Concurrent Lines - Incenter and Circumcircles

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Concurrent Lines - Incenter and Circumcircles

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Source: 2015 Korean Junior MO P5

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1721 posts

#1 h Nov 1, 2015, 1:54 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the incenter of an acute triangle $\triangle ABC$ , and let the incircle be $\Gamma$ .
Let the circumcircle of $\triangle IBC$ hit $\Gamma$ at $D, E$ , where $D$ is closer to $B$ and $E$ is closer to $C$ .
Let $\Gamma \cap BE = K (\not= E)$ , $CD \cap BI = T$ , and $CD \cap \Gamma = L (\not= D)$ .
Let the line passing $T$ and perpendicular to $BI$ meet $\Gamma$ at $P$ , where $P$ is inside $\triangle IBC$ .
Prove that the tangent to $\Gamma$ at $P$ , $KL$ , $BI$ are concurrent.

This post has been edited 1 time. Last edited by rkm0959, Nov 1, 2015, 1:55 PM

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#2 h Nov 1, 2015, 3:57 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Hello.

My solution.

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Suppose that $Q\equiv KL\cap BI$ .Since $ID=IE$ we get $\angle{DBI}=\angle{IBE}\Rightarrow BD=BK$ which gives that

$IB$ is the perpendicular bisector of $DK$ ,hence $\angle{DIQ}=\angle{TIK} \ (1)$ .Suppose that $S\equiv BE\cap CD$ .

Then $\frac{SL}{SK}=\frac{SE}{SD}=\frac{SC}{SB}\Rightarrow KL\parallel BC$ .Hence, $\angle{DIQ}=\angle{DCB}=\angle{DLQ} \ (2)$

yielding that $DILQ$ is cyclic. $(1),(2)\Rightarrow \angle{TIK}=\angle{DLQ}=\angle{TLK}\Rightarrow TILK$ is cyclic.

The latter gives $\angle{KTI}=180^{\circ}-\angle{ILK}=180^{\circ}-\angle{ILQ}=\angle{IDQ}\overset{(1)}\Rightarrow \triangle{DIQ}\simeq \triangle{TIK}$ .

Hence, $\frac{DI}{TI}=\frac{IQ}{IK}\Rightarrow \frac{PI}{TI}=\frac{IQ}{PI} \ (3)$ .

Since $\triangle{PIT},\triangle{PQI}$ also have a common angle ( $\angle{PIQ}$ ),from $(3)$ we deduce that they are similar.

Thus $\angle{QPI}=\angle{PTI}=90^{\circ}$ which gives that $PQ$ is tangent to $\Gamma$ and we are done.

This post has been edited 2 times. Last edited by gavrilos, Nov 1, 2015, 4:00 PM

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#3 h Nov 1, 2015, 9:30 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $X=BI\cap KL$ and $\angle EDC=\angle EBC=\angle EKL$ since $EDKL$ and $EDBC$ are cyclic $\Longrightarrow$ $KL\parallel BC$
$\Longrightarrow$ $DILX$ is cyclic since $\angle IDL=\angle IXL=\angle IBC$ $\Longrightarrow$ $\angle IDL=\angle IXD$ $\Longrightarrow$ $ID^2=IT.IX$
Let $P'$ be a point in incircle of $\triangle ABC$ such that $XP'$ is tangent to incircle of $\triangle ABC$ $\Longrightarrow$ $IP'^2=ID^2=IT.IX$
Then $P'T\perp BI$ $\Longrightarrow$ $P'=P$ $\Longrightarrow$ $KL,BI$ and the tangent of $P$ to incircle of $\triangle ABC$ are concurrent...

This post has been edited 1 time. Last edited by FabrizioFelen, Nov 1, 2015, 9:32 PM

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suli

#4 h Nov 1, 2015, 11:54 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $Q = BI \cap KL$ . By angle chasing we have $\angle TLQ = \angle DEK = \angle TCB$ , so $QL // BC$ . Thus $\angle IQL = \angle IBC = \angle IDC$ , so $DILQ$ is cyclic. Because $IK = IL = ID$ are equal radii of $\Gamma$ , we have $\angle DQI = \angle LQI$ and furthermore
$\angle KIQ = \angle IKL - \angle IQK = \angle ILK - \angle DLI = \angle DLQ.$ Thus $TILK$ is cyclic as well. Then by Power of a Point, $QT \cdot QI = QK \cdot QL$ . Let $P'$ be the intersection point of the tangent from $Q$ to the circle $\Gamma$ with $P'$ in triangle $IBC$ ; then $QP'^2 = QK \cdot QL = QT \cdot QI$ . Hence $QTP'$ and $QP'I$ are similar and so $\angle QTP' = 90^\circ$ , which means $P = P'$ , and so $BI, KL,$ and the tangent from $P$ to $\gamma$ are concurrent.

This post has been edited 3 times. Last edited by suli, Nov 2, 2015, 12:00 AM

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#5 h Nov 2, 2015, 12:54 AM • 2 Y

Y by Adventure10, Mango247

Let the tangent to $\Gamma$ at $P$ cut $BI$ at $Q.$

From $ID = IE$ , we deduce that $\measuredangle DBI = \measuredangle IBE = \measuredangle IBK.$ Together with $ID = IK$ , it follows that $BI$ is the perpendicular bisector of $\overline{DK}.$

By Reim's Theorem, $KL \parallel BC \implies \measuredangle KLT = \measuredangle BCD = \measuredangle BID = \measuredangle KIT \implies I, T, K, L$ are concyclic. $(\star)$

On account of $QI \perp TP$ , it follows that $Q, T$ are inverse points WRT $\Gamma.$ Then under inversion in $\Gamma, \; (\star) \implies Q, K, L$ are collinear. $\square$

This post has been edited 1 time. Last edited by Dukejukem, Nov 2, 2015, 12:54 AM

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#6 h Nov 2, 2015, 11:43 AM • 3 Y

Y by khyeon, Adventure10, Mango247

800th Post!

Denote $l$ as the tangent to $\Gamma$ at $P$ .
Let $Q_1=KL \cap IB$ and $Q_2=l \cap IB$ . It suffices to show that $Q_1=Q_2$ .

From $IP \perp l$ and $IT \perp TP$ , we have $IT \cdot IQ_2 = r^2$ .
From $\angle DCB = \angle DEB = \angle DLK$ , we have $KL \parallel BC$ .
Therefore, $\angle IQ_1L = \angle IBC = \angle IDL = \angle ILT$ .
Now, $\triangle ITL \sim \triangle ILQ$ , so $IT \cdot IQ_1=r^2$ .
We now have $IQ_1=IQ_2$ , so since they both lie on $IB$ , we have $Q_1=Q_2$ as desired. $\blacksquare$

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khyeon

#8 h Oct 2, 2016, 12:37 PM • 2 Y

Y by Adventure10, Mango247

rkm0959 wrote:

800th Post!

Denote $l$ as the tangent to $\Gamma$ at $P$ .
Let $Q_1=KL \cap IB$ and $Q_2=l \cap IB$ . It suffices to show that $Q_1=Q_2$ .

From $IP \perp l$ and $IT \perp TP$ , we have $IT \cdot IQ_2 = r^2$ .
From $\angle DCB = \angle DEB = \angle DLK$ , we have $KL \parallel BC$ .
Therefore, $\angle IQ_1L = \angle IBC = \angle IDL = \angle ILT$ .
Now, $\triangle ITL \sim \triangle ILQ$ , so $IT \cdot IQ_1=r^2$ .
We now have $IQ_1=IQ_2$ , so since they both lie on $IB$ , we have $Q_1=Q_2$ as desired. $\blacksquare$

nice solution!
drawing was hard but solution was easy

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yunseo

#9 h Nov 24, 2019, 7:51 PM • 2 Y

Y by Adventure10, Mango247

We want to show that when inverted around the incircle, N maps to T.
This is analogous to showing that TIKL cyclic. In other words, we wish to show that D is a reflection of K over BI.
This is true because $\triangle IDM$ and $\triangle IEM$ are congruent. Thus, $\angle DBI = \angle IBK$ . We also know that $IK = ID$ .

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#10 h Nov 25, 2019, 2:19 AM • 2 Y

Y by Adventure10, Mango247

Let $KL$ meet $BI$ at $Q$ . Let $BI$ meet $\Gamma$ at $M,N$ . Note that $ID=IE$ implies $BI$ is the angle bisector of $\angle DBE$ . Also, $\Gamma$ is symmetric about $BI$ , so points $D$ and $K$ are symmetric about $MN$ . Hence, $\square DMKN$ is harmonic. Taking a projection wet $L$ onto $BI$ , we get $(Q,T;M,N)=-1$ . Hence, $T$ lies on the polar of $Q$ , and we are done.

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primesarespecial

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primesarespecial

#11 h Jul 20, 2021, 12:22 PM

Y by

Let $H=LK\cap BI$ .
Firstly we have that $\angle TIK=\angle TID=\angle BID=\angle BED=\angle KED=\angle KLD$ ,where the first equality just came from $T$ being the midpoint of $\widehat {DE}$ and $IK=ID$ .
Now we have $TILK$ is cyclic,giving us
Let $P'$ be a point of tangency from $H$ inside $\triangle BIC$ .
Now ,By power of a point
$HP^2=HK.HL=HT.HI$ and this gives $\triangle THP'$ similar to $\triangle P'HI$ , thus giving $\angle P'TH=90$ and hence $P'=P$ .

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