The circles $k_1, k_2$, centered at $O_1, O_2$, meet at two points, one of which is $A$. Let $P, Q$ lie on $AO_1, AO_2$, respectively, so that $PQ \parallel O_1O_2$. The tangents from $P$ to $k_2$ touch it at $X, Y$ and the tangents from $Q$ to $k_1$ touch it at $Z, T$. Show that $X, Y, Z, T$ are collinear or concyclic.
Problem
Source: Serbia TST 2024, P5
Tags: geometry
HyperDunteR
18.05.2024 23:09
Let be $R(\Omega_1, \Omega_2)$ the radical axis of the circles $\Omega_1$ and $\Omega_2$. Claim: $\Gamma_1 = \odot (QZT) \implies O_1 \in \Gamma_1$ and $K$ (the midpoint of $QO_1$) is its center. (analogously with $\Gamma_2$)
Since $QZ, QT$ are tangents to $k_1 \implies \angle QZO_1 = \angle QTO_1 = 90^{\circ} \implies QO_1$ is the diameter of $\Gamma_1. \square$
$L$ is the midpoint of $PY, R \in O_1O_2: AR \perp O_1O_2, M:= KL \cap AR, S:= PQ \cap AR \implies AR = R(k_1,k_2)$ (1)
$a= AO_1, b = AO_2, m= RO_1, n= RO_2 \overset{\text{by Thales}}{\implies} AQ = bk, AP = ak, QS = nk, SP = mk \implies MK = \frac{QS+O_1R}{2}= \frac{nk+m}{2},$ $ ML = \frac{SP+O_2R}{2}= \frac{mk+n}{2}$ and is easy to see that $a^2-m^2=b^2-n^2$ (2).
Final claim: $Pow(M, \Gamma_1) = Pow(M, \Gamma_2)$
$Pow(M, \Gamma_1) = Pow(M, \Gamma_2) \iff KM^2-KO_1^2= LM^2-LO_2^2$
For the LHS: $KM^2-KO_1^2 = \frac{(nk+m)^2}{4} - \frac{(bk)^2 + a^2 -2abk \text{cos} \angle QAO_1}{4}$ (the second substitution is by cosines law on $\triangle O_1QA$) $\implies KM^2-KO_1^2 = \frac{(nk)^2- (bk)^2 + m^2 - a^2 + 2mnk - 2abk \text{cos} \angle QAO_1}{4}$
Similary, the RHS: $LM^2-LO_2^2 = \frac{(mk)^2- (ak)^2 + n^2 - b^2 + 2mnk - 2abk \text{cos} \angle PAO_2}{4}$
By (2) both are equal, then we are done. $\square$
By the last claim, by (1) and $KL \parallel O_1O_2 \parallel PQ$ (Thales): $M \in R( \Gamma_1, \Gamma_2), AM \perp KL \implies AM = R( \Gamma_1, \Gamma_2) \implies $ $R(k_1, k_2) \cap R( \Gamma_1, k_1) \cap R( \Gamma_1, k_2) = AR \cap ZT \cap R( \Gamma_1, k_2)= AM \cap ZT \cap R( \Gamma_1, k_2) = R( \Gamma_1, \Gamma_2) \cap R( \Gamma_1, k_1) \cap R( \Gamma_1, k_2) =$ $ R( \Gamma_1, \Gamma_2) \cap R( \Gamma_1, k_1) \cap R( \Gamma_2, k_1) =P$. Then, this implies that $PX \cdot PY = PZ \cdot PT \implies$ $ XYZT$ is cyclic.
Mahdi_Mashayekhi
19.07.2024 08:04
Claim $1: QZO_1T$ and $PXO_2Y$ are cyclic. Proof $: \angle QZO_1 = \angle QTO_1 = 90$ and $\angle PXO_2 = \angle PYO_2 = 90$. Let $CD$ be the Radical axis of $QZO_1T$ and $PXO_2Y$. Let $E,F$ be midpoints of $QO_1$ and $PO_2$. Let $k_1,k_2$ meet at $A,B$. Claim $2 :AB$ lies on $CD$. Proof $:$ we need to prove $P_{QZO_1T}(A)-P_{PXO_2Y}(A) = P_{QZO_1T}(B)-P_{PXO_2Y}(B) = 0$. Note that $P_{QZO_1T}(A)= AE^2-EO_1^2 = \frac{AQ^2+AO_1^2}{2}-\frac{QO_1^2}{4}-EO_1^2 = \frac{AQ^2+AO_1^2-QO_1^2}{2}$ and similarly $P_{PXO_2Y}(A) = \frac{AP^2 + AO_2^2-PO_2^2}{2}$ so we need to prove $AQ^2+AO_1^2-QO_1^2 = AP^2 + AO_2^2-PO_2^2$ and since $QO_1^2 = AQ^2 + AO_1^2 - 2.AQ.AO_1.\sin{A}$ and $PO_2^2 = AP^2 + AO_2^2 - 2.AP.AO_2.\sin{A}$, we need to prove $2.AQ.AO_1.\sin{A} = 2.AP.AO_2.\sin{A}$ which is true since $PQ \parallel O_1O_2$ so $P_{QZO_1T}(A)-P_{PXO_2Y}(A) = 0$. we prove the other part with same approach. Let $QZO_1T$ meet $k_2$ at $X',Y'$ and let $PXO_2Y$ meet $k_1$ at $Z',T'$. Radical axis on $QZO_1T,PXO_2Y,k_1$ implies that $ZT,Z'T',CD$ are concurrent. Radical axis on $QZO_1T,PXO_2Y,k_2$ implies that $CD,XY,X'Y'$ are concurrent. Radical axis on $QZO_1T,k_1,k_2$ implies that $AB,ZT,X'Y'$ are concurrent. Radical axis on $PXO_2,k_1,k_2$ implies that $AB,Z'T',XY$ are concurrent. Since $AB$ lies on $CD$ we have that $AB,TZ,XY$ are concurrent so $XYZT$ is cyclic or $X,Y,T,Z$ are collinear.