Let $\Gamma$ be a circle, $P$ be a point outside it, and $A$ and $B$ the intersection points between $\Gamma$ and the tangents from $P$ to $\Gamma$. Let $K$ be a point on the line $AB$, distinct from $A$ and $B$ and let $T$ be the second intersection point of $\Gamma$ and the circumcircle of the triangle $PBK$.Also, let $P'$ be the reflection of $P$ in point $A$. Show that $\angle PBT=\angle P'KA$
Problem
Source: 2021 Pan-African Mathematics Olympiad, Problem 2
Tags: geometry, circumcircle, geometric transformation, reflection, PAMO
Jcmoltz875
25.05.2021 07:20
Sorry if this is a bit messy!
[asy][asy]
size(300);
pair P,A,B,K,O,O1,T,P1;
P=(0,0);
A=dir(-35);
B=dir(35);
K=arcpoint(A--B,5/7*abs(A-B));
O=2*circumcenter(P,A,B);
O1=circumcenter(P,K,B);
T=reflect(O,O1)*B;
P1=2*A;
draw(circle(O,abs(A-O)));
draw(circumcircle(P,B,T));
draw(P--B);
draw(T--K);
draw(B--A--T--cycle);
draw(K--P1--A);
draw(P--T--A--cycle);
label("$P$",P,dir(200));
label("$A$",A,SW);
label("$B$",B,dir(70));
label("$K$",K,E);
label("$T$",T,dir(-120));
label("$P'$",P1,SE);
[/asy][/asy]
Note that since $AP'$ is tangent to $\Gamma$ $\angle BTA=\angle BAP'$.
In addition $\angle TBA=\angle BTK=\angle TPK$, and since $PB$ tangent $\Gamma$, $\angle TAB=\angle PBT=\angle PKT$, so $\triangle PTK\sim \triangle BTA$.
Due to the similarity, $\angle PTK=\angle BTA=\angle KAP'$, and $\angle PBT=\angle PKT=\angle BAT=\angle KAT$
Thus we get $\angle PTB+\angle TBK=\angle PTK=\angle BTA=\angle BTK+\angle KTA$ so $\angle BTP=\angle KTA$.
Now by the law of sines, we have
$$\frac{KA}{KT}=\frac{\sin\angle KTA}{\sin\angle KAT}=\frac{\sin\angle BTP}{\sin\angle PBT}=\frac{PB}{PT}=\frac{PA}{PT}=\frac{AP'}{PT}$$Thus by SAS we have $\triangle P'AK\sim\triangle PTK\sim\triangle BTA$, implying $\angle PBT=\angle TAB=\angle P'KA$
\end{document}
math31415926535
25.05.2021 07:31
Jcmoltz875 wrote: Sorry if this is a bit messy!
[asy][asy]
size(300);
pair P,A,B,K,O,O1,T,P1;
P=(0,0);
A=dir(-35);
B=dir(35);
K=arcpoint(A--B,5/7*abs(A-B));
O=2*circumcenter(P,A,B);
O1=circumcenter(P,K,B);
T=reflect(O,O1)*B;
P1=2*A;
draw(circle(O,abs(A-O)));
draw(circumcircle(P,B,T));
draw(P--B);
draw(T--K);
draw(B--A--T--cycle);
draw(K--P1--A);
draw(P--T--A--cycle);
label("$P$",P,dir(200));
label("$A$",A,SW);
label("$B$",B,dir(70));
label("$K$",K,E);
label("$T$",T,dir(-120));
label("$P'$",P1,SE);
[/asy][/asy]
Note that since $AP'$ is tangent to $\Gamma$ $\angle BTA=\angle BTP'$.
In addition $\angle TBA=\angle BTK=\angle TPK$, and since $PB$ tangent $\Gamma$, $\angle TAB=\angle PBT=\angle PKT$, so $\triangle PTK\sim \triangle BTA$.
Due to the similarity, $\angle PTK=\angle BTA=\angle KAP'$, and $\angle PBT=\angle PKT=\angle BAT=\angle KAT$
Thus we get $\angle PTB+\angle TBK=\angle PTK=\angle BTA=\angle BTK+\angle KTA$ so $\angle BTP=\angle KTA$.
Now by the law of sines, we have
$$\frac{KA}{KT}=\frac{\sin\angle KTA}{\sin\angle KAT}=\frac{\sin\angle BTP}{\sin\angle PBT}=\frac{PB}{PT}=\frac{PA}{PT}=\frac{AP'}{PT}$$Thus by SAS we have $\triangle P'AK\sim\triangle PTK\sim\triangle BTA$, implying $\angle PBT=\angle TAB=\angle P'KA$
\end{document}
how did you get $\angle BTA=\angle BTP'?$
Jcmoltz875
25.05.2021 08:31
Oh oops, meant to say $\angle BAP'$, sorry about that!
SerdarBozdag
25.05.2021 15:31
$\angle TAK=\angle TBP$, $ \angle TKA = \angle TPB \implies ATK \sim BTP \implies$ $ \frac{KT}{KA} = \frac{PT}{PB}$ $\implies$ $\frac{AP'}{AK} =\frac{TP}{TK} $. Because we have $\angle P'AK=180-\angle ABP=\angle PTK$, we conclude that $P'AK \sim PTK \implies \angle P'KA=\angle TKP=\angle PBT$. $\square$
rafaello
02.09.2021 00:35
Posted here and here.
Mahdi_Mashayekhi
22.03.2022 16:27
Claim : $KAT$ and $PBT$ are similar. Proof : $\angle TPB = \angle TKA$ and $\angle TBP = \angle TAB$. Claim : $KAP'$ and $ATB$ are similar. Proof : $\angle ATB = \angle P'AK$ and $\frac{P'A}{BT} = \frac{PB}{BT} = \frac{KA}{AT}$ Now we have $\angle P'KA = \angle BAT = \angle PBT$.