Library / /26 of 46

, 2015

Geometry Difficulty 6.3 National Olympiad Prove it Japan

Five points $A$ , $B$ , $C$ , $D$ , $E$ , $P$ lie on a plane. Points $A$ , $B$ , $C$ , $D$ lie on a straight line in this order. Furthermore, it is known that $AB = BC = CD = 6$ , $PB = 5$ and $PC = 7$ are satisfied. Here we denote the length of the line segment $XY$ also by $XY$ . Let $Q$ be the point of intersection, different from $P$ , of the circumcircle of the triangle $AC$ and the circumcircle of the triangle $PBD$ . Determine the length of the line segment $PQ$ .

Solution

$\frac{55\sqrt{7}}{14}$

Let $M$ be the point of intersection of line segments $BC$ and $PQ$ . By using the well-known theorem on the power of a point with respect to a circle for the point $M$ and the two given circles, we get $AM \cdot CM = PM \cdot QM = DM \cdot BM$ . Combining this with the fact $AM + CM = BM + DM = 12$ , we obtain $(12 - CM) \cdot CM = (12 - BM) \cdot BM$ , which reduces to
$(BM - CM)(BM + CM - 12) = 0.$
Since $BM + CM = BC \neq 12$ , we get $BM = CM$ , i.e., $M$ is the mid-point of the line segment $BC$ . Therefore, we have
$PM = \sqrt{\frac{PB^2 + PC^2}{2} - BM^2} = \sqrt{\frac{5^2 + 7^2}{2} - 3^2} = 2\sqrt{7},$
from which we obtain
$QM = \frac{AM \cdot CM}{PM} = \frac{9 \cdot 3}{2\sqrt{7}} = \frac{27\sqrt{7}}{14}.$
Consequently, we obtain
$PQ = PM + QM = 2\sqrt{7} + \frac{27\sqrt{7}}{14} = \frac{55\sqrt{7}}{14}.$

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 reproduced verbatim; metadata (topic, difficulty) added by this project.