In the quadrilateral $ABCD$ $\angle ABC = \angle CDA = 90^\circ$. Let $P = AC \cap BD$, $Q = AB\cap CD$, $R = AD \cap BC$. Let $\ell$ be the midline of the triangle $PQR$, parallel to $QR$. Show that the circumcircle of the triangle formed by lines $AB, AD, \ell$ is tangent to the circumcircle of the triangle formed by lines $CD, CB, \ell$. Proposed by Fedir Yudin
Problem
Source: Ukrainian Mathematical Olympiad 2023. Day 1, Problem 11.3
Tags: geometry, tangency
Philomath_314
05.04.2023 08:38
Let, $l \cap CQ=I$ ,$l \cap CR =J$, $l\cap PQ=E$ ,$l \cap PR= F$ , $l \cap AQ= G$ and $l\cap AR= H$ To prove-$(AGH)$ and $(CIJ)$ are tangent to each other. We define $K$ as the miquel point of the complete quad $QADRCB$ Lemma- $O,P,K$ are collinear. Now, as $\angle QKC= 90$. We can write $\angle IKC = \angle QKC -\angle QKI$ = $90 - \angle QKI$ A simple angle chase gives $QGIK$ as cyclic so $\angle QKI =180- \angle QGI = \angle AGH$ therefore $\angle IKC= 90- \angle AGH= 90-\angle BGJ= \angle BJG= \angle IJC$ Thus we prove $IKJC$ is cyclic. $\angle CKG = \angle CKI +\angle IKG = \angle CKI+\angle GQI = \angle CKI +90- \angle BCQ= \angle CKI + 90- \angle DCR =\angle CJI + \angle DRC = \angle HJR +\angle DRC= \angle DHG$ Thus $AGKH$ is cyclic too. Now, use radical axes or angle chase to prove the centres of $(AGH)$ ,$(CJI)$ and $K$ collinear. And, as $(AGH) \cap (CJI) = K$ and their centres are also collinear with $K$, we say that $(AGH)$ and $(CJI)$ are tangential .$\blacksquare$
Orestis_Lignos
06.04.2023 00:14
This solution is unnecessarily long, but is what I initially came up with. Let $X,Y$ be the midpoints of $PQ,PR$ respectively, and let $XY$ intersect $AB,AD$ at $K,L$ respectively. Main Claim: The perpendiculars from $K$ to $AB$ and from $L$ to $AD$ intersect on $QR$. Proof: Since angles $\angle B, \angle D$ are right, it suffices to prove that $\dfrac{RL}{RD}=\dfrac{BK}{BQ}$. Note that quadrilateral $QDBR$ is cyclic, and so $\dfrac{RD}{BQ}=\dfrac{\sin \angle DBC}{\sin \angle BDC}=\dfrac{CD}{CB}$. Moreover, quadrilateral $KLBD$ is cyclic too, since $\angle LKA=\angle RQA=\angle ADB$. Therefore, $\dfrac{AK}{AL}=\dfrac{AD}{AB}$. Employed with this last relation, Menelaus' theorem on triangle $RAB$ implies that $\dfrac{RL}{BK}=\dfrac{AL}{AK} \cdot \dfrac{SB}{SR},$ and so it suffices to prove that $\dfrac{SB}{SR}=\dfrac{CD \cdot AB}{AD \cdot BC}$. By an easy application of the Ratio Lemma, the last ratio is equal to $\dfrac{BP}{PD},$ and so we are left to prove that $\dfrac{SB}{SR}=\dfrac{BP}{PD}$, i.e. that $SP \parallel AD$. Let the perpendicular from $P$ to $AC$ intersect $RC$ at $T$. Note that $QR \perp AC,$ and so $PT \parallel QR$, which implies that $S$ is the midpoint of $RT$. Moreover, let the foot of the perpendicular from $T$ to $AD$ be $V$. Then, points $A,V,P,T,B$ are concyclic, and so $\angle PVT=\angle DBC=\angle DAC=\angle VTP,$ hence $PV=PT$, and obviously $SV=ST=SR,$ thus $SP$ is the perpendicular bisector of $VT$, which is perpendicular to $AD$. Therefore, $SP \parallel AD$, as desired $\blacksquare$ To the problem, let those two perpendiculars intersect at $U \in QR$. Then, $U \in (AKL)$, and so if that circle intersects $QR$ again at $G$, then $\angle AGU=\angle ALU=90^\circ,$ and so $G$ is the Miquel point of the complete quadrilateral $ABCD.QR$. Let $XY$ intersect $CB,CD$ at points $Z,W$. Similarly, $(CZW)$ passes through $G$ and a point $N \in QR$ such that $NC$ is the diameter of $(CZW)$. If $O_1,O_2$ are the midpoints of $UA$ and $NC$, we are left to prove that $G,O_1,O_2$ are collinear, or equivalently that $AU \parallel CN$, which is an easy angle-chase since $\angle RUA=\angle RUL+\angle LUA=\angle RQC+\angle LKA=\angle BDA+\angle AQC+\angle ADB=$ $=2\angle ADB+\angle AQC=\angle BAC-\angle ACD+2\angle ADB=90^\circ+\angle ACB-\angle ACD,$ where some steps were omitted, oops. Similarly, $\angle RNC=90^\circ+\angle ACB-\angle ACD,$ and we may - finally - conlude.
aleksijt
06.04.2023 15:05
Let $M$ be the Miquel point of the complete quadrilateral, $\ell\cap AB=X$ and $\ell\cap AD=Y$. Claim: $MXAY$ is cyclic. Proof: Since $AC\perp QR$, $M$ is the reflection of $P$ across $\ell$. Thus it suffices to show that $P$ is the orthocenter of $\triangle AXY$. To this end, let $PX\cap QR=Z$ and \[(X,\infty_{PZ};P,Z)=-1=(A,C;P,M)\stackrel{Q}{=}(X,Q\infty_{\perp AD}\cap PZ;P,Z)\]so $XP\perp AD$ and with $AP\perp XY$ we conclude. Similarly, the second circumcircle also contains $M$ and angle chasing shows they're tangent at $M$.
VicKmath7
26.04.2023 18:39
Let $\ell \cap AB, AD=G, H$, $\ell \cap CD, CB=I, J$ and let $AC \cap QR=K$; we will prove that the circles are tangent at $K$. Since $C$ is orthocenter for $\triangle AQR$, then $C$ is incenter for $\triangle BKD$. This combined with the fact that $\ell$ is perpendicular bisector of $PK$ yields that $(BPK)$ passes through $G, J$ and $(CPK)$ passes through $I, H$.
We have that $\angle KIJ=90^{o}-\angle PDC=\angle ADP=\angle KCJ$, so $K \in (CIJ)$ and similarly $K \in (AGH)$. We claim there is a homothety at $K$ taking $(CIJ)$ to $(AGH)$, which will be sufficient to establish the tangency. Indeed, let $(CIJ) \cap KG=L$; we will prove that $CL \parallel AB$. Indeed $\angle KLC =180^{o}-\angle CJK=\angle BGK$ as $BGKJ$ is cyclic. Thus the homothety at $K$ taking $CL$ to $AG$ will take $(CIJ)$ to $(AGH)$, done.
SerdarBozdag
20.01.2024 17:23
$EF$ cuts $QA, QD, RA, RB$ at $I,G,J,H$ respectively. $AC \cap RQ = K$ and $AC \perp RQ$. $-1 = (QP,QR;QA,QD) = (E, QR \cap EF ; I,G) \implies E$ is the midpoint of $IG$ and similarly $F$ is the midpoint of $HJ$. Thus $PIQG$ and $PJRH$ are parallelograms. $JR = PH = HK \implies HJRK$ is an isosceles trapezoid and so is $KGIQ$. $\angle IAK = \angle BRK = \angle HJK \implies AIKJ$ is cyclic and $\angle KHR = \angle KJR = \angle AIK = \angle JGK \implies GCHK$ is cyclic. Note that $K$ and the circumcenters of the circles given in the problem are collinear because $\angle HKJ = \angle HRJ = \angle GQI = \angle GKI$.