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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 ?

enter image description here

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

2 Answers 2

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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.

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

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

enter image description here

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

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

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