PL and QK intersect on circumcircle

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PL and QK intersect on circumcircle

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Source: 2017 Maths Beyond Limits Camp - Olympic Challenge - Older Division - Geometry p3

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#1 h Dec 12, 2020, 11:32 PM

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Let $\omega$ be the circumcircle of $\vartriangle ABC$ . Let the points $D, E$ lie on the sides $AC, AB$ respectively. Let $K = BD\cap \omega$ , $L = CE \cap \omega$ . Tangents to $\omega$ at the points $B, C$ intersect the line $DE$ at the points $P, Q$ respectively. Prove that $PL$ and $QK$ intersect on $\omega$ .

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#2 h Dec 20, 2020, 1:47 PM

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Notice that the problem is purely projective, so the problem still holds after applying a projective transformation.
We want to apply a projective transformation that fixes $\omega$ and makes line $DE$ parallel to $BC$ .
Note that $DE \parallel BC \Leftrightarrow \frac{AD}{DC} = \frac{AE}{EB}$ . If we let $X = BD \cap CE$ and $Y = AX \cap BC$ , then since $AY, BD$ and $CE$ are concurrent cevians of $\triangle ABC$ , by Ceva,
$\frac{AD}{DC} \cdot \frac{CY}{YB} \cdot \frac{BE}{EA} = 1$ and thus $\frac{CY}{YB} = 1$ which gives us $Y$ as the midpoint of $BC$ .
So we can just apply a projective transformation that fixes $\omega$ and sends $Y$ to the center of $\omega$ .
This will make $BC$ as the diameter of $\omega$ and $DE \parallel BC$ .
An angle chase using directed angles $\pmod{\pi}$ gives us
$\measuredangle PBL = \measuredangle BCL = \measuredangle PEL$ and $\measuredangle QCK = \measuredangle CBK = \measuredangle QDK$
Thus $PBEL$ and $QCDK$ are cyclic.
If we let $F = PL \cap \omega$ , we can show that $F, K$ and $Q$ are collinear.
This is just $\measuredangle BKF = \measuredangle BLF = \measuredangle BLP = \measuredangle BEP = \measuredangle EBC = \measuredangle ABC = \measuredangle ACQ = \measuredangle DCQ = \measuredangle DKQ = \measuredangle BKQ$ . And we are done. $\blacksquare$

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#4 h Dec 22, 2020, 7:46 PM

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parmenides51 wrote:

Let $\omega$ be the circumcircle of $\vartriangle ABC$ . Let the points $D, E$ lie on the sides $AC, AB$ respectively. Let $K = BD\cap \omega$ , $L = CE \cap \omega$ . Tangents to $\omega$ at the points $B, C$ intersect the line $DE$ at the points $P, Q$ respectively. Prove that $PL$ and $QK$ intersect on $\omega$ .

If $Z$ is the trilinear pole of the line $DE$ , then $PL\cap QK$ is the $A$ -vertice of cyclocevian triangle of $Z$ wrt anticevian triangle of $Z$ .

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