Let $I$ be the incenter of the triangle $ABC$, and $P$ be any point on the arc $BAC$ of its circumcircle. Points $K$ and $L$ are chosen on the tangent to the circumcircle $\omega$ of triangle $API$ at point $I$, so that $BK = KI$ and $CL = LI$. Show that the circumcircle of triangle $PKL$ is tangent to $\omega$. Proposed by Mykhailo Shtandenko
Problem
Source: Ukrainian Mathematical Olympiad 2023. Day 1, Problem 10.3
Tags: geometry, tangency
ACGNmath
07.04.2023 18:28
[asy][asy]
size(9cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A,B,C);
pair P = dir(130);
pair S = intersectionpoints(A--B,circumcircle(A,P,I))[1];
pair T = intersectionpoints(A--C,circumcircle(A,P,I))[1];
pair K = extension(P,S,I,I+dir(90)*(I-circumcenter(A,P,I)));
pair L = extension(K,I,P,T);
pair D = extension(B,K,C,L);
draw(unitcircle);
draw(A--B--C--A--cycle);
draw(B--D--C);
draw(P--K--L--P--cycle);
draw(S--T);
draw(circumcircle(A,P,I));
draw(B--I--C);
draw(circumcircle(P,D,K),red+dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$I$",I,dir(90));
dot("$D$",D,dir(D));
dot("$S$",S,dir(S));
dot("$T$",T,dir(T));
dot("$P$",P,dir(P));
dot("$K'$",K,dir(315));
dot("$L'$",L,dir(L));
[/asy][/asy]
Let $S$ and $T$ be the intersection of $\omega$ with $AB$ and $AC$ respectively, and let $K'$ and $L'$ be the intersection of $PS$ and $PT$ with the tangent to $\omega$ at $I$. We prove that $BK'=K'I$ and $CL'=L'I$ which shows that $K\equiv K'$ and $L\equiv L'$.
Throughout this proof, we use directed angles modulo $180^\circ$ to avoid configuration issues.
Claim 1: $\triangle BSP\sim \triangle CTP$.
Proof: \[\measuredangle BSP=\measuredangle ASP = \measuredangle ATP = \measuredangle CTP\]\[\measuredangle PBS=\measuredangle PBA=\measuredangle PCA=\measuredangle PTA \]
Claim 2: $\triangle BK'P\sim \triangle CL'T$.
Proof: We know that $\measuredangle BSK'=\measuredangle BSP=\measuredangle CTP=\measuredangle CTL'$, and they must be in fact equal in magnitude because $K'$ and $L'$ both lie on the same side of $ST$. Also, $SI=IT$ since $AI$ bisects $\angle SAT$, therefore $ST\parallel K'L'$.
\[\frac{BS}{SK'}=\frac{BS}{SP}\cdot\frac{PS}{SK'}=\frac{CT}{TP}\cdot\frac{TP}{TL'}=\frac{CT}{TL'}.\]
Let $BK'$ intersect $CL'$ at $D$. Note that since
\[\measuredangle ABD=\measuredangle SBK'=\measuredangle TCL'=\measuredangle ACD,\]this means that $D$ lies on $(ABC)$.
Claim 3: $D, P, K', L'$ are concyclic.
Proof:
\[\measuredangle PK'D=\measuredangle SK'B=\measuredangle TL'C=\measuredangle PL'D.\]
Now, if I were to show $\triangle PAD\sim \triangle PSB$, then we would be done. The result would follow because
\[\measuredangle K'BP=\measuredangle DBP=\measuredangle DAP=\measuredangle BSP\]Therefore, $K'B$ is tangent to $(BSP)$, and
\[K'B^2=K'S\cdot K'P=K'I^2\]and hence $K'B=K'I$.
Finally, taking a homothety from $P$ sending $ST$ to $K'L'$ also sends $\omega$ to $(PKL)$, finishing the problem.
MathSaiyan
07.04.2023 21:29
Let $M_A$, $M_B$ and $M_C$ be the midpoints of arcs $BC,CA,AB$ (not containing the opposite vertex). Recall the fact that $M_BM_C$ is the perpendicular bisector of $AI$, etc. So we can rewrite the problem taking $M_A,M_B,M_C$ as the reference triangle:
Quote:
Let $\triangle ABC$ be a triangle with orthocenter $H$, and let $P$ be a point on its circumcircle. Let $R$ be the second intersection of $AH$ and $(ABC)$. Let the tangent to $(HPR)$ through $H$ meet $AB$ at $K$ and $AC$ at $L$. Prove that $(PLK)$ is tangent to $(PHR)$.
Observe that
\[
\angle PHK = \angle PRH = \angle PRA = 180 - \angle KBP,
\]hence $PBKH$ is cyclic. Similarly, $HCLR$ is also cyclic. Now, we consider the (negative) inversion around $H$ sending $A,B,C$ to the feet of the altitudes. We can now rewrite the problem for a second time (this time i'll mantain the point names):
Quote:
Let $ABC$ be a triangle with incenter $H$, and let $AH$ meet $(ABC)$ again at $R$. Let $P$ be a point on $(ABC)$. If $L\in (AHC)$ and $K\in (AHB)$ such that $H\in LK$ and $RP\parallel LK$, prove that $PL = PK$.
So now there's no weird circles to take care of. Notice that the cyclic quads we proved earlier now translate into $P,C,L$ being collinear and $P,B,K$ being collinear. Now the problem is obvious: if $\ell$ is the bisector of $\angle LPK$, then $\ell$ is perpendicular to $PR$ (as $R$ is the arc midpoint), hence $\ell \perp LK$, implying that $PLK$ is isoscelles in $P$.
lahmacun
07.04.2023 23:22
Was it an IMO style test?
Let $AI$ meet the circumcircle at $M$.
By Pascal's theorem, $BK$ and $CL$ intersect at a point $S$ that lies on the circle $(ABC)$.
Let $P'$ be the second intersection of circles $(SKL)$ and $(SBC)$.
Let $P'D$ meet $BC$ at $D$
In the spiral similarity $\triangle P'KL \sim P'BC$, $$\frac{KI}{IL} = \frac{KB}{CL} = \frac{FB}{FC} = \frac{BD}{DC}$$So we actually have $P'KLI \sim P'BCD$.
Next, $$\measuredangle P'AI= \measuredangle P'AM =\measuredangle P'DB =\measuredangle P'IK$$which means that $P'AI$ is tangent to $KL$ at $I$ and $P'=P$
Now, just note that $PI$ is the angle bisector of $\angle KPL$ due to similarity and by Archimede's lemma, $(PKL)$ is tangent to $(API)$
Fedor Bakharev
19.04.2023 15:48
lahmacun wrote: Was it an IMO style test? No, there are 8=4+4 problems.
VicKmath7
26.04.2023 17:54
Let $M_b, M_c$ be midpoints of the minor arcs $AC, AB$. We have to prove that $PI$ bisects $\angle KPL$. We have that $\angle KM_cI=\angle KM_bI$, so we will be done if we prove that $IKM_cP$ and $ILM_bP$ are cyclic. Indeed, $\angle PIL=\angle PAI=\angle KM_cP$, done.