A tangent line to circumcircle of acute triangle $ABC$ ($AC>AB$) at $A$ intersects with the extension of $BC$ at $P$. $O$ is the circumcenter of triangle $ABC$.Point $X$ lying on $OP$ such that $\measuredangle AXP=90^\circ$.Points $E$ and $F$ lying on $AB$ and $AC$,respectively,and they are in one side of line $OP$ such that $ \measuredangle EXP=\measuredangle ACX $ and $\measuredangle FXO=\measuredangle ABX $. $K$,$L$ are points of intersection $EF$ with circumcircle of triangle $ABC$.prove that $OP$ is tangent to circumcircle of triangle $KLX$. Author:Mehdi E'tesami Fard , Iran
Problem
Source: Iranian geometry olympiad 2014
Tags: geometry, circumcircle, reflection, Iran
drmzjoseph
25.02.2015 04:53
Case 1 $E,F$ and $A$ same semi-plane WRT $OP$ ($EX \perp AB$ and $FX \perp AC$) Lemma $A$ is a external point WRT $(O)$. Draw the tangents from $A$, $AX$ and $AY$; and $B$ is midpoint of $XY$ $ \omega$ is the circle of diameter $BX$ cut to $(O)$ in $X$ and $Z$. Then $XZ$ bisects $AB$ Proof $ZB$ cut to $(O)$ ins $Z$ in $W$ ($YW \perp XY$) $ \angle AXZ = \angle XWZ; \ \ \angle BAX= \angle YXW$ $ \triangle XYW \cup \overleftrightarrow{XZ} \sim \triangle ABX \cup \overleftrightarrow{BW} \Rightarrow ZX$ bisects $AB$ Back to main problem $PX \cap AB=R$ ($AR.AE=AX^2$) $PX \cap AC=S$ ($AS.AF=AX^2$) Then $R,E,F$ and $S$ are cyclic $AX \cap BC=G$ $H$ is midpoint of $PX$ and $AH$ cut to $(ABC)$ in $A$ and $W$ using the lemma $A,W,F$ and $E$ are cyclic. $AG$ is $A$-symmedian of $\triangle ABC$ $\Rightarrow (P,G,B,C)=-1 \Rightarrow (P,X,R,S)=-1 \Rightarrow HX^2=HR.HS \Rightarrow R,W,A$ and $S$ are cyclic Using radical axis on $(EFAW), (REFS)$ and $(RWAS) \Rightarrow F,E$ and $H$ are collinear. $P$ and $X$ are points conjugate WRT $(ABC)$ Then the circle of diameter $PX$ is orthogonal to $(ABC) \Rightarrow KH.LH=HX^2$ Done! Case 2 $E_1,F_1$ same semi-plane WRT $OP$ (but $A$ no) $E_1$ and $F_1$ lies on $AB$ and $BC$ sucht that $\angle RXE_1=\angle EXR \ \ \angle SXF= \angle SXF_1$ $X(E,A,F,H)=X(E_1,G,F_1,H)$ (Same angles) Then are projective rays $ \Rightarrow H,E_1$ and $F_1$ are collinear(Seen from $A$). And the result analogous
TelvCohl
25.02.2015 06:53
My solution: Case 1. $ E, F, A $ are at the same side of $ OP $ : Let $ Y=OP \cap AC, Z=OP \cap AB $ . Let $ G=EF \cap OP $ and $ P^* $ be the reflection of $ X $ in $ G $ . Since $ AX $ is the polar of $ P $ WRT $ \odot (O) $ , so we get $ (Y,Z;X,P)=-1 $ . ... $ (\star) $ Since $ X $ is the projection of $ O $ on $ A- $ symmedian , so $ X $ is the spiral center of $ AC \mapsto BA $ (well-known) $ \Longrightarrow \angle ACX=\angle BAX, \angle ABX=\angle CAX $ , hence combine with $ AX \perp OP $ we get $ E, F $ is the projection of $ X $ on $ AB, AC $, respectively . From $ \angle AFE=\angle AXE=\angle YZA \Longrightarrow E, F, Y, Z $ are concyclic , so $ GY \cdot GZ=GE \cdot GF={GX}^2={GP^*}^2 \Longrightarrow (Y,Z;X,P^*)=-1 $ , hence combine with $ (\star) $ we get $ P \equiv P^* \Longrightarrow G $ is the midpoint of $ PX $ . Since $ \odot (XP) $ is orthogonal to $ \odot (O) $ , so $ {GX}^2=GL \cdot GK \Longrightarrow GX \equiv OP $ is tangent to $ \odot (KLX) $ . Case 2. $ E, F $ are at the different side of $ OP $ WRT $ A $ : Let $ Y=OP \cap AC, Z=OP \cap AB $ . Let $ E^*, F^* $ be the projection of $ X $ on $ AB, AC $, respectively and $ \{ L^*, K^* \}=E^*F^* \cap \odot (ABC) $ . Since $ (A,Y;F,F^*)=-1=(A,Z;E,E^*) $ , so $ EF, YZ \equiv OP, E^*F^* $ are concurrent at $ G $ . From the proof in Case 1 we get $ G $ is the midpoint of $ XP $ and $ \odot (XP) \perp \odot (O) $ , so we get $ GL \cdot GK=GL^* \cdot GK^*=GX^2 \Longrightarrow GX \equiv OP $ is the tangent of $ \odot (KLX) $ . Q.E.D
anantmudgal09
05.09.2016 21:38
Nice use of Steiner and Simson's Line! Solution: Notice that if the $A$ symmedian meets $(O)$ again at $F$ then $X$ is the midpoint of $AF$. It is well-known that $\triangle XAC \sim \triangle XBA$ and so, we get that $\angle EXP=\angle EAX$ (directed angles) and similar argument for $F$ yields that $A,E,F,X$ lie on a circle tangent at $X$ to the line $OP$. We get that $E,F$ are the projections of $X$ onto the sidelines $AB,AC$ respectively. Let $EF$ meet $OP$ at point $T$. We need $TX^2=TK\cdot TL=TO^2-OA^2$ in order to establish the proclaimed assertion. This can be seen to be equivalent to saying that $T$ is the midpoint of $PX$. Indeed, if $P'$ is symmetric to $X$ in $T$ then this implies $TO^2-TX^2=OX\cdot OP'=OA^2=PX\cdot OP$ and so $P \equiv P'$. Now, a homothety about $X$ of ratio $2$ yields that we want point $P$ to be on the line joining the reflections of $X$ in the sidelines $AB,AC$. This lines is infact the Steiner line of $F$ wrt $ABC$ scaled about $A$ with ratio $\frac{1}{2}$. Now we know for a fact that the Steiner line of $F$ passes through the orthocenter of $ABC$ since $F$ lies on $(O)$. Let $M$ be the projection of $H$ onto the median through $A$, the line $AN$ ($N$ is the midpoint of $BC$). We also know that the Steiner line of $F$ passes through $M$ since $M$ is the reflection of $F$ in $BC$. Thus, we obtain the following equivalent, but much easier problem: Let $H$ be the orthocenter and $N$ be the midpoint of side $BC$ in a triangle $ABC$. The tangent to its circumcircle at $A$ meets $BC$ at $P$. Let $E$ be the midpoint of $AH$. Prove that $PE \perp AN$. In essence, the above problem asks for showing that $E$ is the orthocenter of triangle $APN$ since $AE \perp PN$ anyways. We prove it by showing that $NE \perp AP$. Notice a dilation at $H$ of ratio $2$ sends $E$ to $A$ and $N$ to the antipode of $A$. Thus, $EN \parallel AO$ and since $AO \perp AP$, our result follows.
EulerMacaroni
05.09.2016 23:51
First note that $X$ is the midpoint of the $A$-symmedian chord, which is well known to be the spiral center sending segment $\overline{BA}$ to segment $\overline{AC}$. Then $\angle EXP=\angle ACX=\angle BAX=\angle EAX$ so line $PX$ is tangent to $\odot(AEX)$ and similarly to $\odot(AFX)$ hence quadrilateral $AEFX$ is cyclic. Define $\{A, G\} \equiv \odot(ABC) \cap \odot (AEF)$, $\{A, R\} \equiv AX \cap \odot(ABC)$ and suppose the tangent to $\odot(AEF)$ at $A$ intersects $\odot(ABC)$ again at $S$. First I claim that $P, G, S$ are collinear; after inversion about $A$ with power $r^2=AB\cdot AC$ composed with a reflection about the $A$-angle bisector, the claim is equivalent to showing that if $A'$ is the point such that $ABA'C$ is a parallelogram, $E$ is the point on $BC$ such that $EA\perp AA'$, $H$ is the point on $BC$ such that $AA'\perp A'H$, and $D$ is the point on $\odot(ABC)$ such that $AD\parallel BC$, then quadrilateral $EADH$ is cyclic, but this is obvious by symmetry. Therefore, since $RP$ is also tangent to $\odot(ABC)$, $$-1=(A, R; S, G)\stackrel{A}{=}(P, X; P_{\infty, OP}, Y)$$where $Y\equiv AG\cap OP$ and so $AG$ bisects segment $\overline{PX}$. Let $W\equiv PY\cap AB, Z\equiv PY\cap AC$. Then $$-1=(A,R;B,C)\stackrel{A}{=}(P, X;W,Z)$$so $YG\cdot YA=YZ^2=YW\cdot YZ$ and so quadrilateral $AGWZ$ is cyclic. It is easy to see that $EF$ and $WZ$ are antiparallel with respect to angle $A$, and so quadrilateral $EFWZ$ is cyclic as well. Then the radical axis theorem gives us that $EF$ passes through $Y$, and since $YG\cdot YA=YK\cdot YL$, we conclude that $YX$ is the radical axis of $\odot(AEF)$ and $\odot(KLX)$, and the desired conclusion follows.
khina
10.05.2020 03:52
First note that $X$ is the dumpty point, and $E$ and $F$ are its projections onto $AB$ and $AC$, respectively. By projective stuff, it suffices to prove that $Q$ is the midpoint of $XP$.
To do this, say the tangent at $X$ to the circumcircle of $XAB$ meets $AB$ at $B_1$, and define $C_1$ similarly. By spiral similarities, $Q$ lies on $B_1C_1$. Now do some angle chasing to prove that $XB_1 = XP_1$ (in particular, a spiral similarity at $B$ reveals that $B_1P$ and $AC$ are parallel), so this gives us that $B_1C_1$ is the perpendicular bisector of $XP$ and we are done. Yay.
In fact, $B_1C_1$ is the radical axis of $P$ and the circumcircle of $ABC$, but radical axis didn't get me very far oops. I suspect you should also be able to do spiral similarities from the intersection of (AX) and (ABC), by showing that when you move $K$ along to $X$ then $A$, you move $P$ to $Q$ to $X$. But I haven't tried this yet, so whatever.
Another great spiral similarity length problem! Kind of like ISL 2012 G6 in which the intuition of how linear maps works helps a bit.
chirita.andrei
20.08.2020 13:30
It's easy to see that $ X $ is the Dumpty point of triangle $ABC$ .This means that $E$ and $F $ are the orthogonal projections of $ X $ on sides $ AB $ and $ AC $.Also $ OP $ is tangent to the circumcircle of $ AEF $ at point $ X $. Now let $ EF $ meet $ OP $ at point $ Q $.We want to prove that $ XQ^{2}=XK \cdot{XL} $,i.e. $ Q $ lies on the radical axis of the circumcircle of triangle $ ABC $ and triangle $ AEF $. Consider the points where $ OP $ meets the circumcircle of triangle $ ABC $ and call them $ M $ and $ N $.We will prove that $ M,E,F,N $ are cocyclic.Note that this claim finishes the problem because $ Q $ would be the radical center of the circumcircles of $ ABC,AEF $ and $ MEFN $. Consider an inversion centered at $ X $ with radius $ XA $.Then we get the following equivalent sub-problem: Let $ ABC $ be a triangle and $ X $ be the Dumpty point.The perpendicular line at $ A $ on $ AX $ meets the circumcircles of triangles $ AXB $ and $ AXC $ at points $ E $ and $ F $ respectively.The perpendicular line at $ X $ on $ AX $ meets the circumcircle of triangle $ ABC $ at $ M $ and $ N $.Prove that $ M,E,F,N $ are cocyclic. But $ EF||MN $ so we have to prove that $ EFNM $ is an isoscelles trapezoid.But the circumcenter of triangle $ ABC $ is the midpoint of $ MN $ so it sufficies to prove that it is equidistant from $ E $ and $ F $,i.e. $ E $ and $ F $ have equal powers with respect to the circumcircle of $ ABC $. Let the line $ EF $ meet the circumcircle of triangle $ ABC $ again at $ T $.We will show that $ AE \cdot ET = FA \cdot FT $.Since $ EBT,FCT $ and $ ABC $ are similar,we have to prove that $ \frac{EA\cdot {EB}}{FA \cdot {FC}} = (\frac{AB}{AC})^{2}$,which follows from the fact that $ ABE $ and $ CAF $ are similar.
rafaello
28.10.2021 23:53
Killed by ratio lemma, lezgo. Observe that $X$ is the $A$-Dumpty point of $\triangle ABC$. Now, let $S=(AEF)\cap (ABC)$, thus $S$ is the center of spiral sim taking $\overline{BE}$ to $\overline{CF}$. Define $f(\bullet)=\pm\frac{\bullet E}{\bullet F}$, with the choice of signs as usual. Indeed, $$f(X)^2=\frac{XE^2}{XF^2}=\frac{BE}{AF}\cdot \frac{EA}{FC}=\frac{AE}{AF}\cdot \frac{SE}{SF}=f(A)\cdot f(S),$$which yields that $\overline{AS},\overline{EF},\overline{OX}$ are concurrent. Finish by PoP.
hakN
30.10.2021 18:13
Note that $X$ is the polar of $P$ wrt to $(ABC)$, thus $X$ is the midpoint of the symmedian chord, known as the Dumpty point.
It is a well-known property of this point that $\angle BAX = \angle ACX$ and $\angle ABX = \angle CAX$.
By a trivial angle chase, we get $AEXF$ is cyclic and that $90^\circ = \angle AEX = \angle AFX$. Let $Y = XE \cap (KLX)$ and $Z = XF \cap (KLX)$.
By PoP, we get $AXBY$ and $AXCZ$ are cyclic. So $\angle ZAX = \angle ZXC + \angle XAC = \angle FXC + \angle XBA = \angle AXY + \angle XBA = \angle YBX$ , so $Y,A,Z$ are collinear.
Now, $\angle XYZ = \angle XBA = \angle XAC = \angle XEF \implies EF \parallel YZ$.
So, since $EF \parallel YZ$, we get $\angle EXK = \angle FXL$.
Finally, $\angle KLX = \angle EFX - \angle FXL = \angle EXP - \angle EXK = \angle KXP$, implying that $OP$ is tangent to $(KLX)$. $\square$
Tafi_ak
30.09.2023 12:10
This problem is true for any point $X$ inside $\triangle ABC$ and an arbitrary line passing through $X$.
Om245
03.10.2024 13:48
First note that $X$ is $A$-dumpty point. Hence we get $\angle ACX = \angle EAX$. Which give us $A,E,F,X$ lie on same circle and $XO$ is tangent to $(AEF)$ at $X$. Do $\sqrt{bc}$ inversion. Let $X'$ denote image of $X$. Let line through $A$ parallel to $BC$ intersect $E'F'$ at $R$. Also $Z= BC \cap E'F'$. $$(B,C;M,\infty)\stackrel{A}{=}(E',F';X';R)$$Hence $R$ is harmonic conjugate of $X$ in $E'F'$ and $Z$ is midpoint of $X'R$. Which give us $$RX'^2=RE'\cdot RF' = RK'\cdot RL'$$hence $(K'X'L')$ tangent to $E'F'$ at $X'$. If $G = X'K' \cap (AE'F')$ and $H =X'L' \cap (AE'F')$ then, this give us $$\angle L'K'X' = \angle L'X'F' = \angle HX'E' = \angle L'HG$$$GH \parallel E'F'$. Which by simple angle chase give us $$\angle AK'X' - \angle AE'X' = \angle AL'X' - \angle AF'X'$$which is equivalent $$\angle KXE = \angle LXF$$which is enough to prove $XO$ tangent to $(KXL)$.