Consider a $n$-sided regular polygon, $n \geq 4$, and let $V$ be a subset of $r$ vertices of the polygon. Show that if $r(r-3) \geq n$, then there exist at least two congruent triangles whose vertices belong to $V$.
Problem
Source: 2021 Iberoamerican Mathematical Olympiad, P6
Tags: geometry, congruent triangles
Inconsistent
21.10.2021 03:35
Please let me know if this is completely wrong but... assume otherwise, then if two edges in V share side length but not a vertex, let the convex hull of the four vertices be trapezoid ABCD where AB = CD, then ABC is congruent to BCD. So the two edges must share a vertex in $V$. In particular they must be reflections of one another across the line from that vertex to the center of the regular polygon, so there cannot be three equal length edges sharing that vertex. And if two pairs of equal length edges share a vertex, then let the isoceles triangles be ABC, ADE, then ABD is congruent to ACE. Therefore the total number of side lengths that can be taken by edges in $V$ is ${r \choose 2}-r < \frac{n}{2}$ so $r(r-3) < n$. By contrapositive we finish. This solution feels VERY sus since I never use the fact that two congruent triangles can refer to triangles with the same orientation... also it's very short it seems, but I don't see the issue at the moment. EDIT: Oops I probably overestimated the difficulty of the MO, just ignore the remark I suppose.
NTstrucker
21.10.2021 20:49
Be careful: the inequality should be $\tbinom{r}{2}-r \le \lfloor \tfrac{n}{2} \rfloor$, hence more discussion is needed. To be precise, since $r(r-3) \le n$, the equality must attain in the above inequality, which implies that: $n$ is even. Each vertex in $V$ is the common endpoint of exactly one pair equal length edges. Diagonals of length $1,2,\ldots,\tfrac{n}{2}$ (measured in "arc-length") all appear. By (3), there exists an edge $AB$ which is the main diagonal (of length $\tfrac{n}{2}$) of the polygon. But (2) reveals that there are two two equal length edges with common endpoint $A$, say $AC$ and $AD$. Thus $ACB$ and $ADB$ are congruent, which is absurd. The proof ends (finally) here.
PedroBigMan
31.10.2021 22:33
I believe R(R-2) suffices
SnowPanda
15.01.2022 08:30
Draw the $\tfrac{r(r - 1)}{2}$ segments between distinct points in $V$, and label each segment with its length. There are at most $\tfrac{n}{2}$ distinct lengths used.
Claim: If two segments have the same length, they must share a vertex.
Proof. Assume not, so $u_1v_1$ and $u_2v_2$ have the same length for pairwise distinct $u_i$ and $v_i$.
[asy][asy]
unitsize(2cm);
draw(circle((0, 0), 1));
pair U1 = dir(5);
pair U2 = dir(50);
pair V1 = -1/U2;
pair V2 = -1/U1;
dot("$u_1$", U1, dir(U1));
dot("$u_2$", U2, dir(U2));
dot("$v_1$", V1, dir(V1));
dot("$v_2$", V2, dir(V2));
draw(U1--U2--V1--cycle, lightblue);
draw(V2--V1--U2--cycle, lightblue);
[/asy][/asy]
Then the triangles $u_1u_2v_1$ and $v_2v_1u_2$ are congruent. $\blacksquare$
Call a triple $(v_1, u, v_2)$ of distinct vertices a carrot if $v_1u$ and $uv_2$ have the same length, and call $u$ its head.
Claim: Each vertex is the head of at most one carrot.
Proof. Assume not, so $(v_1, u, v_2)$ and $(w_1, u, w_2)$ are carrots. Then $v_1w_1$ and $v_2w_2$ have the same length, contradicting the first claim. $\blacksquare$
So there are at most $r$ carrots. If there are $k_i$ segments with length $\ell_i$, then there are at least $\tfrac{k_i(k_i - 1)}{2} \geq k_i - 1$ carrots with length $\ell_i$. So then $\sum k_i - 1 \leq r$, and there are at least \[\frac{r(r - 1)}{2} - r = \frac{r(r - 3)}{2} \geq \frac{n}{2}\]distinct lengths. Then equality must hold, so two points $u$ and $v$ need to be diametrically opposite (in order to get $\tfrac{n}{2}$ lengths), and $u$ must be the head of a carrot $(w_1, u, w_2)$ (in order to get $r$ carrots). But then $w_1uv$ and $w_2uv$ are congruent, contradiction.