# Problem $5$ of $2011$ USAJMO

This is a problem from $$2011$$ USAJMO

Points $$A$$, $$B$$, $$C$$, $$D$$, $$E$$ lie on a circle $$\omega$$ and point $$P$$ lies outside the circle. The given points are such that (i) lines $$PB$$ and $$PD$$ are tangent to $$\omega$$, (ii) $$P$$, $$A$$, $$C$$ are collinear, and (iii) $$\overline{DE} \parallel \overline{AC}$$. Prove that $$\overline{BE}$$ bisects $$\overline{AC}$$.

I proved it by showing that $$OFBP$$ is cyclic. Then I thought of another way. Let the midpoint of $$AC$$ be $$F$$, and the line $$BF$$ beyond $$F$$ intersects the circle at $$E'$$. If I can show that $$\overline{DE'} \parallel \overline{AC}$$, then it will prove $$E'$$ = $$E$$.

To prove $$\overline{DE'} \parallel \overline{AC}$$, we need to show that $$\angle{FE'D} = \angle{E'FC}$$. But $$\angle{FE'D} = \angle{FE'A} + \angle{AE'D} = \angle{FCB} + \angle{AE'D}$$. If I can show that $$\angle{AE'D} = \angle{FBC}$$ , it will imply that $$\angle{FE'D} = \angle{FCB} + \angle{FBC} = \angle{BFA} = \angle{E'FC}$$ . But I am stuck at showing $$\angle{AE'D} = \angle{FBC}$$ , how can i do that ?

• Back in my math contest days, I always resorted to coordinate geometry for geometry question. Neat question, though. Aug 5 at 19:30
• Does it help to note that tangents are perpendicular to radii? Aug 5 at 21:25
• You can also see solutions here artofproblemsolving.com/community/c5h404355p2254813 Aug 5 at 23:22

This technique that you used of first defining $$F$$ to be the midpoint of $$AC$$ is known in the math olympiad community as phantom points (if you didn't know already). To solve the problem using this method, we can sort of reconstruct the old solution.

Claim: $$(BFODP)$$ are all concyclic.

Proof. First, we have $$\angle OBP = \angle ODP = 90$$ using the tangents, so $$O,B,D,P$$ all lie on a circle. Furthermore, we have $$\angle OFP = 90 = \angle OBP$$, as it is well-known that the line through the center of a circle and the midpoint of a segment is perpendicular to the segment itself. Therefore, $$O,F,B,P$$ all lie on a circle as well.

Now, we deconstruct the angle chasing of the original solution. Using the cyclic quad above, we have $$\angle BFP = \angle BOP$$. Furthermore, $$\angle BFP = 180 - \angle BFC = \angle BCF + \angle CBF = \angle BCF + \angle CDE'$$ and also $$\angle BOP = \frac{\angle BOD}{2} = \angle BCD$$ which means that, $$\angle BCD = \angle BCF + \angle CDE' \Rightarrow \angle ACD = \angle CDE'$$ which also shows that $$AC \parallel DE'$$, as desired.

(1) By tangent properties, we have PBOD is cyclic.

(2) $$\angle 1 = \angle 3= \angle 4$$ shows PBFD is also cyclic.

That means P, B, F, O, D are con-cyclic points.

Since $$\angle = OFP = \angle OBP = 90^0$$, $$AF = FC$$