Track / Stage 5 / 340 of 400 #940 of 1964

Problem 940

AIME late

Geometry Difficulty 5.9 Prove it

Example 4.2.4 From the center $O$ of circle $O$ , draw a perpendicular to line $M N$ , with the foot of the perpendicular being $P$ . From $P$ , draw two secants, intersecting circle $O$ at points $A, B; D, C$ . Let $A C \cap M N=E, B D \cap M N=F$ . Prove: $P$ is the midpoint of $E F$ .

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

Analysis This is a typical scenario of pole and polar in advanced geometry!
Proof Let $A D \cap B C=J, A C \cap B D=I$ .
At this time, by the complete quadrilateral $A B C D I J$ , we know that $I J$ is the polar of $P$ , $I J \perp O P$ [the polar of a point outside a circle with respect to the circle is the line connecting the points of tangency of the two tangents drawn from the point to the circle], $\therefore I J / / E F$ .
By the fundamental theorem, $ID$ , $IP$ , $IA$ , $IJ$ form a harmonic pencil,
i.e., $(I D, I A, I P, I J)=-1$ .
And $(F E P \infty) \bar{\Lambda} I(D A P J)$ ,
$\therefore(F E, P \infty)=-1$ , i.e., $P$ is the midpoint of $E F$ .

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