33

The combined area of the X's is The solution:


31

I think there are three(ish) choices. The simplest one is To find the other solutions, we have to notice that This means we can also start and then we can, of course, start


28

Here's a diagram showing both the parts of the shape that make up each face of the cube in its own colour, and trying to give some indication of how they meet up when folded.


25

Here is the cube I made: Update:


22

I've managed to get As follows The trick here is Counting


19

First, let us define some things: For simplicity, for partial boards presented (with ...), let's consider that the width is equals to or larger than the height. If not, you can just rotate everything 90° to get a board that is like this. Unsolvable board (UB) - One that no matter what you rotates, it is impossible to have all the dominoes with the same ...


17

It's possible EDIT:


17

As this tessellation of the tilted new square neatly matches an overlapped tiling of the unit square, the $w{\small\,\times\,}x$ overlap of unit squares equals the $1 \! - r^2$ difference in the areas of the two types of squares. (The “$ \small\pm $” was deduced to be “$ \small + $” by testing the formula on the case where  $ x = \large{1 \over 3} ...


17

As I don't have a camera handy, I have had to unfold my (pink) cube before I could show it to you. Its sides are Sqrt(52) = Sqrt(4^2+6^2) units long.


16

Here is the answer: More is coming! Here is our original diagram completed with for sure later; the length $|DE|$ is our $x$ and let's put some specific angle that we are going to work with in our main square as below; I call the length of side of the other square as $y$ and as you can see from the $\alpha$ values, right triangles, and hypotenuse as $y$ ...


14

Problem 1: Problem 2: Problem 3: Problem 4: Problem 5:


12

I think this matches up with the other two solutions but uses coordinate geometry which is quicker here.


12

Firstly let's label everything: We have five similar right-angled triangles, which must be enough to get a lot of information about the interrelationships between the quantities $x,y,d,e,f,p,q,r$. Simplifying this expression, the final answer is


9

The number of triangles in my best solution is but I don't know if this is optimal. Addendum: I previously had an incorrect solution, as I used similar triangles instead of congruent ones (i.e. I used some triangles of a different size but the same shape). As requested by @humn, I'll keep that incorrect solution with 12 triangles available below:


9

The answer is Easiest derivation:


8

(No spoilers for most of this; they'd be too clumsy. There are a few "diagrams", if you can call them that, and I've spoilered those. The rest will only spoil things for you if you read it carefully.) Let's say our octahedron's vertices are at the six points in space with one coordinate $\pm1$ and the others $0$. Consider what a horizontal slice $z=...


8

I think the answer is: Reasoning: Update: Based on @Bass's answer, I was curious about the actual calculation. Turns out it's really not that hard.


6

This might not be optimal, but its the best one I could find: here are some of the triangles highlighted for clarity


6

With the following arrangement you can easily stack pieces into the box: I have assumed without loss of generality that $a<b<c$, but as Damien_The_Unbeliever noted in the comments, it also assumes that $a+b>c$. That does not matter however, as this arrangement can be tweaked to insert one more piece: Now it just remains to be proved that it is ...


5

There are four diagonals of a cube, but none of them meet at right angles: image source To see this, observe that The two diagonals define a unique plane containing both of them. In that plane, the two cube diagonals are also the diagonals of a rectangle,


5

Based on your 16-tile example I reduced it down to Sketch of proof why this is minimal:


5

My first answer was wrong for 3 independent reason. Fixing these my new answer is Because


5

Is this the answer you are looking for?


5

Answer Without loss of generality assume the board has 7 rows and 8 columns. Clearly we want all dominoes to be horizontal. Let us say that a state P is optimal if it is not possible to reduce the number of vertical dominoes with legal moves. Clearly the desired end state has zero vertical dominoes and is hence optimal. Suppose P is optimal but has at least ...


4

Problem 1


4

I think the answer is Explanation: First place one of the $4\times4$ squares within the $5\times5$ square. The remaining area (9 units) should be as compact as possible, so let's shift the $4\times4$ square right up to one corner, leaving an L-shape remaining. (I'm not sure how to prove rigorously that this is optimal.) Now we need to place the other two ...


4

Almost complete answer using Jaap's lower bound, i.e. ignore my lower bound and skip the first two blocks: Now that @Jaap Scherphuis has bumped the lower bound to It remains to be shown that Amy can choose in such a way that more becomes impossible.


3

I believe the answer is The question is an instance of Hilbert's Third Problem.


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