an easy geometry from iran tst

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an easy geometry from iran tst

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Source: Iranian TST 2018, third exam day 1, problem 1

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#1 h Apr 18, 2018, 3:32 PM • 2 Y

Y by Adventure10, Mango247

Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A,B$ ,and $O$ lies on $\omega_2$ . Let $S$ be a point on $AB$ such that $OS\perp AB$ . Line $OS$ intersects $\omega_2$ at $P$ (other than $O$ ). The bisector of $\hat{ASP}$ intersects $\omega_1$ at $L$ ( $A$ and $L$ are on the same side of the line $OP$ ). Let $K$ be a point on $\omega_2$ such that $PS=PK$ ( $A$ and $K$ are on the same side of the line $OP$ ). Prove that $SL=KL$ .

Proposed by Ali Zamani

This post has been edited 2 times. Last edited by Etemadi, Apr 21, 2018, 3:42 PM

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#2 h Apr 18, 2018, 4:52 PM • 2 Y

Y by Adventure10, Mango247

Okay so this thing took me like 10 minutes to draw lol. Wording kinda threw me off

After some angle chasing; this reduces to: show that $\angle LKP =45$ deg

We let the intersection point of lines SL and KP be X; and that of KL and SP be Y. This transforms the problem into showing that quadrilateral $SKXY$ is cyclic. Anyone can pick up?

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Yaghi

#3 h Apr 18, 2018, 5:50 PM • 3 Y

Y by brokendiamond, Adventure10, Mango247

Let $PK \cap AB=T$ and let $L'$ be the foot of tangent from $T$ to $w_1$ such that $L',P$ lie on the same side of $AB$ .Obviously, $O,S,L',K,T$ are on a circle with diameter $OT$ ,Also,by POP:
$PK.PT=PS.PO \implies PO=PT$ Again,by POP for $T$ :
$TA.TB=TK.TP=OP.OS=OA^2 \implies TL'=OA=OL' \implies \angle TOL'=\angle L'TO=45=\angle L'SP$ and since $L'$ is on $w_1$ ,we deduce that $L' \equiv L$ .This ends the problem because $L$ is the midpoint of arc $SK$ in $(OSLKT)$ so $LK=LS$ .

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#4 h May 31, 2019, 2:02 AM • 1 Y

Y by Adventure10

Let $C = OK\cap AB$ , draw the line $PC$ intersect $\odot O$ at $L'$ and $D$ , we will prove that $L'$ coincide with $L$ . First, by power of point at $C$ , we have
$CK \cdot CO = CA \cdot CB = CD \cdot CL'$ Thus $K,D,O,L'$ concyclic. Moreover, $PA$ is tangent to $\odot O$ since $O \in \omega_2$ , yielding $PL' \cdot PD = PA^2 = PS \cdot PO$ , thus $D,O,S,L'$ concyclic as well, meaning that $K,D,O,S,L',$ lies on the same circle.

Now remarking that $\angle CKP = \angle CSP = 90^\circ$ and $PK=PS$ , we have $PC$ is the orthogonal bisector of $KS$ . Thus $\angle L'SC = \angle L'KC =\angle L'DO =\angle L'SP$ , implying that $L'S$ bisects the angle $\angle ASL$ , thus $L'$ coincides with $L$ which means $L$ lies $PC$ , the orthogonal bisector of $KS$ .

Attachments:

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#5 h May 31, 2019, 1:13 PM • 2 Y

Y by Adventure10, Mango247

Iran TST #3 2018 P1 wrote:

Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A,B$ ,and $O$ lies on $\omega_2$ . Let $S$ be a point on $AB$ such that $OS\perp AB$ . Line $OS$ intersects $\omega_2$ at $P$ (other than $O$ ). The bisector of $\hat{ASP}$ intersects $\omega_1$ at $L$ ( $A$ and $L$ are on the same side of the line $OP$ ). Let $K$ be a point on $\omega_2$ such that $PS=PK$ ( $A$ and $K$ are on the same side of the line $OP$ ). Prove that $SL=KL$ .

Solution: Let $C \in (O)$ , such, $\angle CSP=\angle ASC=45^{\circ}$ . Let $CP \cap (O)=D$ . Since, $\angle OAP=90^{\circ}$
$-1=( D, ~C ; ~ B, ~A ) \overset{A}{=} ( D, ~C ; AB ~ \cap ~ DP , ~ P) \implies D \equiv L$ Let $L', K'$ be reflections of $L, K$ over $OP$ .
$\angle LSO=\angle CSP=\angle L'SO \implies L' - S - C$ Let $C'$ be the reflection of $C$ over $OP$ $\implies$ $L - S - C'$ and $L' - C' - P$
$\angle LOC=2\angle LL'S=\angle LSC \implies LOSC \text{ and similarly, } L'OSC' \text{ are cyclic}$ Also, $\angle OKP=\angle OK'P=90^{\circ}$ $\implies$ $OK, OK'$ are tangent to $\odot (P)$ with radius $PS=PK$ . Let $LC', L'C$ $\cap$ $\odot (P)$ $=$ $D', D$ , then, $D'$ is the reflection of $D$ over $OS$ and $\angle DSD'=90^{\circ}$ $\implies$ $D - P - D'$ . Let $LP$ $\cap$ $\odot (AOP)$ $=$ $E$ , then by Radical Axes Theorem, $EO$ $\cap$ $AB$ $=$ $G$ lies on tangent at $L$ to $(O)$ . By some simple congruency, $E$ is the center of $\odot (LOSCG)$ . Suppose, $PE$ $\cap$ $AB$ $=$ $M$ $\implies$ $M$ is the orthocenter WRT $\Delta OGP$ ( $OP=OG$ ). Let $GP$ $\cap$ $\odot (AOP)$ $=$ $K'$ $\implies$ $O - M - K'$ , but then, $PS$ $=$ $PK'$ $\implies$ $K' \equiv K$ $\implies$ $K$ lies on $\odot (LOSCG)$ $\implies$ $LP$ is the perpendicular bisector of $SK$

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#6 h Jan 2, 2020, 7:32 PM • 2 Y

Y by AlastorMoody, Adventure10

Let $\omega$ be the circumcircle of $\triangle KSO$ and $D=PK\cap AB.$ First, observe that $PK=PS\implies \triangle SDP\cong \triangle KOP$ by LL, so $ODKS$ is an isosceles trapezoid and $D\in\omega.$ Since $DS\cap OK$ is the radical center of $\omega,\omega_1,\omega_2,$ the perpendicular bisector of $OD$ is the radical axis of $\omega,\omega_1.$ Now note that since the the intersection of the angle bisector of $\angle ASP$ with $\omega$ lies on the perpendicular bisector of $SK,$ it must lie on $\omega_1$ as well so this point must be $L.$ $\blacksquare$

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#7 h Sep 3, 2022, 6:29 PM

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As $OL^2=OA^2=OS \cdot OP$ we have the similarity $\triangle OSL \sim \triangle OLP,$ so $\angle OLP = 135^\circ$ .

Let $K' = (OSL) \cap (OP)$ . Then $\angle OK'L = 45^\circ$ and $\overline{K'L}$ thus bisects $\angle OK'P$ . But as $\angle OLP = 135^\circ= 90^\circ + \frac 12 \angle OK'P,$ $L$ is in fact the incenter of $\triangle OK'P$ . Thus, $\angle K'PL = \angle PLS$ , so $\triangle K'LP \cong \triangle SLP.$ This means $PK' = PS$ , so $K'=K$ ; but this congruence also implies $LK=LS$ , so we are done.

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demmy

#8 h Dec 11, 2023, 7:16 AM • 1 Y

Y by Tung_HP

Coord bash :(

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#9 h Mar 29, 2025, 12:03 PM

Y by

Let $X$ be the intersection of $OK$ and $AS$ . We know $XKP= XSP = 90$ so $XSPK$ is cyclic. $XK$ and $XS$ are tangent to the circle with centred on $P$ and radious $PK$ so $XS=XK$ so $XP$ is the perpendicular bisector of $KS$ . Let $L'$ be the intersection of $XP$ and $w_1$ . Now we have $OAP=90$ and $ASP=90$ so $OA^2=OS.OP=OL'^2$ so $OL'S=OPL= XKS$ so $OSL'K$ Is cyclic. so $L'KO=L'SP=L'KP$ because $L'$ lies on perpendicular bisector of $KS$ . And we have $OKP=90$ so $OKL'=OKP/2=45=L'SP$ so $L'$ is the $L$ and lies on perpendicular bisector of $KS$ .

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