China Western Math Olympiad 2006: Multiple Solutions

Tags: Projective Geometry, Pascal’s Theorem, La Hire’s Theorem, Harmonic Division, Ceva’s Theorem, Manelaus’s Theorem, Co-ordinate Geometry

Problem Description: As in the figure above, $AB$ is a diameter of a circle with center $O$. $C$ is a point on $AB$ extended. A line through $C$ cuts the circle with center $O$ at $D, E. OF$ is a diameter of the circumcircle of $\triangle BOD$ with center $O_{1}$. Line $CF$ intersect the circumcircle again at $G$. Prove that $O,A,E,G$ are concyclic.

Solution 1 ( Due to WONG Chiu Wai ):
Lemma: Let $AE \cap BD = P$ and $AD \cap BE = H$. We will prove that, the polar of point $P$ is through $C$ and $H$. If so, then the extension of $CH$ is perpendicular to $OP$.
From La Hire’s Theorem, we know that, if $X$ lies on the polar of $Y$, then $Y$ too lies on the polar of $X$.
In the above diagram, let the tangents at $A$ and $E$ to the larger circle intersect at $X$ and let the tangents at $B$ and $D$ to the larger circle intersect at $Y$. Since, $AE$ is the polar of $X$ and $BD$ is the polar of $Y$, so the polar of $AE \cap BD = P$ goes through $X$ and $Y$.
Using Pascal’s Theorem on hexagon $BAADEE$, we find that $C = BA \cap CE$ and $H = AD \cap BE$ and $X = EE \cap AA$ are collinear.
Using Pascal’s Theorem on hexagon $ABBEDD$, we find that $C = AB \cap DE$ and $H = AD \cap BE$ and $Y = BB \cap DD$ are collinear.
So, it turns out that $C, Y, H, X$ are all collinear. So, definitely, $CH$ is the polar of $P$.

Let $OP \cap CH = Q$. We will prove that $Q = G$. To show $Q = G$, note that $\angle PQH, \angle PDH$ and $\angle PEH$ are $90°$, which implies $P, E, Q, H, D$ are concyclic. Then $\angle PQD = ∠ PED = ∠ DBO$, which implies $Q, D, B, O$ are concyclic. Therefore, Q = G since they are both the point of intersection (other than $O$) of the circumcircle of $\triangle BOD$ and the circle with diameter $OC$.

Using power of point $P$ with respect to circle $GOBD$, we get, $PG.PO = PD.PB$ and using power of point $P$ with respect to circle $DBAE$, we get, $PD.PB = PE.PA$. Since, $PE.PA = PD.PB = PG.PO$, we can infer that $G, O, A, E$ are concyclic.

Solution 2 ( Due to Me ): My solution is almost analogous to the one above. Except I have my personal proof for the Lemma. Let’s simply the diagram as below. We just need to prove that $AO \perp IJ$.

We will go for a hybrid solution. This is why it’s my favorite. It’s a solution where Euclidean geometry meets Cartesian coordinate geometry.
Let’s define the coordinates of the points in the following way:
$ D = (0, 0)$, $ A = (0, a)$, $ B = (b, 0)$, $ C = (c, 0), H = (0, h), J = (j, 0)$
From here we can deduce that, $ O = (\frac{b+c}{2}, 0)$. Now, let’s find out the coordinates of $H$ and $J$. One would be tempted to solve equations of the lines $BC$ and $EF$ or $BE$ and $CF$. But that would be very cumbersome. Instead, remember that, if we reflect $H$ across $BC$, it’s reflection $H’$ falls on the circumcircle of $\triangle ABC$. So using the power of point $D$,

$|AD| \times |DH'| = |BD| \times |CD|$

Therefore,

$|DH| = |DH'| = \frac{|BD| \times |CD|}{AD} = \frac{-bc}{a}$

So, $H = (0, \frac{-bc}{a})$
Now, let’s find the coordinates of $J$.
Since the cevians $AD, BE, CF$ are collinear. So using Ceva’s Theorem,

$ \frac{AF}{FB} \frac{BD}{DC} \frac{CE}{EA} = 1$

And since $E, F, J$ are collinear. Using Manelaus’s Theorem,

$ \frac{AF}{FB} \frac{BJ}{JC} \frac{CE}{EA} = -1$

Combining both we get,

$ \frac{BJ}{JC} = - \frac{BD}{DC}$

Which means, $J, B, D, C$ are harmonic conjugates.

$ \frac{j-b}{c-j} = - \frac{d-b}{c-d}$

Plugging in $d = 0$ and solving for $j$ yeilds that, $j = \frac{2bc}{b+c}$.
We calculate the slope of $OA$ to be $m_{1} = \frac{a-0}{0 - \frac{b+c}{2}} = - \frac{2a}{b+c}$.
We calculate the slope of $IJ$ to be $m_{2} = \frac{0-\frac{-bc}{a}}{\frac{2bc}{b+c} - 0} = \frac{b+c}{2a}$.
And since, $m_{1} \times m_{2} = -1$, $OA$ and $IJ$ are perpendicular.