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Geometry Difficulty 6.0 National olympiad Prove it

Let Γ\Gamma be a circle and A,B,C,D,EA, B, C, D, E and FF be points on Γ\Gamma. Let P,QP, Q and RR be the points of intersection of (AB)(A B) and (DE)(D E), (BC)(B C) and (EF)(E F), and (CD)(C D) and (FA)(F A), respectively.

Show Pascal's theorem, i.e., that P,QP, Q and RR are collinear.

Solution

Let Γ\Gamma be the unit circle. Let x=af−ab+de−cd+bc−efx=a f-a b+d e-c d+b c-e f. We can use the formula from the previous exercise to calculate the affix of pˉ\bar{p}: we know that p=a+b−abpˉp=a+b-a b \bar{p} and p=d+e−depˉp=d+e-d e \bar{p}, so pˉ=a+b−d−eab−de\bar{p}=\frac{a+b-d-e}{a b-d e}. We proceed similarly for the other points. Thus:

pˉ−qˉ=a+b−d−eab−de−b+c−e−fbc−ef=(b−e)x(ab−de)(bc−ef) \bar{p}-\bar{q}=\frac{a+b-d-e}{a b-d e}-\frac{b+c-e-f}{b c-e f}=\frac{(b-e) x}{(a b-d e)(b c-e f)}

By conjugating this equation, we obtain that:

p−q=(1b−1e)xˉ(1ab−1de)(1bc−1ef)=−abcdef(b−e)xˉ(ab−de)(bc−ef) p-q=\frac{\left(\frac{1}{b}-\frac{1}{e}\right) \bar{x}}{\left(\frac{1}{a b}-\frac{1}{d e}\right)\left(\frac{1}{b c}-\frac{1}{e f}\right)}=-a b c d e f \frac{(b-e) \bar{x}}{(a b-d e)(b c-e f)}

Thus, we obtain that:

p−qpˉ−qˉ=−abcdefxˉx \frac{p-q}{\bar{p}-\bar{q}}=-\frac{a b c d e f \bar{x}}{x}

By cyclic permutation, we obtain that:

q−rqˉ−rˉ=−abcdef(−xˉ)−x=−abcdefxˉx \frac{q-r}{\bar{q}-\bar{r}}=-\frac{a b c d e f(-\bar{x})}{-x}=-\frac{a b c d e f \bar{x}}{x}

Thus:

p−qpˉ−qˉ=q−rqˉ−rˉ \frac{p-q}{\bar{p}-\bar{q}}=\frac{q-r}{\bar{q}-\bar{r}}

This is equivalent to p−qqˉ−rˉ∈R\frac{p-q}{\bar{q}-\bar{r}} \in \mathbb{R}, which means that P,QP, Q and RR are collinear.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.