Let $ABCDE$ be a convex pentagon with $F=BC\cap DE, G=CD\cap EA, H=DE\cap AB, I=EA\cap BC, J=AB\cap CD$, Suppose that the areas of $\triangle AHI, \triangle BIJ, \triangle CJF, \triangle DFG, \triangle EGH$ are all equal. Prove $AF,BG,CH,DI,EJ$ are all concurrent.
A solution I saw to this stated that everything in the problem is preserved by an affine transformation, so it can be assumed without loss of generality that $A,C,D$ are the vertices of a regular pentagon with $AC=AD, \angle CDA=\frac{\pi}{5}$.
I know that an affine transformation maps $(x,y)\rightarrow (ax+by+c,dx+ey+f)$ for real numbers $a,b,c,d,e,f$ with $ae\neq bd$, however could someone explain what makes "everything in the problem preserved by an affine transformation", and why you are allowed to assume $A,C,D$ are three vertices of a regular pentagon? Thanks!
That's from Kedlaya I believe?
First of all, let's mention a useful theorem: "For any two triples of non-collinear points, we can find a unique affine transformation sending the points of the first triple to then corresponding points of the second triple". That's a known theorem.
So, in our case we apply the affine transformation that sends $A,C,D$ (which are non-collinear) to some $A', C', D'$ such that $AC=AD, \angle CDA=\frac{\pi}{5}$. This gives $\angle CAD=\frac{3\pi}{5}$ which is the interior angle of a regular pentagon! So under that specific transformation, $A,C,D$ are sent to three consecuitive vertices of a regular pentagon with side $AC$.
Also, under this transformation every "important" detail of our problem stays invariant. That's because ratios of areas (NOT areas themselves), concurrency and collinearity are preserved by affine transformations. That's another known theorem.
So, if we apply the specific transformation a mentioned before, we end up with our initial problem buy for the special case in which $A,C,D$ are three consecuitive vertices of a regular pentagon.
If we manage somehow to solve the problem for that case then we can apply the inverse transformation, and the conccurency stays invariant. Woala!
Oh, by the way, for the case we have the same source in mind: I believe the given proof is full of typos. E.g. $A,C,D$ here is probably a typo ($A,B,C$ is a possible correction).