symmedian in two circles

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Source: Iran second round- 2013- P4

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32 posts

dan23

#1 h May 4, 2013, 2:03 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $P$ be a point out of circle $C$ . Let $PA$ and $PB$ be the tangents to the circle drawn from $C$ . Choose a point $K$ on $AB$ . Suppose that the circumcircle of triangle $PBK$ intersects $C$ again at $T$ . Let ${P}'$ be the reflection of $P$ with respect to $A$ . Prove that
$\angle PBT = \angle {P}'KA$

This post has been edited 5 times. Last edited by dan23, May 25, 2013, 5:07 PM

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76 posts

duanby

#2 h May 4, 2013, 2:30 PM • 2 Y

Y by Adventure10, Mango247

my solution (may be it is complicated):
Let K' on AB AK'=BK
Let K'' on line AB,AK''=AK'
Then we have $\angle{P'K''K}=\angle{PTB}$
So need to prove P'K''K similar to PTB
sufficient to provePK''/KK''=PT/TB
that is PK/AB=PT/TB
by PTK simialr to BTA
done.

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339 posts

leader

#3 h May 4, 2013, 3:00 PM • 2 Y

Y by thunderz28, Adventure10

$PTB\sim TKA$ so $PT/TK=PB/KA=AP'/AK$ so $P'AK\sim PTK$ done.

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76 posts

#4 h May 4, 2013, 3:37 PM • 1 Y

Y by Adventure10

we need to prove that $AT$ is parallel to $PK$ and because $A$ is midpoint of $PP'$ then we need to prove that $AT$ goes through midpoint of $PK$ so now let $AT$ intersect $PK$ at $L$ then we have $\angle LKT= \angle PBT= \angle TAK \Rightarrow LK^2=LT.LA$
in same way we can prove that $LP^2=LT.LA$ so now $LP=LK$ and we are done

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1 post

#5 h Jul 5, 2013, 3:14 PM • 2 Y

Y by Adventure10, Mango247

lemma:suppose a tiangle ABC and point T on the BC. so we would have: BT/CT=(sin(BAT)*BA)/(sin(CAT)*CA)

in the triangle AKP , continue the line AT to meet PK at the point S.since the PTKB is embedded in a circle we will easily find it out that $\angle TPK=\angle TAP$ and $\angle TKP = \angle KAT$
now use the lemma in two triangles $%Error. "triagles" is a bad command. AKP$ and $%Error. "triagles" is a bad command. TKP$ for the point S and we will prove that SP=PK.
so the two lines AT and kp' are parallel and $\angle P'KA=\angle KAT =\angle TBP$
and we are done

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239 posts

#6 h Dec 15, 2015, 3:52 PM • 2 Y

Y by Adventure10, Mango247

First , $AT \cap \circ BKP$ = ${X}$ where $X$ is distinct from $T$ and also let $AT \cap KP=Z$ . By an angle chase $AKXP$ is a $||^{gm}$ .Diagonals of a $||^{gm}$ bisect each other and $P{P}'$ is bisected at $A$ . So , $ZA || K{P}'$ which (again after an angle chase) yields us $\angle PBT = \angle {P}'KA$

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1079 posts

#8 h Feb 13, 2021, 12:13 AM

Y by

Restated Iran Second Round 2013 P4 wrote:

Let $P$ be a point out of circle $\omega$ . Let $PA$ and $PB$ be the tangents to the circle drawn from $C$ . Choose a point $C$ on $AB$ . Suppose that the circumcircle of triangle $PBC$ intersects $\omega$ again at $K$ . Let $Q$ be the reflection of $P$ with respect to $A$ . Prove that
$\angle DBP = \angle QCA$

Angle chase.
$\measuredangle PAD=\measuredangle ABD=\measuredangle CBD=\measuredangle CPD$ and $\measuredangle DAC=\measuredangle DAB=\measuredangle DBP=\measuredangle DCP.$
By this, we conclude that $D$ is the $A$ -Humpty point of $\triangle APC$ .
By Humpty properties, we have $AD$ bisecting $PC$ , hence $AD\parallel CQ$ .
Thus, $\measuredangle DBP=\measuredangle DAB=\measuredangle DAC=\measuredangle QCA$

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Mahdi_Mashayekhi

696 posts

Mahdi_Mashayekhi

#9 h Jan 2, 2022, 3:32 PM

Y by

step 1: TAK and TBP are similar.
well easy to see ∠TAK = ∠PBT and ∠TKA = ∠TPB.

step 2: AP'K and TBA are similar.
from step1 we have AT/AK = BT/BP = BT/AP = BT/AP' and ∠P'AK = 180 - ∠BAP = 180 - ∠ABP = ∠KTP = ∠ATB.

now from step2 we have ∠P'KA = ∠BAT = ∠PBK.
we're Done.

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