Competition problem — Geometry — Maths Olympiad Prep

Library / /1 of 10

Geometry Difficulty 4.5 AIME Prove it Austria

Let $ABCDE$ be a convex pentagon having a circumcircle and satisfying $AB = BD$ . The point $P$ is the intersection of the diagonals $AC$ and $BE$ . The lines $BC$ and $DE$ intersect in point $Q$ .
Show that the line $PQ$ is parallel to the diagonal $AD$ .

Solution

Figure 2: Problem 6

Solution:

We denote the circumcircle of the pentagon $ABCDE$ by $k$ , see Figure 2. By assumption, the triangle $ABD$ is isosceles, which implies that the tangent $t_B$ to $k$ in $B$ is parallel to $AD$ .

We apply Pascal's theorem to the inscribed hexagon $BEDACB$ : The intersection point of the opposite sides $BE$ and $AC$ is $P$ , the intersection point of the opposite sides $ED$ and $CB$ is $Q$ , and the intersection point of the parallel opposite sides $BB$ (i.e., $t_B$ ) and $DA$ is the point at infinity corresponding to direction $AD$ . Therefore, $PQ$ is parallel to $AD$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.