Point $I_a$ is the $A$-excircle center of $\triangle ABC$ which is tangent to $BC$ at $X$. Let $A'$ be diametrically opposite point of $A$ with respect to the circumcircle of $\triangle ABC$. On the segments $I_aX, BA'$ and $CA'$ are chosen respectively points $Y,Z$ and $T$ such that $I_aY=BZ=CT=r$ where $r$ is the inradius of $\triangle ABC$. Prove that the points $X,Y,Z$ and $T$ are concyclic.
Problem
Source: Saint Petersburg MO 2020 Grade 9 Problem 5
Tags: geometry, circumcircle, inradius
a1267ab
07.05.2020 23:02
[asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(-30); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair Z = B+I-F; pair T = C+I-E; pair Ia = 2*circumcenter(B, I, C)-I; pair Y = Ia+I-D; pair X = foot(Ia, B, C); pair P = D+E-I; pair Q = Y+E-I; draw(A--B--C--cycle); draw(I--D ^^ I--E ^^ I--F); draw(B--Z--I ^^ C--T--I); draw(Ia--X); draw(circumcircle(X, Y, Z), red); draw(B--Ia--Q); draw(D--P ^^ Y--Q, dashed); draw(P--I); draw(D--T--Y, blue); draw(P--C--Q, blue); string[] names = {"$A$", "$B$", "$C$", "$I$", "$D$", "$E$", "$F$", "$Z$", "$T$", "$I_a$", "$Y$", "$X$", "$P$", "$Q$"}; pair[] pts = {A, B, C, I ,D, E, F, Z, T, Ia, Y, X, P, Q}; pair[] labels = {A, B, C, dir(90), D, E, F, dir(-45), dir(225), Ia, dir(135), dir(225), dir(45), Q}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Let $I$ be the incenter and $DEF$ be the intouch triangle. We claim that $DXYZT$ is cyclic with diameter $DY$. It is clear that $\angle DXY=90^{\circ}$, so it suffices to show that $\angle DTY=90^{\circ}$. Construct $P$ and $Q$ so that $\overrightarrow{DP}=\overrightarrow{YQ}=\overrightarrow{IE}=\overrightarrow{TC}$. Then $IDPE$ is a rhombus, so $P$ lies on $CI$. Also, $\overrightarrow{I_aQ}=\overrightarrow{DE}$, so $Q$ lies on $CI_a$. Therefore \[\angle DTY=\angle PCQ=\angle ICI_a=90^{\circ}.\]
zuss77
08.05.2020 14:55
I guess it need to be stated in the problem that point $A'$ is on the arc $BC$. Otherwise it's not working.
dolly33
08.05.2020 16:27
Another proof... let $D=(I)\cap BC$ let $M$ be the midpoint of arc $BC$. $M$ is the center of $(YTXZ)$ first, note that $I, Z, A', M, T$ are cyclic with diameter $IA'$. then we obtain $\triangle MZB\sim \triangle MTC$. since $BZ=TC=r$, $MZ=MT$ meanwhile, since $CT=ID$, $CM=IM$, and $\angle MCT=\angle A'AM=\angle MID$, $\triangle MID\equiv \triangle MTC$ $\therefore MT=MD=MX=MZ=MY$
Andrei246
31.05.2020 14:43
Here is a solution using angle chasing and the law of cosines.
Claim $M$ is the circumcentre of $\triangle YTZ$
ProofFrom cyclic quadrilater $BA'MC$ leads us to $\angle A'BM=\angle A'CM$.So let $ \angle ZBM=\angle TCM=x$
It remains to prove $\angle YI_aM=x$
But from $I_aX\parallel IQ$ (where $ Q\in BC, IQ\perp BC$).It remains to prove $\angle MII_A=x$
$(BA'MA) $ is cyclic implies $\angle A'AM=\angle A'BM=x$ $ OA=OM$ implies $\angle OAM=\angle OMA=x$ .But $OM\parallel IQ $(because $OM,IQ\perp BC$) So $\angle MIQ=x$
Finally,from the law of cosines applied for $\triangle BZM,\triangle CTM$ and$\triangle MYI_a$ using $MI_A=MB=MC=p_a$(from incenter lemma)
we obtain that $MY^2=MZ^2=MT^2=r^2+p_a^2-2rp_a cosx$.So $ MY=MZ=MT$
Claim $MX^2=MY^2$
Proof From the law of cosines applied for $\triangle I_aYM $and $\triangle XMI_a$.It remains to prove
$MX^2=(r+XY)^2+p_a^2-2(r+XY)p_a cosx=MY^2=r^2+p_a ^2-2rp_a cosx$
Reducing the common factors we obtain $cosx=\frac{XY+2r}{2MI_a}$. or this let $Q_1\in(I_aX$ such that $ XQ_1=r$. Easily $ Q_1IQX$ is rectangle .So in $\triangle Q_1I_aI$ results
$ cosx=\frac{Q_1I_a}{II_a}=\frac{XY+2r}{2MI_a}$ which completes the proof .
Mathematicsislovely
08.06.2020 15:30
Let $I$ be the incentre of $ABC$ and $M$ is the mispoint of $II_a$.It is known that $M$ is the midpoint of arc $BC$ not containing $A$.Let $D$ be the point where incircle touches $BC$. Claim:$M$ is the midpoint of $DY$. proof: $ID=YI_a=r$ and $ID||YI_a$.So $IDI_aY$ is a parallelogram.So midpoint of $YD$ is same as midpoint of $II_a$ i.e. $M$. Now as $\angle YXD=90^{\circ}$ and $M$ is the midpoint of hypotenuse $YD$ so $MD=MX=MY$ Claim:$MZ=MT$ Proof.As $M$ is the midpoint of arc $BC$ ( not containing $A$, so $MB=MC$.We have $\angle MBZ= \angle MBA'= \angle MCT'= \angle MCT$. And $BZ=CT=r$.So $MBZ$ and $MCT$ are congruent.Hence, $MZ=MT$ Claim:$IBZD$ is a issoceles trapizium. Proof.As $ID=BZ$ it is enough to show $IBZD$ is cyclic.In order to show that let $F$ be the point where incircle touch $AB$.Note that $IF=ZB=r$ and $\angle IFB=\angle ZBF=90^{\circ}$. So $IFBZ$ is a rectangle.So $\angle IZB=90^{\circ}$.Also $\angle IDB=90^{\circ}$.So $IBZD$ is concyclic. Claim:$MZ=MD$ Proof. It is well known that $MB=MI$ and from the last claim we have $ZDIB$ issoceles trapizium.So $Z$ is symmetric image of $D$ in perpendicular bisector of $BI$ which pass through $M$.It means $MZ=MD$. SO $MY=MX=MD=MZ=MT$.SO $Y,X,D,Z,T$ lies on a circle with centre $M$.$\blacksquare$