly placing one type of diamond on two sides adjacent. With this, 7 of each diamond is used. Doc1.docx (Link leads to image of outer edge). Next, I can eliminate that layer and use the next. The next will have 6 less spots for diamonds to be put in. So, instead of 7, 5 will be placed onto the two sides. Then it's 3, and 1. So, each number of diamonds used is 7+5+3+1. This equals 16, the number of each diamond. To conclude, there can be an even number of red, green, and blue diamonds in the hexagon.

I agree. I might be able to decipher it, but it might be wrong. Lindafirsthenry@gmail.com The last part is right, but I am not sure if the first part is different or not.