Track / Stage 6 / 101 of 400 #1101 of 1964

Problem 1101

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Japan Mathematical Olympiad Initial Round · Japan

On the circumference of a circle, 6 points $A$ , $B$ , $C$ , $D$ , $E$ , $F$ are placed in this order counter-clockwise, and three lines $AD$ , $BE$ and $CF$ intersect at a single point. If
$AB = 1,\ BC = 2,\ CD = 3,\ DE = 4,\ EF = 5,$
find the value of $FA$ . Here we denote the length of a line segment $XY$ also by $XY$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$\boxed{\frac{15}{8}}$
Let $P$ be the point of intersection of lines $AD$ , $BE$ and $CF$ . Then, we have $\angle PBA = \angle PDE$ , since they are subtended by the same arc $\text{EA}$ of the circle at the points $B$ and $D$ on the circumference. We also have $\angle BPA = \angle DPE$ so that the triangles $BPA$ and $DPE$ are similar. Consequently, we get
$(1) \qquad PA : PE = BA : DE = 1 : 4.$
In the same way, we see that the triangles $CPB$ and $EPF$ are similar, and therefore, we get
$(2) \qquad PE : PC = EF : CB = 5 : 2.$
From (1) and (2) above, we get $PC : PA = 8 : 5$ . Also, from the similarity of the triangles $FPA$ and $DPC$ , we obtain $PC : PA = DC : FA$ , from which it follows that $FA = \frac{5}{8} \cdot DC = \frac{15}{8}$ , which is the desired answer.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.