Let $S$ be the set of all convex cyclic heptagons in the plane. Define a function $f:S \rightarrow \mathbb{R}^+$, such that for any convex cyclic heptagon $ABCDEFG,$ $$f(ABCDEFG)=\frac{AC \cdot BD \cdot CE \cdot DF \cdot EG \cdot FA \cdot GB} {AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FG \cdot GA}. $$ a) Show that for any $M \in S$, $f(M) \geq f(\prod)$, where $\prod$ is a regular heptagon. b) If $f(M)=f(\prod)$, is it true that $M$ is a regular heptagon?
Problem
Source: Serbia IMO TST 2024, P3
Tags: algebra
EthanWYX2009
18.05.2024 17:08
It can be done by doing inversion on $A$ I think.
scannose
18.05.2024 21:15
The problem statement is false. Consider a degenerate heptagon with equal side lengths in the form of a triangle with side lengths $3, 3, 1$, respectively. Represent this degenerate heptagon as $\Delta$; then, $f(\Delta) = 2^4 \cdot \frac{1}{3} \cdot (\frac{\sqrt{60}}{6})^2 = \frac{80}{9} = 8.8888889$.
On the other hand, with the help of a calculator we get $f(\prod) = (2 \cos \frac{\pi}{7})^7 = 61.6848428$.
I was trying to prove this problem with a trig argument, but then arrived at a Jensen's inequality pointing the wrong way which showed that $M = \prod$ actually maximizes $f(M)$ for all heptagons $M$ with equal side lengths.
Edit: I managed to achieve $f(M) = 80$ with another choice of a degenerate heptagon $M$ (square with three midpoints), so it's probably more than just the inequality sign / fraction flipped. nvm it's actually 40
Edit 2: consider a regular hexagon with a corner cut off; the cut off part can be arbitrarily small
fjukh7872
19.05.2024 17:32
In the original statement the heptagon is inscribed.
gvole
19.05.2024 19:52
Statement on heptagons doesn't imply statement on hexagons. Title misleading.
gvole
19.05.2024 19:58
Now for real, the function is a product of cross-ratios so it's preserved under projectivity. This kills (b). For (a), the local minimum of the function under movement of one vertex at a time is when all the lines $A_iA_i$, $A_{i-1}A_{i+1}$,$A_{i-2}A_{i+2}$ are concurrent. It follows by Pascal that $A_{i-3}A_{i+3}$ also passes through this point and that all 7 such points of concurrency are collinear (as is expected from the regular heptagon case). Now a homography fixing the circumcircle and sending these intersection to infinity gives us a regular heptagon so all local minima are equivalent to the regular heptagon.
StefanSebez
19.05.2024 21:40
gvole wrote: Now for real, the function is a product of cross-ratios so it's preserved under projectivity. This kills (b). For (a), the local minimum of the function under movement of one vertex at a time is when all the lines $A_iA_i$, $A_{i-1}A_{i+1}$,$A_{i-2}A_{i+2}$ are concurrent. It follows by Pascal that $A_{i-3}A_{i+3}$ also passes through this point and that all 7 such points of concurrency are collinear (as is expected from the regular heptagon case). Now a homography fixing the circumcircle and sending these intersection to infinity gives us a regular heptagon so all local minima are equivalent to the regular heptagon. You forgot to mention who came up with the idea for local minimum. Also im pretty sure this part (a) solution is worth 0 points.