Intersection point on circle

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Intersection point on circle

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Source: VII Caucasus Olympiad, Senior, Day1 P4.

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#1 h Mar 13, 2022, 7:37 PM • 1 Y

Y by farhad.fritl

Let $\omega$ is tangent to the sides of an acute angle with vertex $A$ at points $B$ and $C$ . Let $D$ be an arbitrary point onn the major arc $BC$ of the circle $\omega$ . Points $E$ and $F$ are chosen inside the angle $DAC$ so that quadrilaterals $ABDF$ and $ACED$ are inscribed and the points $A,E,F$ lie on the same straight line. Prove that lines $BE$ and $CF$ intersectat $\omega$ .

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#2 h Mar 13, 2022, 7:37 PM • 1 Y

Y by farhad.fritl

Again easy P4 (Like lsast year's P4). But anyway, it's better than last year's P4.
Invert around circle $(A,AB)$ and assume $X'$ is image of $X$ under this inversion. We get $D'=AD\cap \omega, E'=AE\cap CD', F'=AF\cap BD'$ and let $T=BE\cap CF$ .
Since $AE\cdot AE'=AC^2$ we get $(CEE')$ tangents to $AC \implies \angle E'EC=\angle E'CA=\angle D'BC \implies BCEF'$ is cyclic. Similarly $BCFE'$ is cyclic. So $\angle TEF=\angle BEF'=\angle BCF'$ and $\angle F'FC=\angle F'CA \implies \angle BTC=\angle TEF + \angle EFT = \angle BCF'+\angle F'CA=\angle BCA \implies T\in \omega$ . So we are done!

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1388 posts

#3 h Mar 13, 2022, 7:37 PM

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Well, almost trivial. Note that there is a spiral similarity taking $DEF$ to $DBC$ (more formally, we have that triangles $DBE$ and $DCF$ are also similar, so if $BE$ intersects $CF$ at $P$ , then $BCDP$ is cyclic).

This post has been edited 1 time. Last edited by VicKmath7, Mar 13, 2022, 7:41 PM

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#4 h Apr 2, 2022, 10:41 AM

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Easy for P4...
$\angle DBC = \angle DCF = \angle 180 - \angle ACD = \angle 180 - \angle AED = \angle DEF$ and $\angle BCD = \angle 180 - \angle ABD = \angle EFD$ so $DBC$ and $DEF$ have spiral similarity. Let $BE$ meet $CF$ at $X$ . It's well known that $DEXF$ and $DBCX$ are cyclic so $X$ lies on $\omega$ .

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#5 h May 19, 2024, 3:32 PM

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Equivalent, but without the use of spiral similarity basic trickery (I know it is easier to figure out with it, but just for the sake of inexperienced people).

We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$ ,
similarly $\angle AED = \angle ACD = \angle DBC$ . Hence $\triangle DBC \sim \triangle DEF$ . Hence $\frac{BD}{DC} = \frac{DE}{DF}$ , i.e. $\frac{BD}{DE} = \frac{DC}{DF}$ and also $\angle BDC = \angle EDF$ , i.e. $\angle BDE = \angle CDF$ . Hence $\triangle BDE \sim \triangle CDF$ , so $\angle DBE = \angle DCF$ , thus after extending $BE$ and $CF$ to intersect $\omega$ , the respective arcs would be equal and so the intersection points would coincide, done.

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1239 posts

#6 h Jun 1, 2024, 11:54 PM

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Plot twist: only angle chasing suffices! (Credits to @africanboy)

We have $\angle DFE = 180^{\circ} - \angle AFD = \angle ABD = \angle BCD$ from the cyclic $ABDF$ and the tangency around $\omega$ , similarly $\angle AED = \angle ACD = \angle DBC$ and hence $\angle BDC = \angle EDF$ . Now define $T = BE \cap \omega$ . Then $\angle DTE = \angle BTD = \angle BCD = \angle DFE$ , hence $DFTE$ is cyclic. From here $\angle ETF + \angle CTE = (180^{\circ} - \angle FDE) + \angle BTC = (180^{\circ} - \angle BDC) + \angle BTC = 180^{\circ}$ , so $CE$ passes through $T$ and we are done.

This post has been edited 1 time. Last edited by Assassino9931, Jun 1, 2024, 11:55 PM

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