Let$ ABCD $be a cyclic quadrilateral, let the tangents to$ A$ and $C$ intersect
in $P$, the tangents to$ B $and$ D$ in$ Q$. Let$ R$ be the intersection of$ AB $and$ CD $and let
$S $be the intersection of$ AD$ and$ BC$. Show that$ P,,R, S $are collinear.
Let$ ABCD $be a cyclic quadrilateral, let the tangents to$ A$ and $C$ intersect
in $P$, the tangents to$ B $and$ D$ in$ Q$. Let$ R$ be the intersection of$ AB $and$ CD $and let
$S $be the intersection of$ AD$ and$ BC$. Show that$ P,,R, S $are collinear.
$\textbf{Solution 55}$
Applying Pascal's theorem on $AABCCD$ we have that $R,P,S$ are collinear. Again applying pascal on $ABBCDD$, we get $R,Q,S$ are collinear.
Therefore, points $P,Q,R,S$ are collinear. $\blacksquare$
Re: Geometry Marathon : Season 3
Posted: Sun Mar 14, 2021 11:12 pm
by ~Aurn0b~
$\textbf{Problem 56}$
We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$.
Re: Geometry Marathon : Season 3
Posted: Mon Mar 15, 2021 5:48 pm
by Asif Hossain
how to attach picture??
Re: Geometry Marathon : Season 3
Posted: Tue Mar 16, 2021 9:27 pm
by Dustan
Click "Full editor and preview " then "attachment "
We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$.
Solution(It would be kind if somebody confirm it ):
$\Rightarrow$
Let the reflection of $D$ $D'$ which is in the circumcircle of $ABC$.Let us denote $D'Q \bigcap PA \equiv M$,$PQ \bigcap AD \equiv N$ and $D'P \bigcap AQ \equiv T$.
Claim: $(AQMN)$ and $(D'PMN)$ is cyclic.
Proof:Since $PDAQ$ is cyclic notice $\angle PD'Q =\angle PDQ=\angle PAQ$ and $\angle MAN=\angle PQD=\angle D'QP=\angle MQN$ $\Rightarrow (AQMN)$ and $(D'PMN)$ is cyclic.$\square$
Claim:$(TD'NQ)$ and $(TANP)$ is cyclic
Proof: It is easy to see since $\angle PD'N=\angle PMN=\angle AQN$ $\Rightarrow (TD'NQ)$ is cyclic.
Similarly, $\angle NAQ=\angle NMQ=\angle D'PN$ $\Rightarrow (TANP)$ is cyclic.
Now Let us denote $(TD'NQ) \equiv \omega_1$ and $(TANP) \equiv \omega_2$
Now notice $Pow_{\omega_1}(P)=PN.QN=PD'.TD'$ and $Pow_{\omega_2}(Q)=PN.QN=TA.AQ$ $\Rightarrow TD'.D'P=TA.AQ \Rightarrow (D'PAQ)$ is cyclic.
Consequently,$(D'AQDP)$ is cyclic. $\Rightarrow \angle PDQ=\angle PAQ=\angle A=90^0$
Now, $\angle BD'C=90^0=\angle PD'Q$ which implies, $\angle BD'P=\angle QD'C$ so $D'PB$ is similar to $D'QC$
Now by symmetry it is easy to see that $D'AQN$ is a parellelogram.
Now,$\angle D'CA=\angle 180^0 -\angle D'AC-\angle CD'A=\angle AQP-\angle AD'C=\angle AD'N-\angle AD'C=\angle QD'C$
$\Rightarrow D'Q=QC \Rightarrow D'P=PB \Rightarrow BP=PD \Rightarrow AB=AC$
$\square$
Note:
there could be cruxy proof of this by taking an extreme choice of $D$ as shown on aops.
$\Leftarrow$
Now assume $\angle A=90^0$ and $AB=AC$
$\Rightarrow PB=PD=PD'$ it is also easy to do by some trivial angle chasing that $\angle D'PB=\angle D'QC$ which implies $\angle BD'P=\angle QD'C$ since $\angle PDQ=\angle PD'Q=90^0 \Rightarrow \angle BD'C=90^)$ So, $D'$ is on the circumcircle.$\square$
Screenshot from 2021-03-19 07-36-02.png (96.99KiB)Viewed 12972 times
Thanks.But currently i ran out of problems so you may post one which can be solved without projec or inversive geo....
Problem 57
Posted: Sun Mar 21, 2021 9:02 pm
by Anindya Biswas
Does there exists $6$ points in the plane such that the distance between any two of them is an integer?
Edit : I forgot to mention that no $3$ points collinear.
Source :