If #4 is true, it is provable:

If #4 is true, changing the color of a single node (except to forbidden colors on edges) cannot change the parity of the trichromatic triangle count, and this would be checkable through a finite case analysis of graphs of size <=7. Given that lemma, we can recolor one corner to red, the remainder of one large edge blue and the remaining nodes green, producing the odd count 1.

#8:

We are looking for a surface [0,1]²->[0,1] whose intersection with the plane x=y does not contain a function of t. It suffices to show that the intersection looks like the letter s, with exactly the endpoints reaching t=0 or t=1. It suffices to show that the intersection can be any continous function of x including the points t=x=y=0 and t=x=y=1. Within each plane of constant x, define the surface as 0 for small t, 1 for large t, and rapidly rising through the plane x=y wherever we want the intersection.

I'm late, but I'm quite proud of this proof for #4:

Call the large triangle a graph and the $n^2$ triangles simply triangles. First, note that for any size, there is a graph where the top node is colored red, the remaining nodes on the right diagonal are colored green, and all nodes not on the right diagonal are colored blue. This graph meets the conditions, and has exactly one trichromatic triangle, namely the one at the top (no other triangle contains a red node). It is trivial to see that this graph can be changed into an arbitrary graph by re-coloring finitely many nodes. This will affect up to six triangles; we say that a triangle has changed iff it was trichromatic before the recoloring but not after, or vice versa, and we shall show that re-coloring any node leads to an even number of triangles being changed. This proves that the number of trichromatic triangles never stops being odd.

Label the three colors $x,y$, and $z$. Let $v$ be the node being recolored, wlog from $x$ to $y$. Suppose first that $v$ has six neighbors. It is easy to see that a triangle between $v$ and two neighbors has changed if and only if one of the neighbors has color $z$ and the other has color $x$ or $y$. Thus, we must show that the number of such triangles is even. If all neighbors have color $z$, or if none of them do, then no triangles have changed. If exactly one node has color $z$, then the two adjacent triangles have changed. Otherwise, let $j$ and $k$ denote two different neighbors of color $z$. There are two paths using only neighbors of $v$ between $j$ and $k$. Both start and end at a node of color $z$. By the 1-D Sperner Lemma (assuming the more general result), it follows that both paths have an even number of edges between nodes of color $z$ and $\{x,y\}$; these correspond to the triangles that have changed.

If $v$ is a node on one of the graph's boundaries changing color from $x$ to $y$, it has exactly 4 neighbors and three adjacent triangles. The two neighbors that are also on the boundary cannot have color $z$, so either none, one, or both of the ones that aren't do. If it's none, no triangle has changed; if it's one, the two neighboring triangles have changed; and if it's both, then the two triangles with two nodes on the graph's boundary have changed.

Some preliminary thoughts on q9:

As Jessicata pointed out, in 3-dimensions or higher, our n-hedra don't split up as nicely as in the 2d case.

That isn't the only issue: many of the surfaces of connected blocks may not correspond to the type of 2d grid that we just proved this result for and it doesn't seem trivial to figure out how to characterise what kind of grids we need to extend our result too (it will be even worse in higher dimensions)

I've found two results: firstly that if you remove a triangular face and replace it with three others (imagine allowing the surface to jut out of the page), then the trichromatic parity will be preserved. Secondly, that we can replace two triangle with four triangles

Given what I've discussed above, I'd be keen for a hint as learning enough geometry to make progress on this problem would seem to be taking me pretty far afield from maths useful for ai-risk.

Inappropriately highbrow proof of #4 (2d Sperner's lemma):

This proves a generalization: any number of dimensions, and any triangulation of the simplex in question. So, the setup is as follows. We have an n-dimensional simplex, defined by n+1 points in n-dimensional space. We colour the vertices with n+1 different colours. Then we triangulate it -- chop it up into smaller simplexes -- and we extend our colouring somehow in such a way that the vertices on any face (note: a face is the thing spanned by any subset of the vertices) of the big simplex are coloured using only the colours from the vertices that span that face. And the task is to prove that there are an odd number of little simplexes whose vertices have all n+1 colours.

This colouring defines a map from the vertices of the triangulation to the vertices of the big simplex: map each triangulation-vertex to the simplex-vertex that's the same colour. We can extend this map to the rest of each little simplex by linear interpolation. The resulting thing is continuous on the whole of the big simplex, so we have a continuous map (call it f) from the big simplex to itself. And we want to prove that we have an odd number of little simplices whose image under f spans the whole thing. (Call these "good" simplices.)

We'll do it with two ingredients. The easy one is induction: when proving this in n dimensions we shall assume we already proved it for smaller numbers of dimensions. The harder one is homology, a standard tool in algebraic topology. More precisely we'll do homology mod 2. It associates with each topological space X and each dimension d an abelian group Hd(X), and the key things you need to know are (1) that if you have f : X -> Y then you get an associated group homomorphism f* : Hd(X) -> Hd(Y), (2) that Hd(a simplex) is the cyclic group of order 2 if d=0, and the trivial group otherwise, and (3) that Hd(the boundary of a simplex) is the cyclic group of order 2 if d=0 or d = (dimension of simplex - 1) and the trivial group otherwise. Oh, and one other crucial thing: if you have f : X -> Y and g : Y -> Z then (gf)* = g*f*: composition of maps between topological space corresponds to composition of homomorphisms between their homology groups.

(You can do homology "over" any commutative ring. The groups you get are actually modules over that ring. It happens that the ring of integers mod 2 is what we want to use. A simplex is, topologically, the same thing as a ball, and its boundary the same thing as a sphere.)

OK. So, first of all suppose not only that the number of good simplices isn't odd, but that it's actually zero. Then f maps the whole of our simplex to its boundary. Let's also consider the rather boring map g from the boundary to the whole simplex that just leaves every point where it is. Now, if the thing we're trying to prove is true in lower dimensions then in particular the map gf -- start on the boundary of the simplex, stay where you are using g, and then map to the boundary of the simplex again using f -- has an image that, so to speak, covers each boundary face of the simplex an odd number of times. This guarantees -- sorry, I'm eliding some details here -- that (gf)* (from the cyclic group of order 2 to the cyclic group of order 2) doesn't map everything to the identity. But that's impossible, because (gf)*=g*f* and the map f* maps to Hn(whole simplex) which is the trivial group.

Unfortunately, what we actually need to assume in order to prove this thing by contradiction is something weaker: merely that the number of good simplices is even. We can basically do the same thing, because homology mod 2 "can't see" things that happen an even number of times, but to see that we need to look a bit further into how homology works. I'm not going to lay it all out here, but the idea is that to build the Hd(X) we begin with a space of things called "chains" which are like linear combinations (in this case over the field with two elements) of bits of X, we define a "boundary" operator which takes combinations of d-dimensional bits of X and turns them into combinations of (d-1)-dimensional bits in such a way that the boundary of the boundary of anything is always zero, and then we define Hd(x) as a quotient object: (d-dimensional things with zero boundary) / (boundaries of d+1-dimensional things). Then the way we go from f (a map of topological spaces) to f* (a homomorphism of homology groups) is that f extends in a natural way to a map between chains, and then it turns out that this map interacts with the boundary operator in the "right" way for this to yield a map between homology groups. And (getting, finally, to the point) if in our situation the number of good simplices is even, then this means that the map of chains corresponding to f sends anything in n dimensions to zero (essentially because it means that the interior of the simplex gets covered an even number of times and when working mod 2, even numbers are zero), which means that we can think of f* as mapping not to the homology groups of the whole simplex but to those of its boundary -- and then the argument above goes through the same as before.

I apologize for the handwaving above. (Specifically, the sentence beginning "This guarantees".) If you're familiar with this stuff, it will be apparent how to fill in the details. If not, trying to fill them in will only add to the pain of what's already too long a comment :-).

This is clearly much too much machinery to use here. I suspect that if we took the argument above, figured out exactly what bits of machinery it uses, and then optimized ruthlessly we might end up with a neat purely-combinatorial proof, but I regret that I am too lazy to try right now.

Rough approach for qu 6:

Join the center to each of the corners and color each segment a different color and arbitrarily coloring each ambiguous point. Radially extend the colored sections to infinity.

To prove f(x) has a fixed point consider g(x) = f(x) - x which can take values outside of the triangle. To find a fixed point, we simply need to show that g(x) will map at least one point to the center. It is easy to prove that each corner will map to the color opposite and each edge can only map to points of a different color (unless it passes through the center, in which case we would have obtained out proof). At this point the problem reduces to 5 assuming that we construct a big enough circle.

Strategies that I've found helpful:

If something doesn't seem tractable, try flipping between algebraic and geometric interpretations of a problem. Problems 1 and 3 fell to this approach.

Specific solutions (or suggestive handwaving):

Problem 1:

I thought of it like parity - going left to right, each unichromatic edge doesn't change the color, while each bichromatic edge does. So to have an overall change, we need either 1 bichromatic edge, or 3 (1 and 2 that cancel), or 5 (1 and 4 that cancel)...

Problem 2:

I couldn't understand this one at first. After checking Wikipedia, I think that $R_n$ refers to the space that each point in the sequence lies within. An example of a finite sequence in $R_2$ would then be $(1,2),(3,1),(5,3)$

Problem 3:

Consider the unit square. We need to draw one continuous line, going from left to right, that covers the entire vertical extent of the square. No matter how you do that, you need to cross the diagonal line from the bottom left to the top right.

Why? Because you need to touch the top and the bottom edges. You can't do that at the bottom-left or top-right corners, since then you'd touch the diagonal line. But then the point where you touch the top edge is entirely within the top triangle, and it cannot touch the bottom edge without entering the bottom triangle. Switching between triangles is identical to crossing the diagonal line.

As for why this isn't true if the set is open rather than closed: if we exclude the edges from our consideration of "does it intersect the diagonal", then it's fairly trivial to construct a curve that stays inside one triangle and has a codomain of (0,1). $f\left(x\right)=x^2$ should work.

Ex 1

Let $V=\{v_0,...,v_n\}$ and $e_k:=(v_{k-1},v_k)$. Given an edge $e$, let ${\varphi }_e$ denote the map that maps the color of the left to that of the right node. Given a $k\in \{1,...,n\}$, let ${\varphi }_k={\varphi }_{e_k}\circ ...\circ {\varphi }_{e_1}$. Let $b$ denote the color blue and $g$ the color green. Let $c\left(k\right)$ be 1 if edge $e_k$ is bichromatic, and 0 otherwise. Then we need to show that ${\varphi }_n\left(b\right)=g⟹\left({\sum }_{k=1}^{n}c\right(k\left)\right)\text{is odd}$. We'll show $\left({\sum }_{k=1}^{n}c\right(k\left)\right)\text{is even}⟺{\varphi }_n\left(b\right)=b$, which is a striclty stronger statement than the contrapositive.

For $n=1$, the LHS is equivalent to $c\left({\varphi }_1\right)=1$, and indeed $\varphi \left(e_1\right)\left(b\right)$ equals $g$ if $e_1$ is bichromatic, and $b$ otherwise. Now let $n>1$ and let it be true for $n-1$. Suppose ${\sum }_{k=1}^{n}c\left(k\right)\text{is even}$. Then, if $c\left(e_n\right)=1$, that means $\left({\sum }_{k=1}^{n-1}c\right(k\left)\right)\text{is odd}$, so that ${\varphi }_n\left(b\right)={\varphi }_{e_n}\left({\varphi }_{n-1}\right(b\left)\right)={\varphi }_{e_n}\left(g\right)=b$, and if $c\left(e_n\right)=0$, then $\left({\sum }_{k=1}^{n-1}c\right(k\left)\right)\text{is even}$, so that ${\varphi }_n\left(b\right)={\varphi }_{e_n}\left({\varphi }_{n-1}\right(b\left)\right)={\varphi }_{e_n}\left(b\right)=b$. Now suppose ${\sum }_{k=1}^{n}c\left(k\right)\text{is odd}$. If $c\left(e_n\right)=1$, then $\left({\sum }_{k=1}^{n-1}c\right(k\left)\right)\text{is even}$, so that ${\varphi }_n\left(b\right)={\varphi }_{e_n}\left({\varphi }_{n-1}\right(b\left)\right)={\varphi }_{e_n}\left(b\right)=g$, and if $c\left(e_n\right)=0$, then $\left({\sum }_{k=1}^{n-1}c\right(k\left)\right)\text{is odd}$, so that ${\varphi }_n\left(b\right)={\varphi }_{e_n}\left({\varphi }_{n-1}\right(b\left)\right)={\varphi }_{e_n}\left(g\right)=g$. This proves the lemma by induction.

Ex 2

Ordinarily I'd proof by contradiction, using sequences, that $inf\{x\in [0,1\left]|f\right(x)>0\}$ can neither be greater nor smaller than 0. I didn't manage a short way to do it using the two lemmas, but here's a long way.

Set $x_0=0$. Given $n\in N_{+}$, let $G_n$ be a path graph of $n+1$ vertices $\{v_{n,0},...,v_{n,n}\}$, where $v_{n,k}=\frac{k}{n}$. If for any $n$ and $k$ we have $f\left(v_{n,k}\right)=0$, then we're done, so assume we don't. Define $c(v,v^{\prime })$ to be 1 if $f\left(v\right)$ and $f\left(v^{\prime }\right)$ have s different sign, and 0 otherwise. Sperner's Lemma says that the number of edges $e$ with $c\left(e\right)=1$ are odd; in particular, there's at least one. Let the first one be denoted $(v,w)$, then set $x_n=max\{v,x_{n-1}\}$.

Now consider the sequence $(x_n{)}_{n\in N}$. It's bounded because $x_k\in [0,1]$. Using the Bolzano-Weierstrass theorem, let $(x_n^{\prime }{)}_{n\in N}\mapsto x^{\ast }$ be a convergent subsequence. Since $f\left(x_k\right)>0$ for all $k\in N_{+}$, we have $f\left(x^{\ast }\right)\ge 0$. On the other hand, if $f\left(x^{\ast }\right)>0$, then, using continuity of $f$, we find a number $x^{\prime }>x^{\ast }$ such that $f(x^{\ast },x^{\prime })>0$. Choose $n$ and $k$ such that $v_{n,k}\in (x^{\ast },x^{\prime })$, then $c\left(v_{n,j}\right)=0$ for all $j<k$, so that $x_n\ge v_{n,k}>x^{\ast }$ and then $x_{\ell }\ge x_n$ for all $\ell \ge k$, so that $x^{\ast }\ge x_n\ge v_{n,k}>x^{\ast }$, a contradiction.

Ex 3

Given such a function $f$, let $g:[0,1]\mapsto [-1,1]$ be defined by $g\left(x\right)=f\left(x\right)-x$. We have $g\left(0\right)=f\left(0\right)-0\ge 0\ge f\left(1\right)-1=g\left(1\right)$. If either inequality isn't strict, we're done. Otherwise, $g\left(0\right)>0>g\left(1\right)$. Taking for granted that the intermediate value theorem generalizes to this case, find a root $x^{\ast }$ of $g$, then $f\left(x^{\ast }\right)=f\left(x^{\ast }\right)-x^{\ast }+x^{\ast }=g\left(x^{\ast }\right)+x^{\ast }=x^{\ast }$.

#8 actually comes up in physics:

in the field of nonlinear dynamics (pretty picture, actual wikipedia). The fact that continuous changes in functions can lead to surprising changes in fixed points (specifically stable attractors) is pretty darn important to understanding e.g. phase transitions!

Does this work for #7? (and question) (Spoilers for #6):

I did #6 using 2D Sperner's lemma and closedeness. Imagine the the destination points are colored [as in #5, which was a nice hint] by where they are relative to their source points - split the possible difference vectors into a colored circle as in #5 [pick the center to be a fourth color so you can notice if you ever sample a fixed point directly, but if fixed points are rare this shouldn't matter], and take samples to make it look like 2d Sperner's lemma, in which there must be at least one interior tri-colored patch. Define a limit of zooming in that moves you towards the tri-colored patch, apply closedness to say the center (fixed) point is included, much like how we were encouraged to do #2 with 1D Sperner's lemma.

To do #7, it seems like you just need to show that there's a continuous bijection that preserves whether a point is interior or on the edge, from any convex compact subset of R^2 to any other. And there is indeed a recipe to do this - it's like you imagine sweeping a line across the two shapes, at rates such that they finish in equal time. Apply a 1D transformation (affine will do) at each point in time to make the two cross sections match up and there you are. This uses the property of convexity, even though it seems like you should be able to strengthen this theorem to work for simply connected compact subsets (if not - why not?).

EDIT: (It turns out that I think you can construct pathological shapes with uncountable numbers of edges for which a simple linear sweep fails no matter the angle, because you're not allowed to sweep over an edge of one shape while sweeping over a vertex of the other. But if we allow the angle to vary slightly with parametric 'time', I don't think there's any possible counterexample, because you can always find a way to start and end at a vertex.)

Then once you've mapped your subset to a triangle, you use #6. But.

This doesn't use the hint! And the hints have been so good and educational everywhere I've used them. So what am I missing about the hint?

Here's a solution to 4:

We will first prove a lemma that all connected groups of green blocks that are completely surrounded by red or blue blocks will produce an even number of trichromatic triangles. We will then augment the triangle by adding an extra blue row on the bottom and an extra red side on the right such that we now have would what be a triangle if it weren't missing the bottom right corner. This will mean that all green blocks will now be surrounded so we'll have an even number of trichromatic triangles, but we have added exactly one additional such triangle, so that the original has an odd number.

------

Proof of Lemma:

A trivial variant of 1-d Sperner's Lemma is that if we start and finish on the same colour, we get an even number of bichromatic edges. For any block that is completely surrounded by red and blue blocks, we apply this variant to show that there is an even number of bichromatic edges on the outside that then translates to an even number of trichromatic triangles.

EDIT: Actually, there is one case we can run into which is tricky and that is something like:

R R R R R

R G G G R

R G R B R

R G G G R

R R R R R R

To see how to solve this case, read how to solve tricky interior cases below.

END EDIT

Thick interior blocks work similarly, but we can run into weird scenarios such as a blue and a red surrounding by green block:

g g g

g b r g

g g g

We might also run into something like this (ie. a three-spoked shape that is blue at the center and red on the sides surrounded by green).

g g g

g r g g

g b r g

g r g g

g g g

For these weird shapes we can still trace a path around this interior section, we just include going both down and up a spoke in our path (thanks Hoagy!). So the interior sections are also even.

Ex 5 (fixed version)

Let $D$ denote the triangle. For each $n\in N$, construct a 2-d simplex $S_n$ with $(n+1{)}^2$ nodes in $D$, where the color of a point corresponds to the place in the disk that $f$ carries that point to, then choose $x_n$ to be a point within a trichromatic triangle in the graph. Then $(x_n{)}_{n\in N}$ is a bounded sequence having a limit point $x^{\ast }$. Let $t$ be the center of the disc; suppose that $f\left(x^{\ast }\right)\ne t$. Then there is at least one region of the disc that $f\left(x^{\ast }\right)$ doesn't touch. Let $ϵ$ be the distance to the furthest side, that is, let $ϵ=\text{max}\left\{d\right(f\left(x^{\ast }\right),G),d(f\left(x^{\ast }\right),R),d(f\left(x^{\ast }\right),B\left)\right\}$. Since the sides are closed regions, we have $ϵ>0$. Using continuity of $f$, choose $\delta \in R_{+}$ small enough such that $B_d(x^{\ast },\delta )\subset B_d\left(f\right(x^{\ast }),ϵ)$. Then choose $n\in N$ large enough so that (1) all triangles in $S_n$ have diameter less than $\frac{\delta }{2}$ and (2) $d(x_n,x^{\ast })<\frac{\delta }{2}$. Then, given any other point $y$ in the triangle around $x_n$ in $S_n$, we have that $d(y,x^{\ast })\le d(y,x_n)+d(x_n,x^{\ast })<\frac{\delta }{2}+\frac{\delta }{2}=\delta$, so that $d\left(f\right(y),f(x^{\ast }\left)\right)<ϵ$. This proves that the triangle in $S_n$ does not map points to all three sides of the disc, contradicting the fact that it is trichromatic.

Ex 6

(This is way easier to demonstrate in a picture in a way that leaves no doubt that it works than it is to write down, but I tried to do it anyway considering that to be part of the difficulty.)

(Assume the triangle $T=\overline{ABO}$ is equilateral.) Imbed $T$ into $R^2$ such that $O=(0,0)$, $A=(-\frac{1}{2},\frac{1}{2}\sqrt{3})$, $B=(\frac{1}{2},\frac{1}{2}\sqrt{3})$. Let $f:T\mapsto T$ be continuous. Then $h:T\mapsto R^2$ given by $h:x\mapsto f\left(x\right)-x$ is also continuous. If $h\left(x\right)=O=(0,0)$ then $(0,0)=h\left(x\right)=f\left(x\right)-x⟹f\left(x\right)=x$. Let $R$ be the circle with radius 2 around $O$; then $h\left(T\right)\subset R$ because $||h\left(x\right)||=||f\left(x\right)-x||\le ||f\left(x\right)||+||x||\le 2$ (it is in fact contained in the circle with radius 1, but the size of the circle is inconsequential). We will use exercise 5 to show that $h$ maps a point to the center, which is $O$, from which the desired result follows. For this, we shall show that it has the needed properties, with the modification that points on any side may map precisely into the center. It's obvious that weakening the requirement in this way preserves the result.

Rotate the disk so that the red shape is on top. In polar coordinates, the green area now contains all points with angles between ${60}^{\circ }$ and ${180}^{\circ }$ , the blue area contains those between ${180}^{\circ }$ and ${300}^{\circ }$, and the red area those between $0^{\circ }$ and ${60}^{\circ }$ and those between ${300}^{\circ }$ and ${360}^{\circ }$. We will show that $h\left(\overline{AB}\right)$ does not intersect the red area, except at the origin. First, note that we have

$h\left(\overline{AB}\right)=\left\{f\right(x)-x|x\in \overline{AB}\}\subset \{y-x|x\in \overline{AB},y\in T\}=\{T-x|x\in \overline{AB}\}=T-\overline{AB}$

Since both $T$ and $\overline{AB}$ are convex combinations of finitely many points, it suffices to check all combinations that result by taking a corner from each. This means we need to check the points

$A-A$ and $B-A$ and $C-A$ and $A-B$ and $B-B$ and $C-B$.

Which are easily computed to be

$(0,0)$ and $(-1,0)$ and $(\frac{1}{2},-\frac{1}{2}\sqrt{3})$ and $(1,0)$ and $(0,0)$ and $(-\frac{1}{2},-\frac{1}{2}\sqrt{3})$

Two of those are precisely the origin, the other four have angles ${270}^{\circ }$ and ${150}^{\circ }$ and ${90}^{\circ }$ and ${210}^{\circ }$. Indeed, they are all between ${60}^{\circ }$ and ${300}^{\circ }$.

Now one needs to do the same for the sets $T-\overline{OA}$ and $T-\overline{BO}$, but it goes through analogously.