In an acute triangle $ABC$ with $\overline{AB} < \overline{AC}$, angle bisector of $A$ and perpendicular bisector of $\overline{BC}$ intersect at $D$. Let $P$ be an interior point of triangle $ABC$. Line $CP$ meets the circumcircle of triangle $ABP$ again at $K$. Prove that $B, D, K$ are collinear if and only if $AD$ and $BC$ meet on the circumcircle of triangle $APC$.
Problem
Source: KJMO 2021 P4
Tags: geometry, angle bisector, perpendicular bisector, circumcircle
BVKRB-
13.11.2021 15:06
All you need to do is invert!
Let $AD \cap BC = L$ and let $X'$ denote the inverse of $X$ under $\sqrt{bc}$ inversion followed by reflection across angle bisector
$E$ goes to a random point $E'$ outside the triangle $ABC$ and $B \iff C$ and $D \iff L$
Now $F' = CE' \cap \odot(ABE')$ after inversion
Now all remains is to angle chase
For the if part $$\measuredangle E'BA = \measuredangle AF'C = \measuredangle ALB = \measuredangle DBA \implies B-D-E'$$The only if part is just the reverse of the above angle chase, and so we are done $\blacksquare$
Fakesolver19
13.11.2021 16:04
Let $AD \cap BC=M$ It suffices to show that $\angle APC=\angle AMC$ $\angle APC+\angle APK=\angle APC+\angle ABK=\angle APC+180-(\angle B+\angle \frac{A}{2})=180 \implies \angle APC=\angle B+\angle \frac{A}{2}$ $\angle AMC=\angle BMD=180 -(\angle C+\angle \frac{A}{2})=\angle B+\angle \frac{A}{2}$ and therefore we are done. Edit:-Oops read the question wrong but we can use this chase to prove the other direction also
TheCollatzConjecture
13.11.2021 21:15
Firstly, let $AD$ and $BC$ intersect at $E$.
Claim 1$ABDC$ is cyclic
ProofLet $D'$ be the unique point on the perpendicular bisector of $BC$ such that $D'$ lies on $\odot ABC$. So, $\angle BD'C= 180-A$, and $BD'=D'C$. So $\angle D'BC = \frac{A}{2} = \angle D'BC$. So $D'$ lies on he angle bisector of $A$. Do $D' \equiv D$ and we're done $\square$
Firstly, let us prove that if $B,D,K$ are collinear then $E$ is on $\odot APC$. Firstly $\angle DBA= 180- C-\frac{A}{2}$. So $\angle KBA= \angle KPA =C +\frac{A}{2}$. So $\angle APC= \angle AEC= 180- C-\frac{A}{2}$, so we are done with 'if' part.
Now we prove the 'only if part' . So we need to show that if $E$ lies on $\odot APC$ then $B,D,K$ are collinear. But we can just reverse engineer the angle chase and prove that $\angle BDK=180$ if $APEC$ is cyclic. $\blacksquare$.
REYNA_MAIN
22.12.2021 19:37
REYNA_MAIN wrote: BVKRB- wrote: All you need to do is invert! sar really sar haw so ors max? like i couldnt even imagine inverting this even if i was used to inversion again, pleex gib teepz
Let $AD \cap BC = E$.
if partWe have that $APEC$ is cyclic.
$\measuredangle ABD = \measuredangle AEC = \measuredangle APC = \measuredangle APK = \measuredangle ABK \implies B - D - K$ are collinear.
else if partWe have that $B - D - K$ are collinear.
$\measuredangle ABK = \measuredangle ABD$.
$$\measuredangle ABK = \measuredangle APK = \measuredangle APC$$$$\measuredangle ABD = \measuredangle AEC$$
$\implies \measuredangle APC = \measuredangle AEC \implies APEC$ is cyclic.
and we are done .
Attachments:
Mogmog8
05.03.2022 04:30
[asy][asy] /* Coordinates of P from Geogebra */ size(7cm); import geometry; pair A,B,C,D,E,P,K; A=(-5,8); B=(-7,0); C=(4,0); D=intersectionpoints(circle(A,B,C),bisector(B,C))[0]; E=extension(A,D,B,C); P=(-4.5,1.9); K=intersectionpoints(line(C,P),circle(A,B,P))[1]; draw(A--B--C--cycle); draw(circle(A,B,C)); draw(circle(A,B,P)); draw(circle(A,C,P),dotted); draw(K--D,dashed); draw(A--D); draw(E--P--A); draw(C--K); dot(A^^B^^C^^D^^E^^P^^K); label("$A$",A,N); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,NE); label("$P$",P,S); label("$K$",K,NW); [/asy][/asy] Let $E=\overline{AD}\cap\overline{BC}.$ Notice $B,D,$ and $K$ being collinear is equivalent to \begin{align*}&\measuredangle DBA=\measuredangle ANK\\&\iff\measuredangle DAB+\measuredangle BDA=\measuredangle KPA\\&\iff\measuredangle CPA=\measuredangle CAE+\measuredangle BCA=\measuredangle CEA,\end{align*}which is in turn equivalent to $ACEP$ being cyclic. $\square$
Jishnu4414l
24.10.2023 10:43
We first prove the "only if" part... Claim: $ABCD$ is a cyclic quadrilateral. Proof: We will use phantom points. Suppose $AI$ meets $(ABC)$ at $D'$. Drop a perpendicular from $D'$ Drop a perpendicular from $D'$ onto $BC$ and name the foot of perpendicular as $X$. Due to $Fact-5$, we have $DB=DC$. Now notice that $\triangle BDX \cong \triangle CDX$, from which the conclusion follows. Now let us first proof the "only if" part. Suppose $B,D,K$ are collinear.Now let $\angle BAD = \angle CAD= \angle DCB = \angle CBD =\theta$. Let $\angle BAP= \phi \implies \angle BKP = \phi \implies \angle PAD= \theta-\phi$. Also looking at $\triangle KBC$, we get that $\angle KCB=\theta-\phi$ too.We thus proved that $APXC$ is cyclic. The if part can be proved by a similar method too.(We need $\angle ABD+ \angle ABK =180$)