AP and IM meet on circumcircle

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AP and IM meet on circumcircle

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3 posts

#1 h Feb 9, 2015, 12:01 PM • 2 Y

Y by Adventure10, Mango247

In $\triangle ABC$ , $AB>AC$ . In the circumcircle $(O)$ of $\triangle ABC$ , $M$ is the midpoint of arc $BAC$ . The incircle $(I)$ of $\triangle ABC$ touches $BC$ at $D$ , the line through $D$ parallel to $AI$ intersects $(I)$ again at $P$ . Prove that $AP$ and $IM$ intersect at a point on $(O)$ .

This post has been edited 1 time. Last edited by boblimb, Nov 10, 2015, 12:46 PM
Reason: Fix latex

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2312 posts

#2 h Feb 9, 2015, 12:30 PM • 2 Y

Y by AdithyaBhaskar, Adventure10

My solution:

Let $\ell_P$ be the tangent of $\odot (I)$ through $P$ .
Let $\ell$ be a line passing through $I$ and perpendicular to $AI$ .
Let $B'=\ell_P \cap AC, C'=\ell_P \cap AB$ and $T$ be the tangent point of $A-$ mixtilinear circle with $\odot (ABC)$ .

Since $\ell_P$ is the reflection of $BC$ in $\ell$ ,
so $\ell_P$ is anti-parallel to $BC$ WRT $\angle BAC \Longrightarrow \triangle AB'C' \sim \triangle ABC$ ,
hence $AP$ is the isogonal conjugate of $A-$ Nagel line WRT $\angle BAC \Longrightarrow T\in AP$ .

Since It's well-known that $T \in MI$ (see incenter of triangle) ,
so we get $T$ is the intersection of $AP$ and $IM$ which is lie on $\odot (ABC)$ .

Q.E.D

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9775 posts

jayme

#3 h Feb 10, 2015, 8:13 AM • 2 Y

Y by AdithyaBhaskar, Adventure10

Dear Mathlinkers,

With another regard…

1. N the antipole of M wrt (O)
A*, X the second point of intersection of MI, DN wrt (O)
U the second point of intersection of the parallel to BC with (O)
2. according to the Reim’s theorem, A*, X, D, I are concyclic on (2)
3. P’ the second point of intersection of A*A with (2)
4. A*M is the A*-inner bissector of A*AU (http://jl.ayme.pagesperso-orange.fr/Docs/Mixtilinear1.pdf, p. 28)
In consequence, P’ is on (I)
5. according to the Reim’s theorem, DP’ // AIN and P’=P.

Sincerely
Jean-Louis

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#4 h Sep 9, 2015, 3:17 AM • 3 Y

Y by AdithyaBhaskar, Adventure10, Mango247

Let $I_a,I_b,I_c$ be the excenters of $\triangle ABC$ . Let the incircle touch $AC,AB$ at $E,F$ , $H$ be the orthocenter of $\triangle DEF$ , let $J$ be the midpoint of $DH$ , $G$ the midpoint of $EF$ , $(N)$ the nine-point circle of $\triangle DEF$ .

It is well-known that $IG//JH$ , $IG=JH$ hence $IGHJ$ is a parallelogram $\Rightarrow IJ//GH$ .
Now $M$ is the second intersection of the nine-point circle of $\triangle I_aI_bI_c$ $\odot(ABC)$ with $I_bI_c$ , hence $M$ is the midpoint of $I_bI_c$ . We now consider the homothety carrying $\triangle DEF$ to $\triangle I_aI_bI_c$ , which also carries $H\rightarrow I$ , $G\rightarrow M$ . Hence under the homothety we have $HG//IM\Rightarrow J,I,M$ collinear.

Note that $P,H$ are reflections of each other across $EF\Rightarrow IGPJ$ is a isoceles trapezium and hence cyclic.

We now consider the inversion with respect to the incircle, note that the nine point circle $(N)$ of $\triangle DEF$ is sent to $\odot(ABC)$ . Thus $G$ is sent to $A$ . $IGJP$ cyclic implies $J$ , which lies on $(N)$ , is sent to a point on $\odot(ABC)$ which lies on line $AP$ , as well as $IJ\equiv IM$ .

This post has been edited 1 time. Last edited by buzzychaoz, Sep 9, 2015, 3:21 AM
Reason: Typo

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#5 h Sep 9, 2015, 4:06 AM • 2 Y

Y by AdithyaBhaskar, Adventure10

Invert about the incircle, and denote inverses with a $'$ . Suppose the intouch triangle is $DFG$ ; since the tangents from $F$ and $G$ intersect at $A$ , $A'$ is the midpoint of $FG$ , i.e. $(ABC)$ inverts to the nine-point circle of $(DFG)$ . Note that $MI$ hits the circumcircle again at the tangency point of the $A$ -mixtilinear incircle (see #3 here). $\angle IAM=\angle IM'A'=90^{\circ}$ and since $T$ lies on $IM$ , then $T'$ is the antipode of $A'$ with respect to said nine-point circle, which is the midpoint of $DH$ , where $H$ is the orthocenter of the intouch triangle. Thus, we want to prove that $PA'IT'$ is cyclic. This is equivalent to the following problem:

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$ . Suppose the intersection of $AH$ with the circumcircle is $K$ , the midpoint of $BC$ is $E$ , and the midpoint of $AH$ is $J$ . Prove that $JOEK$ is cyclic.

Invert about the circumcircle, and use complex numbers. We set $(ABC)$ as the unit circle. Note that $e'=\frac{2bc}{b+c}$ , $k=-\frac{bc}{a}$ , and $j'=\frac{1}{\frac{1}{a}+\frac{\frac{1}{b}+\frac{1}{c}}{2}}=\frac{2abc}{ab+ac+2bc}$ . Then we just have to show that $e', k, j'$ are collinear which is relatively easy.

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1979 posts

#6 h Sep 9, 2015, 9:33 PM • 2 Y

Y by Adventure10, Mango247

No $\sqrt bc$ inversion yay!

Let $X$ be the second intersection point of $MI$ with the circumcircle, $D'$ be the antipodal of $D$ in the incircle and $E,F$ be the touch points of the incircle with $AC,AB$ respectively. Let $AP$ meet $EF$ at $T$ and $AD'$ meet $EF$ at $Y$ . Notice that $X$ is the touch point of the $A$ mixitilinear incircle with the circumcircle. It is well known that $AX$ and $AD'$ are isogonal rays. By some angle chase we can see that $\triangle FPE$ is congruent to $\triangle ED'F$ . which gives $AP$ and $AD'$ isogonal so $A,P,T,X$ (4869 lol) are collinear. Done.

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851 posts

#7 h Sep 10, 2015, 12:31 AM • 2 Y

Y by Adventure10, Mango247

anantmudgal09 wrote:

No $\sqrt bc$ inversion yay!

Lol, I (and it looks like leeky), both inverted about the incircle. I think $\sqrt bc$ makes this harder, because then you have to deal with a mixtilinear excircle

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123 posts

#8 h Sep 10, 2015, 6:23 AM • 2 Y

Y by Adventure10, Mango247

hint ; let AH is the altitude and z is the intewrestion AH and PD and Q bethe symetric point of d betwwn i prove that the feodbakh point passes from pQ

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228 posts

ATimo

#9 h Sep 10, 2015, 4:35 PM • 3 Y

Y by Mehdijahani1998, Adventure10, Mango247

My solution:
Suppose that the $A$ -mixtilinear touches $\odot O$ at $Q$ . Then $IM$ passes trough $Q$ . So we must say that $AP$ passes trough $Q$ . Suppose that $X$ is the intersection point of $A$ excircle and $BC$ . Then we have $\angle CAQ=\angle BAX$ . So we must say that $\angle CAP=\angle BAX$ . Suppose that $K$ is the intersection point of $\odot I$ and $DI$ . Then $AX$ passes trough $K$ . Suppose that $E$ and $F$ are the intersecting points of $\odot I$ with $AC$ and $AB$ respectively. $DP$ is the altitude in triangle $\triangle EDF$ , so $\angle FDK=\angle EDP$ . Now from $\triangle AEP=\triangle AFK$ we get that $\angle FAK=\angle EAP$ . And we are done.

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Generic_Username

#10 h May 27, 2018, 12:00 AM • 2 Y

Y by Adventure10, Mango247

It is well known that $IM$ intersects the circumcircle again at the $A$ -mixtilinear touch point; call this $X.$ We wish to show that $A,P,X$ are collinear.

Let $D'$ be the antipode of $D$ on $\odot (I).$ Then $PD' \perp AI$ and $IP=ID'$ so $\triangle PAI\cong \triangle D'AI$ and $AP,AD'$ are isogonal. But it is well known that $AD'$ is the $A$ -extouch cevian in $\odot (O)$ and that this is isogonal to $AX,$ from which the conclusion follows.

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#11 h May 27, 2018, 12:30 AM • 2 Y

Y by Adventure10, Mango247

This follows directly from the fact that circumcevian triangle of the orthocenter of the intouch triangle wrt incircle is homothetic to $\Delta ABC$

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277 posts

#12 h Oct 28, 2024, 5:13 PM

Y by

Let $\alpha=\frac{\angle A}{2}$ , and define $\beta, \gamma$ similarly.
Note that $\angle PDC = \alpha + 2 \beta$ , so $\angle PDI = \gamma - \beta$ , so $\angle (PI, BC)=90^\circ - 2(\gamma-\beta) = 90^\circ - 2\gamma + 2\beta$ .
Also, $\angle (AO,BC)=90^\circ - 2\gamma + 2\beta = \angle (PI, BC)$ , so $AO // PI$ .

It is not hard to see that $O$ and $I$ lie on the same side of line $AP$ (since $\angle CAP < \angle CAI < \angle CAO < 90^\circ$ ), so segments $AP,OI$ do not intersect.
Now let $X$ be the centre of homothety taking $A$ to $P$ , and $O$ to $I$ . Note that this is a positive homothety.
Since $A$ lies on $\odot(O)$ and $P$ lies on $\odot(I)$ , $X$ is in fact the exsimilicenter of $\odot(O)$ and $\odot(I)$ .
It is then well-known that $APX$ passes through the $A$ -mixtilinear intouch point $T$ . But $IM$ is also known to pass through $T$ , hence we are done. $\square$

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#13 h Oct 28, 2024, 10:41 PM

Y by

It is well-known that $IM\cap(ABC)$ is the $A$ -mixti touch, now reflect about $AI$ , we see that $P$ is sent to $D'$ , $D$ -antipode in the incircle by rectangle property. We want $AD'$ to be the Nagel Line but this is well-known.

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