Iranian geometry olympiad 2014(3)

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Source: Iranian geometry olympiad 2014

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237 posts

#1 h Feb 24, 2015, 6:23 PM • 3 Y

Y by HWenslawski, Adventure10, Mango247

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

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#2 h Feb 25, 2015, 2:53 AM • 3 Y

Y by Infinityfun, Adventure10, Mango247

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

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#3 h Feb 25, 2015, 4:53 AM • 5 Y

Y by drmzjoseph, K.N, Infinityfun, Adventure10, Mango247

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

This post has been edited 1 time. Last edited by TelvCohl, Mar 2, 2015, 3:41 AM

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#4 h Sep 5, 2016, 6:38 PM • 3 Y

Y by guptaamitu1, Adventure10, Mango247

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.

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#5 h Sep 5, 2016, 8:51 PM • 2 Y

Y by Adventure10, Mango247

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.

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khina

#6 h May 10, 2020, 12:52 AM

Y by

sketch

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#7 h Aug 20, 2020, 10:30 AM • 1 Y

Y by minta15

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.

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1079 posts

#8 h Oct 28, 2021, 8:53 PM • 1 Y

Y by HarryVu

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.

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hakN

#9 h Oct 30, 2021, 3:13 PM • 1 Y

Y by HarryVu

Solution

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#11 h Sep 30, 2023, 9:10 AM

Y by

This problem is true for any point $X$ inside $\triangle ABC$ and an arbitrary line passing through $X$ .

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Om245

#12 h Oct 3, 2024, 10:48 AM

Y by

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 $(KXL)$ .

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