Assume this is not the case. Let $Y$ be the number of monochromatic isosceles triangles, and $X$ the number of isosceles triangles with two vertices of one color and the last vertex of a different color. Let $a$, $b$, $c$ be the number of vertices of each color. On the one hand, we have \[ X + 3Y = 3\left( \binom a2 + \binom b2 + \binom c2 \right) \]just by double-counting the triangles: the conditions $\gcd(n,6) = 1$ imply that exactly three isosceles triangles use any given edge. (To be precise, we are counting pairs $(\triangle, e)$, where $\triangle$ is isosceles and has edge $e$ with matching colors. The left-hand side counts by $\triangle$ while the right-hand side counts by $e$). On the other hand, we have \[ X + Y = \binom n2 \]since an isosceles triangle is determined by its base (again since $\gcd(n,6) = 1$). Therefore we have the following equality modulo $2$: \[ \binom n2 \equiv \binom a2 + \binom b2 + \binom c2 \pmod 2. \]Doubling and expanding we get \begin{align*} n^2-n &\equiv (a^2-a) + (b^2-b) + (c^2-c) \pmod 4 \\ &\equiv a^2+b^2+c^2 - n \pmod 4. \end{align*}But since $n$, $a$, $b$, $c$ are odd this is impossible. Remark: I think the conditions $\gcd(n,6) = 1$ and $a$, $b$, $c$ odd are huge give-aways that this will be ``global obstructions modulo $2$''.

Assume the opposite. Let two colors be red and green. Let $a$ be the number of red vertices and let $b$ be the number of green vertices. Let $X$ be the number of triangles which contains both red and green vertices. Since $\gcd(n,6)=1$, every pair of vertices belongs to $3$ isosceles triangles, once as base and twice as leg. Also every isosceles triangle which contains both red and green vertices, has two pairs of red-green vertices. Hence $2X=3ab$, but $ab$ is odd, contradiction.

Also German TSTST #5.

Korea TST #1, but we only had $n \ge 7$ with $n$ being odd.

Wouldn't it be possible to weaken the problem statement to just $n$ being odd? We modify the proof as follows. Suppose the colors are red, green, and blue (call them $R, G, B$). Let $r, g, b$ be the number of vertices for each respective color. We look at the number of edges with endpoints $G, B$. Since $g, b$ are odd, the number of edges is $gb$ which is odd. We will now count up the number of $GB$ edges in a different way. Denote the blue vertices as $U_1, U_2, \ldots, U_b$ and the green vertices as $V_1, V_2, \ldots, V_g$. Observe that since $n$ is odd and the fact that there are no $RBG$ isosceles triangles, each $GB$ edge corresponds to an isosceles triangle with a well-defined apex, which must be one of the $U_i$'s or $V_j$'s. In this manner, the blue and green vertices partition all the $GB$ edges into classes. Consider an apex $V_i$ and the $\frac{n - 1}{2}$ isosceles triangles that it forms. For these triangles, the other two vertices can be one of $GG, GB, BB, RR, GR$. From the perspective of $V_i$, since the number of blue vertices must be odd, we see that the cases $GG, BB, RR, GR$ each contribute an even number of blue vertices while $GB$ contributes one blue vertex. Therefore, of those $\frac{n-1}{2}$ triangles, an odd number of them have a base of the form $GB$. Similarly, from the perspective of an arbitrary $U_i$, there must also be an odd number of bases of the form $GB$. The $U_i$'s and $V_j$'s partition all the $GB$ edges, with each $U_i, V_j$ class having an odd number of $GB$ edges. But since $g, b$ are both odd, this would imply the number of $GB$ edges is $b + g = 0 \mod 2$, a contradiction.

We use only the fact that $n$ is odd. Let the colors be $a, b, c$. For each vertex $v$ of color $a$ consider the $\frac{n - 1}{2}$ pairs of vertices that form an isosceles triangle with peak $v$. If any of this pairs is colored $b, c$ we are done. Assume this never happens, then every pair that contains an odd number of endpoints colored $b$ is of the form $a, b$. As $b$ is used an odd number of times it follows that an odd number of these pairs exist opposite to each vertex of color $a$. Similarly this happens for each vertex of color $b$. Thus, as an even number of vertices are colored either $a$ or $b$, and every pair of vertices has exactly one isosceles triangle of which it is the base, it follows that the total number of segments with endpoints colored $a, b$ that are opposite to a vertex of color $a, b$ is even. Since there exists an odd number of total segments of this type, as every color was used an odd number of times, it follows that there exists a segment with endpoints $a, b$ opposite to a vertex of color $c$. These three vertices form our desired triangle.

Are there any solution using $n$-th or $2n$-th roots of unity ?

We prove an easily generalisable claim: Generalisation to IMOSL 2016 C3 wrote: Let $n$ be an odd positive integer. We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of atleast two colors. Show that the number of isosceles triangles whose three vertices are of different colours are odd.

Nice problem. Here's my solution: Suppose there are $x,y,z$ vertices of each color, where $x,y,z$ are odd numbers satisfying $x+y+z=n$. Suppose there are $A$ isosceles triangles having two vertices of same color, and third vertex of another color. Also, suppose there are $B$ polychromatic isosceles triangles. As $\gcd(n,6)=1$, so each edge is part of three isosceles triangles, once as base and twice as side-edges. Then each polychromatic edge (i.e. edge connecting two vertices of different colors) is part of an isosceles triangle. Also, each polychromatic triangle uses a polychromatic edge thrice, while a triangle having exactly two vertices of same color uses a polychromatic edge twice. Thus, we get $$2A+3B=3(xy+yz+zx) \Rightarrow 2 \mid (xy+yz+zx)-B$$But, as $x,y,z$ are odd, so $xy+yz+zx$ is also odd. This means that $B$ must also be odd, proving that $B \geq 1$. Hence, done. $\blacksquare$

Really nice, especially after the semi-number theoretic C2. cjquines0 wrote: Let $n$ be a positive integer relatively prime to $6$. We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours. Assume on the contrary that we have no isosceles triangles all of whose vertices are of different colors. Let $T$ denote the set of isosceles triangles that are not monochromatic. Let $a,b,c$ be the number of points of each color. The key observation is that any edge joining two distinct colored points is a part of exactly three isosceles triangles. To see this, notice that $n$ is odd so exactly one isosceles triangle has this edge as its base. Also, two (isosceles) triangles have this edge as one of the equal sides. These two triangles don't coincide since $(n,3)=1.$ To finish, we count the number of pairs $(\Delta, E)$ in two ways, where $\Delta \in T$ is a triangle and $E \in \Delta$ is an edge of $\Delta.$ If we first choose $\Delta,$ then we find that there are $2|T|$ such pairs, since exactly two sides of $\Delta$ are not monochromatic. If we first choose $E,$ then there are three possibilities for $\Delta$ as discussed above. The number of pairs by counting in this way is thus $3(ab+bc+ca).$ So $$2|T|=3(ab+bc+ca)$$This is a contradiction since $a,b,c$ are all odd. $\blacksquare$

Isn't the following even stronger statement also true? Modified 2016 C3 wrote: Let $n$ be a odd positive integer. We paint the vertices of a regular $n$-gon with three colours so that there is at least one vertex of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours.

JNEW wrote: Isn't the following even stronger statement also true? Modified 2016 C3 wrote: Let $n$ be a odd positive integer. We paint the vertices of a regular $n$-gon with three colours so that there is at least one vertex of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours. No. [asy][asy] size(5cm); draw(unitcircle); for(int i=0; i<25; ++i){ if (i % 5 != 0) { dot(dir(360/25*i), red); } } dot(dir(0), blue); for(int i=1; i<5; ++i){ dot(dir(360/5*i), green); } [/asy][/asy]

Let $A,B,C$ be the number of vertices of each color. Suppose there are $M$ isosceles triangles with 2 colors, and $N$ with one color. Assume that every isosceles triangle has at most two colors. We will get equations with this assumption to get a contradiction. Firstly, $M+N$ is the total number of isosceles triangles. We count another way; the key idea is that since $n$ is odd, any pair of vertices we choose is the base of a unique isosceles triangle. So counting the isosceles triangles by choosing their base gives us $\tbinom{A+B+C}{2}$ isosceles triangles. Therefore, \[ M+N = \binom{A+B+C}{2}. \]Now, consider choosing an isosceles triangle by first choosing two vertices of the same color, and then choosing the third vertex. There are $\tbinom{A}{2}+\tbinom{B}{2}+\tbinom{C}{2}$ ways to choose the two vertices of the same color, and 3 ways to choose which two vertices in the triangle these are. But this is also $M+3N$, since we overcount the monochromatic triangles 3 times in the count. Therefore, \[ M+3N = 3\left[\binom{A}{2} + \binom{B}{2} + \binom{C}{2}\right]. \]Now we solve for $N$: \begin{align*} 2N &= 3\binom{A}{2} + 3\binom{B}{2} + 3\binom{C}{2} - \binom{A+B+C}{2} \\ &=A^2+B^2+C^2-AB-BC-CA-A-B-C \\ &=A(A-1) + B(B-1) + C(C-1) - (AB+BC+CA) \\ &\equiv 0+0+0-1 \equiv 1 \pmod2, \end{align*}contradiction.

Let the colors be $1,2,3$. There are $a,b$ vertices colored $1,2$ respectively. Now suppose there is no such polychromatic triangle. Let $A$ be the number of isosceles triangles such that the base is $1--2$ or $2--1$ and the peak is $1$. Let $B$ be the number of isosceles triangles such that the base is $1--2$ or $2--1$ and the peak is $2$. For each $1--2$ or $2--1$ base the peak vertex is either $1$ or $2$. So $A+B=ab \implies A+B \equiv 1 \mod 2$. OTOH, fix a peak vertex colored $1$. So its bases that contain $2$ are either $2--2$ or $2--1$ or $1--2$. The point is since $b$ is odd the number of bases of type $1--2$ or $2--1$ is also odd. Every peak vertex determines unique isosceles triangles (since $(n,3)=1$). Since there are an odd number of peak vertices colored $1$ and for each an odd number of type $A$ triangles $\implies A \equiv 1 \mod 2$. Similarly $B\equiv 1 \mod 2$. So $A+B \equiv 0 \mod 2$. Contradiction.

Solved with yusufsheikh2207. Claim: Every edge of the regular $n$-gon is an edge of $3$ isosceles triangles. Proof: Consider a polygon with vertices $P_1, P_2, \dots P_n.$ Also consider edge $P_iP_{i+1} = K.$ Then, $K$ is a edge of $\triangle P_iP_{i+1}P_{i+2},$ $P_{i-1}P_iP_{i+1},$ and $P_iP_{i+1}P_{i+\lfloor n/2 \rfloor}.$ $\square$ For the sake of contradiction, assume there are no triangles with all different color vertices. Let the number of single color triangles be $P$ and the rest of the traingles be $Q$, both isosceles. Also let $a, b, c$ denote the number of vertices of each color. All are odd. Note that $P+Q =\tbinom{n}{2}.$ To see this note that each pair of every pair of vertices form the base of an isosceles triangle, not necessarily the ones we found in the first claim. Also, we can have isosceles triangles with $2$ same color vertices, and one other vertice (possibly the same color). There are $\tbinom{a}{2}+\tbinom{b}{2}+\tbinom{c}{2}$ ways to choose to vertices of the same color, and then $\tbinom{3}{2} = 3$ ways to which of the three vertices in the triangle they are. However, note that the triangles that belong to $P$ are counted $3$ times. So, $$3P+Q = 3\left(\binom{a}{2}+\binom{b}{2}+\binom{c}{2}\right).$$ Thus, since $P+Q \equiv Q+3P \pmod2,$ $$\binom n2 \equiv \binom a2 + \binom b2 + \binom c2 \pmod 2 $$$$\implies \equiv a^2+b^2+c^2 - n \pmod 4,$$which is impossible because odd square $\mod 4$ have residue $1$, so the statement would imply that the value is even. $\blacksquare$

Solved with nukelauncher. The \(\gcd(n,6)=1\) condition implies there are no equilateral triangles, and for every chord \(AB\), there is a unique vertex \(C\) with \(CA=CB\). Let there be \(A\), \(B\), \(C\) vertices of the three colors, respectively. Let \(X_A\) be the number of triangles with exactly two \(A\)'s, and \(Y\) the number of monochromatic triangles colored \(A\). Assume for contradiction all isosceles triangles contain at most two distinct colors, so \(X_A+X_B+X_C+Y_A+Y_B+Y_C=\binom n2\). Of the \(\binom A2\) segments with both endpoints colored \(A\), each determines three isosceles triangles: it is the base of one and the leg of two; moreover, for each of these \(3\binom A2\) triangles, at least two vertices are colored \(A\). Double counting, we can verify \[X_A+3Y_A=3\binom A2.\]In particular, \[\binom n2\equiv\binom A2+\binom B2+\binom C2\pmod2.\]Note that for odd \(x\), we have \(\binom x2\equiv\frac{x-1}2\pmod2\), so \[\frac{n-1}2\equiv\frac{A-1}2+\frac{B-1}2+\frac{C-1}2\equiv\frac{n-3}2\pmod2,\]contradiction.

Let $X$ be the number of triangles with all different coloured vertices. Let $Y$ be the number of triangles with exactly 2 vertices of the same colour. Let $a, b, c$ be the number of vertices of the three colours, respectively. Let $D$ be the number of edges with endpoints of different colours that are $\tfrac{n}{3}$ vertices apart (form one side of a possible equilateral triangle). Let's be edgy (pun intended) and only use a $\gcd(n, 2) = 1$ condition. Note that this allows for equilateral triangles to be present. Claim : $X$ is odd. Proof: We proceed by double counting. For each edge with different coloured endpoints, let us count the total possible number of triangles $T$. As each edge has three possible isosceles triangles except for the $D$ edges which only have one, we obtain the total $$T = 3(ab+bc+ca)-2D$$where we subtract to account for the $2D$ triangles being overcounted. However, observe that each of the $X$ all-differently coloured triangles has 3 edges with different-coloured endpoints, so they are triple-counted by $T$. Also, each of the $Y$ triangles with exactly 2 of the same colour have 2 edges with different-coloured endpoints, so they are double-counted by $T$. Thus we get $$T = 3X+2Y$$ Thus, equating both results, we obtain \begin{align*} 3X+2Y &= 3(ab+bc+ca)-2D\\ 3X &\equiv 3(ab+bc+ca) \pmod{2}\\ X &\equiv ab+bc+ca \pmod{2}\\ X &\equiv 1 \pmod{2} \end{align*}So $X$ is odd, as desired. $\blacksquare$ As $X$ is odd, it clearly cannot be 0.

rkm0959 wrote: Korea TST #1, but we only had $n \ge 7$ with $n$ being odd. I think that's enough for this ,we don't need to know gcd(n,6)=1 we just need to know that n is an odd number

cjquines0 wrote: Let $n$ be a positive integer relatively prime to $6$. We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours. Surprising! Let us say that there does not exist an isosceles triangle whose vertices are of three different colors. Let $a, b, c$ be the number of vertices colored in the colors $C_1, C_2$ and $C_3$ respectively. Let $A$ be the number of monochromatic isosceles triangles and let $B$ be the number of isosceles triangles colored in exactly $2$ colors. Then $A + B = \dbinom{n}{2}$ (observe that since $n$ is odd, an isosceles triangle is uniquely determine by its base) and $\dfrac{3A + B}{3} = \dbinom{a}{2} + \dbinom{b}{2} + \dbinom{c}{2}$ because exactly three isosceles triangles use a given edge. Therefore, $2 \mid \dbinom{n}{2} - \dbinom{a}{2} - \dbinom{b}{2} - \dbinom{c}{2} \implies 4 \mid n^2 - a^2 - b^2 - c^2$ which is impossible since $n, a, b$ and $c$ are all odd.

Suppose otherwise. Let $m$ be the number of triangles with all three vertices of the same color and $n$ denote the number of triangles with only 2 colors for all its vertices. Let $2a+1, 2b+1, 2c+1$ denote the number of vertices of each color. We then count the number of edges with the same colored endpoints in two ways. Namely, for each monochromatic triangle, there are 3 such edges and for each 2-colored triangle there is one such edge. Moreover, since $\gcd(n,6) = 1,$ each edge is part of exactly 3 triangles. Hence, $$3m+n = 3\left(\binom{2a+1}{2} + \binom{2b+1}{2}+\binom{2c+1}{2}\right).$$Additionally, by assumption $m+n = \binom n 3.$ Letting $k = a+b+c,$ expanding we have $$\binom{2k+3}{3} = (2k+3)(k+1)(2k+1)\equiv 2(a^2+b^2+c^2) + k \pmod 2,$$However, it is easy to verify that for both odd and even $k,$ this cannot be true. Contradiction. $\blacksquare$

Very noice ! FTSOC assume that such an isosceles triangle doesn’t exist. Let there be $m$ instances of monochromatic isosceles triangles, and $n$ instances where the triangle has two vertices of the same color and one different. We double count on $S=(\text{isosceles triangles, monochromatic edges})$. Let their be $a,b,c$ vertices of the respective Colors. Fixing the triangles, we see that $S = 3m + n$. Fixing monochromatic edges, we get $S= 3( \sum{\binom{a}{2}})$ because each edge is involved in $3$ isosceles triangles. Hence $3m+n = 3( \sum{\binom{a}{2}})$ We all see that since $(n,6)=1$, every edge is the base of exactly one isosceles triangle, so in total we have $m+n=\binom{a+b+c}{2}$ isosceles triangles. After this, mod 2 brings a contradiction

Not too bad. FTSOC, assume none exists. Let the three colors be $r,g,b$, and define a colorful triangle as a triangle with 3-colors as its vertices. If a vertex $v$, with color $r$, does not have a colorful isosceles triangle with $v$ as it's head, then, there must be an odd number of pairs of vertices $A, B$ such that $A$ is red, $B$ is green, and $vA = vB$ (otherwise, there must clearly be a green-blue pair $A',B'$ such that $vA' = vB'$). Now, consider all red-green pairs of points; there are an odd number of them. For each pair $(A,B)$, there is exactly one point $C$ such that $CA = CB$ (since $n$ is odd). If $C$ was blue, we're done, so assume $C$ is red or green. Then, there are either an even amount of $(A,B)$ pairs such that $C$ is green or an even amount of $(A,B)$ pairs such that $C$ is red; wlog let there be even amount such that $C$ is red. Now, for each red vertex $v$, there are an odd number of such $v$s, and each $v$ has an odd number of red-green pairs $(E,F)$ such that $vE = vF$. However, this means there are an odd number of triples $(E,F,v)$ such that $vE = vF$, $(E,F)$ is a red-green pair, and $v$ is red. Contradiction. We conclude there exists an isosceles triangle with three distinct colors.

Connect every diagonal. Let there be $a$ of the first color, $b$ of the second, and $c$ of the third. Then, let $x$ be the number of isosceles triangles with all three colors the same, let $y$ be the number of isosceles triangles with two colors the same and the last one different. Suppose there are no isosceles triangle with all different colors. $~$ Consider any edge. Note that there are three isosceles triangles with this as an edge. With this edge as the base, there is only one, and with it as a leg, there are two that don't overlap due to the lack of equilaterals. Thus, consider all the edges that connects two vertices of the same color, and every monochromatic triangle will be counted three times and every non-monochromatic triangle will be counted once, so \[x+3y=3\binom{a}{2}+3\binom{b}{2}+3\binom{c}{2}.\]However, we also have $x+y=\binom{a+b+c}{2}.$ Now, $4y=n^2-n+3(a^2-a)+3(b^2-b)+3(c^2-c)=n^2-4n+3(a^2+b^2+c^2).$ Note that we can calculate the RHS mod $4$ to be $1-0+3(3)\equiv 2\pmod 4$ so that's contradiction.

.

The key idea is to prove that there are an odd number of tricolored isosceles triangles. Note that $\gcd(n,6)$ implies that every chord of the circle lies on exactly three different isosceles triangles, acting as the base once and each side once. Suppose there are $a,b,c$ vertices of each color, so $a+b+c=n$. There are $\tbinom{n}{2}$ isosceles triangles in total. If we "invalidate" an isosceles triangle once per monochromatic edge it has, every non-tricolored isosceles triangle is invalidated either once or three times, both of which are odd, so the number of tricolored isosceles triangles has the same parity as $$\binom{n}{2}-\binom{a}{2}-\binom{b}{2}-\binom{c}{2}.$$Now, taking cases on how many of $a,b,c$ are $1 \pmod{3}$, we find that the above quantity is indeed always odd, as advertised. $\blacksquare$

Let $\Delta_k$ denotes number of isosceles triangles with $k$ different colors of vertices. It's suffice to prove that $\Delta_3$ is odd. Let $a,b,c$ denote numbers of vertices of each color. Since $\gcd (n,6)=1$ every isosceles triangle is uniquely defined by it's base, implying $\Delta_1+\Delta_2+\Delta_3=\binom n2.$ Moreover, every segment is edge of exactly three different isosceles triangles, so we can doubly count pairs of one-colored vertices as $$\Delta_1+3\Delta_2=3\left( \binom a2 +\binom b2 +\binom c2 \right) \implies \Delta_1+\Delta_2\equiv \binom a2 +\binom b2 +\binom c2 \pmod 2.$$Therefore $\Delta_3\equiv \binom n2-\binom a2-\binom b2-\binom c2=ab+bc+ca\equiv 1 \pmod 2$ $\blacksquare$

Let $a, b, c$ be the number of red, blue, and black vertices, respectively. Assume for the sake of the contradiction that the conclusion is not true. The key is double count pairs $(e, \Delta)$ where $\Delta$ is an isosceles triangle that contains an edge $e$ between two vertices of different colors. On one hand, the number of isosceles triangles that contain $e$ is $3$ (two triangles that contain $e$ as a leg, and one triangle that contains it as a base.) However, for all isosceles triangles which are not distinctly colored, the number of such edges $e$ is even. Then, $3(ab+bc+ca)$ is even, but this is an obvious contradiction because $a, b, c$ are all odd.

Let, there be $r$ red vertices, $b$ blue vertices, and $g$ green vertices. Now FTSOC let's assume that there aren't any isosceles triangles with all vertices of different color. Notice that all isosceles triangles are uniquely determined by the bases. Now we will count the number of pairs of vertices $(A, B)$ s.t. $A, B$ are of different colors in two ways to show a contradiction. and this is the same as counting the number of isosceles triangles with different colors on it's base. Let $R$ be a red vertex. Consider pair of vertices $(A, B)$ s.t. $RA = RB$. We are interested in counting pairs with exactly one red vertex. Now, $r-1$ is even and there are $r-1$ red vertices other than $R$ they all will be present in a pair of two red vertices or in a pair of exactly one red and one other vertex. So, for $R$ we get an even number of pair $(A, B)$ with exactly one red vertex. So we are getting an even number of triangles with $R$ as it's top vertex and the colors on the base different. Summing this for all vertex gives that the total number of triangles with different colour on it's base is even. But clearly the number of pair with different color is $rg+gb+br$ which is odd. So this gives the contradiction. $\blacksquare$

Assume otherwise. Let there be $r$ red vertices and $g$ green vertices. Let $T$ be the set of isosceles triangles containing a red and green vertex. We double count the following: $$\sum_{t\in T}f(t)$$Where $f(t)$ is the number of edges of $t$ that are between a red and green vertex. By the assumption, this is $2|T|$. However, we can also write this sum as $$\sum_{\text{red-green edges}}\#(\text{Triangles containing edge})$$Since there are exactly three isosceles triangles containing a given edge, this is $3rg$. Thus $3rg=2|T|$ which is a parity contradiction.

For the sake of contradiction suppose not. Let there be $a$ red vertices, $b$ blue vertices, and $c$ vertices triangles. Let $S$ be the number of monochrome triangles and $T$ the number of triangles with exactly two vertices of the same color. There are a total of $\binom{n}{2}$ isoceles triangles. Then, it follows by double counting on the number of triangles and monochrome edges with triangle that \begin{align*} S + T &= \binom{n}{2} \\ S + 3T &= 3\left(\binom{a}{2} + \binom{b}{2} + \binom{c}{2}\right) \end{align*}This implies that \[ \binom{a+b+c}{2} \equiv \binom{a}{2} + \binom{b}{2} + \binom{c}{2} \pmod{2} \]which is equivalent to $2 \mid ab + bc + ca$, contradiction.

Let there be $a$, $b$, and $c$ vertices colored red, blue, and black, respectively, so that $a$, $b$, $c$ are odd with $a+b+c=n$. The total number of isosceles triangles induced by the vertices is $\tbinom{n}{2}$. Note that by the $\gcd(n, 6)=1$ restriction, all isosceles triangles are either colored with $2$ or $3$ distinct colors, and there are $\tbinom{a}{2}$ triangles with exactly $2$ colors being red, and analogous counts hold for the other two color. Thus the number of triangles with $3$ distinct colors is \[ M = \binom{n}{2}-\binom{a}{2}-\binom{b}{2}-\binom{c}{2}. \]We instantly kill the problem by showing that $M$ is odd. Note that $M$ rewrites as \[ M = ab+bc+ac \equiv 1 \pmod{2}, \]as desired.

We count the number of triples $(T,v_1,v_2)$ such that $v_1v_2$ is an edge which is a part of an isosceles triangle $T$ and vertices $v_1$ and $v_2$ are colored different colors. First, each isosceles triangle has exactly 0 (all vertices are of the same color) or 2 (two vertices of the same color and the other of one of the remaining colors). Thus, this count is clearly even. Further, each edge is included in exactly 3 isosceles triangles (every edge is the base side of exactly 1 isosceles triangle and one of the other sides of 2). Now, say we have $r$ red vertices, $b$ blue vertices and $k$ black vertices. Thus, the number of such triples is clearly \[3(rb+bk+kr)\]which is an odd number since all of $r,b$ and $k$ are odd. But this is a very clear contradiction. Thus, there must exist some isosceles triangle with all three vertices of different colors as desired.

Suppose for the sake of contradiction there does not exist an isosceles triangle with all $3$ colors. Let there be $b$ blue vertices, $l$ black vertices, and $r$ red vertices respectively. Suppose there are $M$ monochromatic isosceles triangles and $N$ isosceles triangles with $2$ colors. Then $M+N= \tbinom{n}{2}$, as for each isosceles triangle there are $n$ choices for the vertex and $\tfrac{n-1}{2}$ choices for the base because $\gcd(6,n)=1$ gaurentees there are no equilateral triangles, and $\tfrac{n-1}{2}$ choices for the base for every vertex. Due to counting the number of isosceles triangles with at least two of each color and correcting for overcounting we obtain $3(\tbinom{b}{2}+\tbinom{l}{2}+\tbinom{r}{2})= 3M+N$ Now $$3(b(b-1)+l(l-1)+r(r-1))-n(n-1) = 4M$$$$3(b^2+l^2+r^2)-2(b+l+r)-(b+l+r)^2 \equiv 0 \pmod{4}$$$$2(b^2+l^2+r^2-b-l-r-bl-lr-br) \equiv 0 \pmod{4}$$$$b^2+l^2+r^2 \equiv b+l+r+bl+lr+br \pmod{2}$$$$1^2+1^2+1^2 \equiv 1+1+1+1+1+1 \pmod{2}$$$$1 \equiv 0 \pmod{2}$$Which isn't true, a contradiction therefore such a triangle must exist.

Sketch: Claim 1: For every pair of vertices of distinct colors A, B we have 3 distinct isosceles triangles involving that pair using colors either A or B. Proof: $\frac{a + b}{2}$, $2a - b$, $2b - a$ are distinct modulo $n$ for $n \equiv 1, 5 \pmod{6}$. Claim 2: For every isosceles triangle with exactly two colors it has 2 pairs of distinctly colored vertices => There are 3/2 as many isosceles triangles with exactly two colors as number of pairs of vertices with distinct colors. Since odd number each vertex odd pairs so 3/2 takes it to a fraction (!). Solved with puffypundo

Over all pairs of differently colored vertices, count the number of isosceles triangles having those two as vertices. Each pair of vertices counts exactly $3$ triangles by the gcd condition. However, overall each triangle with three of the same color is counted $0$ times, each triangle with two of one color and one of another is counted $2$ times, and each triangle with three distinct colors is counted $3$ times. Taking everything $\pmod 2$, we get that there are an odd number of isosceles triangles whose vertices have different colors, so there is at least one.

Assume this isn't the case. Let there be $a$ red vertices, $b$ blue vertices and $c$ black vertices. We count the number of pairs (bicolor isosceles triangle, heterochromatic edge in that triangle). On the one hand, since each heterochromatic edge is joined to three such triangles (because of no tricolour triangles), this is thrice the number of heterochromatic pairs of vertices, i. e. $3(ab+bc+ca) \equiv 1 \pmod{2}$. But on the other hand, each bicolor triangle has two such heterochromatic edges. So this quantity is also twice the number of bicolour isosceles triangles, which is even. This is a contradiction.

storage

Suppose that $A$, $B$, and 0 are the numbers of isosceles triangles with 1, 2, and 3 distinct colors of vertices, and let the number of vertices colored red, blue, and black be $x$, $y$, and $z$. Using the condition on $n$, we see that each isosceles triangle is fixed by choosing its base, so \[A+B = \binom n2 = \binom{x+y+z}{2}.\] Notice that each edge is part of exactly 3 isosceles triangles. Since the triangles in $A$ contain 3 monochromatic edges while the triangles in $B$ contain just 1, so \[3A+B = 3 \cdot \left(\binom x2 + \binom y2 + \binom z2\right).\] Solving, we get \[2B = 3(xy+yz+zx) = \text{odd},\] contradiction. $\blacksquare$