IMO ShortList 2002, geometry problem 7

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3647 posts

orl

#1 h Sep 28, 2004, 1:00 PM • 14 Y

Y by narutomath96, Davi-8191, valsidalv007, A-Thought-Of-God, samrocksnature, mathematicsy, donotoven, jhu08, Adventure10, mathmax12, Mango247, Rounak_iitr, Funcshun840, drago.7437

The incircle $\Omega$ of the acute-angled triangle $ABC$ is tangent to its side $BC$ at a point $K$ . Let $AD$ be an altitude of triangle $ABC$ , and let $M$ be the midpoint of the segment $AD$ . If $N$ is the common point of the circle $\Omega$ and the line $KM$ (distinct from $K$ ), then prove that the incircle $\Omega$ and the circumcircle of triangle $BCN$ are tangent to each other at the point $N$ .

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orl

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orl

#2 h Sep 28, 2004, 1:00 PM • 6 Y

Y by samrocksnature, Adventure10, jhu08, sabkx, dxd29070501, Mango247

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

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grobber

7849 posts

grobber

#3 h Sep 29, 2004, 1:20 AM • 12 Y

Y by eshan, Anar24, naw.ngs, Supercali, BobaFett101, samrocksnature, hakN, Adventure10, jhu08, agwwtl03, Mango247, MS_asdfgzxcvb

I don't like this solution, but I couldn't find a better one this late at night (or this early in the morning; it's 4:15 AM here

).

Let $S=KA\cap \Omega$ , and let $T$ be the antipode of $K$ on $\Omega$ . Let $X,Y$ be the touch points between $\Omega$ and $CA,AB$ respectively.

The line $AD$ is parallel to $KT$ and is cut into two equal parts by $KS,KN,KD$ , so $(KT,KN;KS,KD)=-1$ . This means that the quadrilateral $KTSN$ is harmonic, so the tangents to $\Omega$ through $K,S$ meet on $NT$ . On the other hand, the tangents to $\Omega$ through the points $X,Y$ meet on $KS$ , so $KXSY$ is also harmonic, meaning that the tangents to $\Omega$ through $K,S$ meet on $XY$ .

From these it follows that $BC,XY,TN$ are concurrent. If $P=XY\cap BC$ , it's well-known that $(B,C;K,P)=-1$ , and since $\angle KNP=\angle KNT=\frac{\pi}2$ , it means that $N$ lies on an Apollonius circle, so $NK$ is the bisector of $\angle BNC$ .

From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$ , we get $B'C'\|BC$ , so there's a homothety of center $N$ which maps $\Omega$ to the circumcircle of $BNC$ .

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darij grinberg

6555 posts

darij grinberg

#4 h Sep 30, 2004, 7:58 AM • 4 Y

Y by samrocksnature, Adventure10, jhu08, Mango247

Grobber, I like your solution! Just to clarify a few points which took me some time to understand:

grobber wrote:

and since $\angle KNP=\angle KNT=\frac{\pi}2$ ,

This is because the segment KT is a diameter of $\Omega$ .

grobber wrote:

From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$ , we get $B'C'\|BC$ ,

Is this trivial? The only explanation I have is to use the intersecting secant and tangent theorem, which yields $BB^{\prime} \cdot BN = BK^2$ and $CC^{\prime} \cdot CN = CK^2$ , from what we conclude $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BK^2}{CK^2}$ , but since the line NK bisects the angle BNC, we have $\frac{BK}{CK} = \frac{BN}{CN}$ , so that we get $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BN^2}{CN^2}$ , and thus $\frac{BB^{\prime}}{CC^{\prime}} = \frac{BN}{CN}$ , what immediately implies B'C' || BC.

As for another solution of the problem, see http://www.mathlinks.ro/Forum/viewtopic.php?t=14741 .

Darij

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sprmnt21

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sprmnt21

#5 h Sep 30, 2004, 9:27 AM • 4 Y

Y by samrocksnature, Adventure10, jhu08, Mango247

I like very mutch Grobber's solution too.

Another way to see that B'C'//BC, once we know that <BNK = <CNK is the following: <B'C'K = <B'NK = <C'NK = <C'KC.

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darij grinberg

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darij grinberg

#6 h Sep 30, 2004, 11:19 AM • 4 Y

Y by samrocksnature, jhu08, Adventure10, Mango247

Indeed, I was stupid...

Darij

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Agr_94_Math

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Agr_94_Math

#7 h May 1, 2010, 10:46 AM • 3 Y

Y by samrocksnature, jhu08, Adventure10

An alternate solution using some computation and inversion :
Let us prove that for a circle through $B<C$ tangent to the incircle of triangle $ABC$ at point $N'$ , $K, M, N'$ are collinear.
Now, it is enough if we prove that angles $BKM$ and $BKN'$ are equal.
After a few computations, we get $\tan {BKM} = \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$
Now apply an inversion with center $K$ and radius $BK$ .
Thereagain, after a few computations, we get $tan(BKN') = tan(BKM) = \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$ .
So we are done.

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jayme

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jayme

#8 h May 1, 2010, 1:05 PM • 4 Y

Y by vsathiam, samrocksnature, jhu08, Adventure10

Dear Mathlinkers,
this problem was already posted, but where?
In order to have a complete synthetic proof, we can observe that NK goes through the A-excenter...
Sincerely
Jean-Louis

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Agr_94_Math

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Agr_94_Math

#9 h May 1, 2010, 2:26 PM • 3 Y

Y by samrocksnature, jhu08, Adventure10

Dear jayme,
I too on trying for a synthetic solution, tried by the same collinearity.
That is the midpoint of the altitude , tangency point of the incircle with the corresponding side and the corresponding excenter are collinear.
This is true by considering the diametrically opposite point of the tangency point of the excircle with $BC$ and drawing a parallel through it to $BC$ and using homothety and semiprojection result.

But I was not able to finish the problem using this synthetic idea.
So, could you please tell your complete solution?

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Virgil Nicula

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Virgil Nicula

#10 h May 2, 2010, 2:00 AM • 4 Y

Y by hatchguy, samrocksnature, jhu08, Adventure10

I observed now this old and nice problem. I''ll search its synthetical proof and return soon. Yes Jayme,

your remark is very interesting. Indeed, if denote the point $L$ where the $A$ -exincircle touches the side $[BC]$ ,

then $M\in I_aK\cap IL$ because $\{\begin{array}{c} \frac {KD}{KL}=\frac {s-a}{a}=\frac {h_a}{2r_a}=\frac {MD}{LI_a}\\\\ \frac {LD}{LK}=\frac sa=\frac {h_a}{2r}=\frac {MD}{IK}\end{array}$ .

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Luis González

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Luis González

#11 h May 2, 2010, 9:11 PM • 8 Y

Y by eziz, amar_04, samrocksnature, jhu08, kamatadu, Adventure10, Mango247, MS_asdfgzxcvb

Let $N'$ be the tangency point of the circle $\omega$ passing through $B,C$ with $\Omega.$ Let $U$ denote the antipode of $K$ WRT $\Omega$ and $V$ the tangency point of the A-excircle $(I_a)$ with $BC.$ According to this topic, $N'U,N'K$ bisects $\angle AN'V$ internally and externally. Let $N''$ be the image of $N'$ under the homothety with center $A$ that takes $\Omega$ and $(I_a)$ into each other. Then $UN' \parallel VN''$ $\Longrightarrow$ $N'K \perp N''V$ $\Longrightarrow$ $\triangle N'VN''$ is isosceles with apex $N',$ which implies that $I_a \in N'K,$ due to $I_aV=I_aN''=r_a.$ But since $M \in KI_a,$ we deduce that $N \equiv N'.$

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ThinkFlow

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ThinkFlow

#12 h Dec 30, 2011, 1:49 AM • 5 Y

Y by samrocksnature, Bubu-Droid, jhu08, Adventure10, Mango247

Here is a somewhat longer solution...

Solution

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SnowEverywhere

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SnowEverywhere

#13 h Jan 18, 2012, 11:43 PM • 4 Y

Y by samrocksnature, jhu08, Adventure10, Mango247

Let $I$ be the incenter and $I_A$ be the $A$ -excenter of triangle $\triangle{ABC}$ . Let $\omega$ be the incircle, $\Gamma$ be the $A$ -excircle and let $\Omega$ be the circle with diameter $AD$ . The homothety with center $A$ sending $\omega$ to $\Gamma$ takes $K$ to $K'$ where the tangent at $K'$ to $\Gamma$ is parallel to $BC$ , perpencular to $AD$ and hence parallel to the tangent to $\Omega$ at $A$ . Hence the negative homothety taking $\Omega$ to $\Gamma$ takes $A$ to $K'$ and therefore has center on $AK'$ . The center of this negative homothety also lies on the common tangent $BC$ to $\Omega$ and $\Gamma$ and therefore the center of the negative homothety is $K$ . This implies that $M$ , $I_A$ and $K$ are collinear. Now let the circle $\gamma$ with diameter $II_A$ intersect $\omega$ at $X$ and $Y$ . Note that this implies that $XI_A$ and $YI_A$ are tangent to $\omega$ and hence that $XY$ is the polar of $I_A$ with respect to $\omega$ . Now let $XY$ intersect $BC$ at $T$ . Let $N'$ denote the point on $\omega$ such that $TN'$ is tangent to $\omega$ and $N' \neq K$ . Now note that since $T$ lies on the polar of $I_A$ with respect to $\omega$ , $I_A$ lies on the polar of $T$ with respect to $\omega$ , which is $N'K$ . Hence $I_A$ , $N'$ , $K$ and $M$ are collinear and thus $N=N'$ . Since $BI_A$ and $CI_A$ bisect the exterior angles of the triangle at $B$ and $C$ , respectively, $\gamma$ passes through $B$ and $C$ . Hence $BCYX$ is cyclic and, by power of a point with respect to $\gamma$ and $\omega$ , $TN^2 = TX \cdot TY = TB \cdot TC$ . This implies that the circumcircle of $\triangle{BCN}$ is tangent to $\omega$ at $N$ , as desired.

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Zhero

2043 posts

Zhero

#14 h Mar 21, 2012, 12:44 AM • 3 Y

Y by samrocksnature, jhu08, Adventure10

WLOG, let $AB<AC$ (the case $AB=AC$ is trivial.) Let $a=BC$ , $b=CA$ , $c=AB$ . Let $I$ be the the center of $\Omega$ , let $r$ be its radius, let $\Omega$ be tangent to $AB$ and $AC$ at $X$ and $Y$ , respectively, and let $Z$ be the point where $XY$ meets $BC$ . Let $\omega_A$ be the $A$ -excircle of $ABC$ , let $I_A$ be its center, $r_A$ be its radius, and let $J$ be its tangency point with $BC$ , We first claim that $M$ , $K$ , and $J$ are collinear.

Let $J'$ be its antipode in $\omega_A$ . A homothety centered at $A$ taking $\Omega$ to $\omega_A$ sends $K$ to $J'$ . Thus, a homothety centered at $K$ mapping $\Omega$ to $\omega_A$ must send $A$ to $J'$ as well. Line $AD$ is sent to a line parallel to it passing through $J'$ , i.e., line $JJ'$ . Since $J,D$ both line on $BC$ , the homothety must map $D$ to $J$ . Thus, the homothety must map the midpoint of $AD$ to the midpoint of $JJ'$ , so $M$ , $K$ , and $I_A$ must be collinear.

Now let $P$ be the midpoint of $ZK$ . Because $(Z,B,K,C)$ is harmonic, we must have $PK^2 = PB \cdot PC$ (this can also be verified by computing the lengths directly, given $ZB$ , which can be found through Menelaus.) Consequently, if the tangent to $\Omega$ from $P$ distinct from $PB$ is tangent to $\Omega$ at $N'$ , we have $PN'^2 = PK^2 = PB \cdot PC$ , whence $PN'$ is tangent to the circumcircle of $\triangle N'BC$ , i.e., $N'$ is the tangency point of the circle through $B,C$ tangent to $\Omega$ . We wish to show that $N'$ , $K$ , and $I_A$ are collinear.

We claim that $\triangle PIK \sim \triangle I_a KJ$ . The result would then follow, for we would have $\angle I_a KJ = \angle ZIK = 90^{\circ} - \angle N'KI = \angle N'KD$ . Since both triangles are right, it suffices to prove $PK/KI = I_aJ / KJ \iff PK \cdot KJ = r r_A$ . Now, $rr_A = \frac{K^2}{s(s-a)} = (s-b)(s-c)$ , where $K$ is the area and $s$ is the semiperimeter of $\triangle ABC$ . Also, $KJ = a - BK - CJ = a - 2(s-b) = b-c$ , and
$\frac{ZB}{ZB + a} = \frac{KB}{KC} = \frac{s-b}{s-c} \implies ZK = ZB + s-b = \frac{(s-b)(a+b-c)}{b-c}.$
Thus,
$PK \cdot KJ = \frac{ZK}{2} \cdot KJ = \frac{(s-b)(s-c)}{b-c} \cdot (b-c) = (s-b)(s-c) = r r_A,$
so we are done.

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v_Enhance

6870 posts

v_Enhance

#15 h Aug 20, 2012, 6:45 PM • 45 Y

Y by Amir Hossein, sjaelee, aopsqwerty, Einstein314, eshan, AlgebraFC, Delray, vsathiam, Durjoy1729, Pluto1708, thczarif, valsidalv007, amar_04, Nymoldin, srijonrick, anonman, Wizard0001, like123, Gaussian_cyber, HamstPan38825, samrocksnature, blackbluecar, hsiangshen, jhu08, SSaad, lneis1, kac3pro, myh2910, David-Vieta, channing421, yee5487, sabkx, dili96, Stuffybear, Mogmog8, Adventure10, Mango247, Ritwin, fearsum_fyz, Frank25, MS_asdfgzxcvb, CreyJonhson, and 3 other users

Let $I_A$ be the $A$ -excenter with tangency points $X_A$ , $X_B$ , and $X_C$ to $BC$ , $CA$ and $AB$ , respectively. Define $P$ to be the midpoint of $KI_A$ . Let $r$ be the radius of the incircle and $R$ the radius of the $A$ -excircle.

It is well-known that $M$ , $K$ and $I_A$ are collinear. We claim that $NBPC$ is cyclic; it suffices to prove that $2BK \cdot KC = 2KP \cdot KN = KN \cdot KI_A$ . On the other hand, by Power of a Point we have that $I_AK \left( I_AK + KN \right) = II_A^2 -r^2 \implies KN \cdot KI_A = II_A^2 - r^2 - I_AK^2$ Now we need only simplify the right-hand side using the Pythagorean Theorem; it is $\left( (r+R)^2 + KX_A^2 \right) - r^2 - \left( R^2 + KX_A^2 \right) = 2Rr$ . So it suffices to prove $Rr = (s-b)(s-c)$ , which is not hard.

Now, since $P$ is the midpoint of minor arc $\widehat{BC}$ of $(NBC)$ (via $BK=CX_A$ ), while the incircle is tangent to segment $BC$ at $K$ , the conclusion follows readily.

$[asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-4.18; real xmax=3.94; real ymin=-2.38; real ymax=2.34; /* Initialize Objects */ pair A = (-1.5, 2.0); pair B = (-2.0, 0.0); pair C = (0.5, 0.0); pair I = incenter(A,B,C); pair D = foot(A,B,C); pair M = midpoint(A--D); pair K = foot(I,B,C); pair N = (2)*(foot(I,M,K))-K; pair I_A = (2)*(IntersectionPoint(circumcircle(A,B,C),Line(A,I,lisf),1))-I; pair X_B = foot(I_A,A,B); pair X_C = foot(I_A,A,C); path CircleNBC = circumcircle(N,B,C); pair X_A = foot(I_A,B,C); path NI_A = N--I_A; pair P = IntersectionPoint(CircleNBC,NI_A,1); /* Draw objects */ draw(A--B, rgb(0.0,0.6,0.6)); draw(B--C, rgb(0.0,0.6,0.6)); draw(C--A, rgb(0.0,0.6,0.6)); draw(incircle(A,B,C), rgb(0.0,0.8,0.0) + linewidth(1.2)); draw(A--D, dotted); draw(CircleNBC, rgb(0.2,0.8,0.0)); draw(CirclebyPoint(I_A,X_A), rgb(0.0,0.8,0.0) + linewidth(1.2) + dashed); draw(B--X_B, rgb(0.6,0.6,0.6) + linetype("4 4")); draw(C--X_C, rgb(0.6,0.6,0.6) + linetype("4 4")); draw(NI_A, rgb(0.0,0.0,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(I); dot(D); dot(M); dot(K); dot(N); dot(I_A); dot(X_B); dot(X_C); dot(X_A); dot(P); /* Label points */ label("$A$", A, lsf * dir(135)); label("$B$", B, lsf * dir(135)); label("$C$", C, lsf * dir(45)); label("$I$", I, lsf * dir(45)); label("$D$", D, lsf * dir(135)); label("$M$", M, lsf * dir(45)); label("$K$", K, lsf * dir(45)); label("$N$", N, lsf * dir(115)); label("$I_A$", I_A, lsf * dir(45)); label("$X_B$", X_B, lsf * dir(170)); label("$X_C$", X_C, lsf * dir(45)); label("$X_A$", X_A, lsf * dir(45)); label("$P$", P, lsf * dir(45)); /* Clip the image */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

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OronSH

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OronSH

#111 h Apr 28, 2024, 2:26 AM • 2 Y

Y by bjump, Zhaom

By a well known lemma $KM$ passes through the $A$ -excenter $J.$ If $P$ is the midpoint of $KN$ then $\angle IPJ=90^\circ$ where $I$ is the incenter so $B,I,C,J,P$ are concyclic by fact 5. Thus $BK\cdot CK=PK\cdot JK=NK\cdot XK$ where $X$ is the midpoint of $JK$ so $B,N,C,X$ are concyclic. But then $X$ is the bottommost point on this circle since $K$ and the foot from $J$ to $BC$ are equidistant from the midpoint of $BC.$ Thus the homothety at $N$ sending $K$ to $X$ must send the incircle to circle $(BNCX),$ done.

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Zhaom

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Zhaom

#112 h Apr 30, 2024, 8:16 PM • 1 Y

Y by OronSH

It is well known that $\overline{KM}$ passes through $I_A$ the $A$ -excenter. Now, consider the circle $\omega$ through $B$ and $C$ orthogonal to $\Omega$ . This contains $B'$ and $C'$ the inverses of $B$ and $C$ with respect to $\Omega$ . Then, the center of $\omega$ is the intersection of $\ell_1$ the perpendicular bisector of $\overline{BC}$ and $\ell_2$ the perpendicular bisector of $\overline{B'C'}$ . Homothety at $K$ with factor $2$ sends $\ell_1$ to the perpendicular from $I_A$ to $\overline{BC}$ and $\ell_2$ to $\overline{AI}$ , which intersect at $I_A$ , so the center of $\omega$ is the midpoint of $\overline{KI_A}$ . Therefore, inversion around $\omega$ swaps $(BKC)$ and $(BNC)$ fixing $\Omega$ . Since $(BKC)$ is tangent to $\Omega$ , we see that $(BNC)$ is tangent to $\Omega$ .

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Pyramix

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Pyramix

#113 h Jun 17, 2024, 6:11 PM

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Let $S$ be the antipode of $K$ in $\Omega$ and $T=EF\cap BC$ where $E,F$ are the $B-,C-$ touchpoints of $\Omega$ .

Claim. $S,T,N$ are collinear.
Proof. Projecting $(A,D;M,\infty_{AD})=-1$ from $K$ onto $\Omega$ gives $(AK\cap\Omega,K;N,S)=-1$ . Let $L=AK\cap\Omega$ . So, $LNKS$ is harmonic quadrilateral. So, tangents at $L,K$ meet on line $SN$ . But since tangents to $E,F$ meet on line $LK$ , it means $LEKF$ is harmonic quadrilateral. Hence, $T=EF\cap BC$ is the meeting point of tangents at $L,K$ . So, $T\in SN$ , as claimed. $\blacksquare$

Since $KN\perp NS$ , it means $N$ is the foot from $K$ to $ST$ . Extend $NK$ to meet $(BCN)$ again at $P$ . Note that $(T,K;B,C)=-1$ which means $(BCN)$ is the Apollonius circle for $KT$ as $\angle TNK=90^\circ$ . Hence, $\frac{BN}{NC}=\frac{BK}{KC}$ , which means $NK$ is the angle-bisector of $\angle BNC$ . So, $NK$ extended to meet the circle $(BNC)$ is the midpoint of arc $BC$ not containing $N$ . Since $\Omega$ is tangent to $BC$ , $\Omega$ is tangent to $(BNC)$ by Shooting Lemma.

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fearsum_fyz

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fearsum_fyz

#114 h Jun 22, 2024, 9:43 AM • 1 Y

Y by GeoKing

We can remove $AD$ by replacing $KM$ with $KI_A$ , where $I_A$ is the $A$ -excenter.

By shooting lemma, it would suffice to show that $TK \cdot TN = TB^2 = TC^2$ where $T$ is the midpoint of arc $\widehat{BC}$ of $(BCN)$ . We will show this using phantom points.

Let $T'$ be the intersection of the perpendicular bisector of $BC$ and line $KI_A$ . Let $A'$ be the midpoint of $BC$ and $X$ be the $A$ -extouch point. Since $T'A'$ and $IK$ are both perpendicular to $BC$ , they are parallel, and hence $\angle{NKI} = \angle{KT'A'} = \angle{T'I_AX} = x$ (say).
By the converse of midpoint theorem in $\Delta{KI_AX}$ , $T'$ is the midpoint of $KI_A$ . This yields:

$KT' \cdot KN = \frac{1}{2} KI_A \cdot KN = \frac{1}{2} \cdot KI_A \cdot \frac{\sin{(180^{\circ} - 2x)}}{\sin{x}} \cdot r = KI_A \cdot \cos{x} \cdot r = r_a \cdot r = (s - b) \cdot (s - c) = KB \cdot KC$

implying that $N, B, C, T'$ are concyclic as desired.

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bjump

989 posts

bjump

#115 h Jul 3, 2024, 10:43 PM • 1 Y

Y by OronSH

Xoink phone write ups
Let $MK$ meet the perpendicular bisector of $BC$ at $E$ note that by the shooting lemma $E \in (BCN)$ . Let $O$ be the circumcenter of $(BCN)$ . Then note that since $ON=OE$ and $IK=IN$ and $\angle NKI = \angle NEO$ . Since $\triangle KNI \sim \triangle ENO$ and $N$ , $K$ , and $E$ are collinear. This implies $O$ , $I$ , $N$ collinear, and we are finished.

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MELSSATIMOV40

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MELSSATIMOV40

#116 h Jul 24, 2024, 7:53 PM

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MELSSATIMOV40

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MELSSATIMOV40

#117 h Jul 24, 2024, 7:54 PM

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MELSSATIMOV40 wrote:

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Eka01

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Eka01

#118 h Aug 27, 2024, 5:40 PM • 2 Y

Y by Sammy27, SuperBarsh

By the midpoints of altitudes lemma, $KN$ passes through the $A$ excenter $I_A$ so now we remove the extraneous altitude. Now we take the tangent to the incircle at $N$ and show it is tangent to $(BCN)$ as well. Let the tangent intersect $BC$ at $T$ and let $IT \cap KN=X$ . Since $IT$ is the perpendicular bisector of $KN$ , X is the midpoint of $KN$ and it lies on the circle with diameter $II_A$ which is $(BIC)$ .
Since $\Delta IKT$ is right angled at $T$ and $X$ is foot of altitude from $K$ to the hypotenuse, it follows by similarity that $TK^2=TX.TI$ .
Now by Power of Point,
$TN^2=TK^2=TX.TI=TB.TC$ So $TN$ is indeed tangent to the circle $(BCN)$ .

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mcmp

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mcmp

#119 h Nov 5, 2024, 7:22 AM

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Someone PLEASE stop me from becoming too reliant on projective stuff :noo:

Relabel several of the points to match with the diagram.
Funny diagram
Let $A’=(DEF)\cap\overline{AD}\neq D$ , $T’$ the pole of $\overline{A’D}$ and $T$ the midpoint of $AD$ . I claim that $\overline{KT}$ tangent to $(DEF)$ , and $KT^2=TB\cdot TC$ which will finish. First notice that $\overline{KD’}$ passes through $T$ , since $(A’D;KD’)\stackrel{D}{=}(AX;M\infty_{AX})=-1$ , however since $A$ is the pole of $\overline{EF}$ , the polar of $A$ which is just $\overline{EF}$ also passes through $T’$ , the pole of $\overline{A’D}$ . Hence by the midpoints of harmonic bundles lemma $TD^2=TD\cdot TT’=TB\cdot TC$ . So it suffices to show $KT=TD$ .

Now $\measuredangle D’KD=90^{\circ}$ , so $\triangle T’KD$ has circumcentre $T$ . Hence we indeed have $KT=DT$ . Since $K\in(DEF)$ , we must also have that $KT$ tangent to $(DEF)$ and $(BKC)$ as desired. :yoda:

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Aiden-1089

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Aiden-1089

#120 h Nov 12, 2024, 8:24 AM

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Let $I$ be the incentre, $\Delta KPQ$ be the intouch triangle, and let $R=PQ \cap BC$ .
Let $K'$ be the antipode of $K$ in $\Omega$ .
Let $\Omega_A$ be the $A$ -excircle with centre $I_A$ , $L$ be the $A$ -extouch point, $L'$ be the antipode of $L$ in $\Omega_A$ .
Recall that $A-K-L'$ and $A-K'-L$ . Since $I_A$ is the midpoint of $LL'$ , there exists negative homothety centred at $K$ takes $D$ to $L$ , $A$ to $L'$ , and so $M$ to $I_A$ . Thus $M-K-I_A$ .

Note that $AK$ is the polar of $R$ wrt $\Omega$ , so $AK \perp IR$ . It follows that $\Delta IKR \sim \Delta KLL'$ .
Let $T$ be the midpoint of $KR$ , then we have $\Delta IKT \sim \Delta KLI_A$ , so $IT \perp KI_A$ , and this implies that $KI_A$ is the polar of $T$ wrt $\Omega$ .
Now $N$ lies on $KI_A$ , so $TN$ is tangent to $\Omega$ at $N$ . Then since $(B,C;K,R)=-1$ , we have $TB \cdot TC = TK^2 = TN^2$ .
Thus $TN$ is also tangent to $(BCN)$ , so we are done. $\square$

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Thelink_20

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Thelink_20

#121 h Nov 19, 2024, 3:06 PM

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Lemma 1: If $P'$ is the midpoint of the arc $\overarc{BC}$ in $(BNC)$ , then $(BNC)$ is tangent to $\Omega\iff N-K-P'$ .
Proof: Inverting by $P'$ fixing $B$ and $C$ sends $BC\longleftrightarrow (BNC)$ and $\Omega$ is tangent to $(BNC)\iff\Omega\longleftrightarrow\Omega\iff K\longleftrightarrow N\iff N-K-P' \$ $_{\blacksquare}$

Lemma 2: $M-K-I_A$ .
Proof: If $K''$ is the antipode of $K'$ on the excircle, then the homothety from $A$ sending the incircle to th excircle yields $A-K-K''$ thus the homothety from $K$ mapping $AD\mapsto K'K''$ maps $M\mapsto I_A\Rightarrow M-K-I_A \$ $_{\blacksquare}$

Lemma 3: If $P=(BNC)\cap MK$ , then $KP=I_AP$ .
Proof: Let $Q=(BIC)\cap MK$ . $\angle IQK=\angle IQI_A=\angle IBI_A=90^{\circ}\implies NQ=KQ$ . But $NK\cdot KP=BK\cdot CK=KQ\cdot KI_A=\frac{NK}{2} \cdot KI_A\implies KP=\frac{KI_A}{2}=I_AP \$ $_{\blacksquare}$

Now notice that $P$ lies on the perp. bissector of $KK'$ because $K'I_A\perp BC$ , thus it lies on the perp. bissector of $BC\implies \boxed{P=P'}$ , but that means we are done because clearly $N-K-P' \$ $_{\blacksquare}$ .

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Scilyse

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Scilyse

#122 h Dec 22, 2024, 9:54 AM

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If $AB = AC$ then the conclusion is obvious since $N$ lies on the perpendicular bisector of $\overline{BC}$ ; thus without loss of generality assume that $AB < AC$ . Let $I_A$ be the $A$ -excentre, which is known to lie on line $MK$ , let $I$ be the incentre, and let the tangent to $\Omega$ at $N$ intersect line $BC$ at $T$ . Further let $M_A$ be the midpoint of arc $BC$ of $(ABC)$ not containing $A$ , let $\ell$ be the tangent to $(ABC)$ at $M_A$ , let $K'$ be the foot from $I_A$ to $BC$ and let $L$ be the midpoint of $\overline{AC}$ . As $T$ is the pole of line $NK$ in $\Omega$ , we have $TI \perp KI_A$ .

Orient $BC$ parallel to the $x$ -axis; now $\operatorname{slope}(TI) = \frac{IK}{TK} = \frac{r}{TK}$ and $\operatorname{slope}(KI_A) = \frac{I_A K}{-KK'} = \frac{\operatorname{dist}(I_A, \ell) + M_A L}{-KK'} = -\frac{\operatorname{dist}(I, \ell) + M_A L}{2KL} = -\frac{IK + 2M_A L}{2(BL - BK)} = -\frac{r + 2BL \tan \angle M_ABC}{2(\frac a2 - (s - b))} = -\frac{r + a \tan(\frac{\alpha}{2})}{b - c}.$ Since $\operatorname{slope}(TI) \cdot \operatorname{slope}(KI_A) = -1$ , we have $-\frac{r}{TK} \cdot \frac{r + a \tan(\frac{\alpha}{2})}{b - c} = -1 \implies TK = \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c}.$ Let $E$ be the foot from $I$ to $AC$ ; now $\tan\left(\frac{\alpha}{2}\right) = \tan(\angle IAE) = \frac{IE}{AE} = \frac{r}{s - a}$ . As such,
$\begin{align*} TK &= \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c} \\ &= \frac{r^2(1 + \frac{a}{s - a})}{b - c} \\ &= \frac{(\frac{A}{s})^2(\frac{s}{s - a})}{b - c} \\ &= \frac{\frac{s(s - a)(s - b)(s - c)}{s^2} \cdot \frac{s}{s - a}}{b - c} \\ &= \frac{(s - b)(s - c)}{b - c}. \end{align*}$ Now we can calculate
$\begin{align*} TK &= \frac{(s - b)(s - c)}{b - c} \\ TK((s - c) - (s - b)) &= (s - b)(s - c) \\ TK^2 &= TK^2 + TK((s - c) - (s - b)) - (s - b)(s - c) \\ TK^2 &= (TK - (s - b))(TK + (s - c)) \\ TK^2 &= (TK - BK)(TK + CK) \\ TK^2 &= TB \cdot TC. \end{align*}$ But $TN = TK$ , so $TN^2 = TB \cdot TC$ , and therefore $TN$ is tangent to $(NBC)$ . Since $TN$ is by definition also tangent to $\Omega$ , we're done.

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Captainscrubz

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Captainscrubz

#123 h Feb 27, 2025, 8:15 AM • 1 Y

Y by MrdiuryPeter

hehe nice problem :-D

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cursed_tangent1434

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cursed_tangent1434

#124 h Mar 23, 2025, 1:00 PM • 1 Y

Y by ihategeo_1969

Extremely dissatisfying solution but it's the best I've got. We denote by $I_a$ the $A-$ Excenter and by $D$ , $E$ and $F$ the intouch points of $\triangle ABC$ (this overrides a point definition in the problem). It is well known by the Midpoint of the Altitude Lemma that the midpoint of the $A-$ altitude , $D$ and $I_a$ are collinear. Thus, it suffices to show that points $D$ , $N$ and $I_a$ lie on the same line which we shall now do.

Let $P$ and $Q$ be the reflections of $D$ across points $B$ and $C$ respectively. We start off with the following key claim.

Claim : Quadrilaterals $BFQI_a$ and $CEPI_a$ are cyclic.

Proof : We only provide the proof for one since the other is entirely similar. Let $X= PD \cap AI$ . Note,
$\measuredangle I_aXP = \measuredangle CPF + \measuredangle (BC,AI) = 90 + \measuredangle IBA + \measuredangle ACB + \measuredangle IAC = 90 + \measuredangle ICB = \measuredangle PCI_a$ which implies that points $P$ , $X$ , $C$ and $I_a$ are concyclic. Further note that since $PF \perp FD \perp BI$ lines $PF$ and $BI$ are parallel. Thus, $\triangle AFX \sim \triangle ABI$ . This then gives us,
$AX \cdot AI_a = \frac{AX\cdot AC \cdot AB}{AI} = AC \cdot AF = AC \cdot AE$ which implies that points $X$ , $E$ , $C$ and $I_a$ are also concyclic, which in conjunction with the previous observation will imply the claim.

Now, let $N = (I_aBF) \cap (I_aCE) \ne I_a$ . We first note that,
$\measuredangle ENF = \measuredangle I_aNF+ \measuredangle ENI_a = \measuredangle NI_aB + \measuredangle EI_aN = \measuredangle EI_aF = \measuredangle EDF$ which implies that $N$ lies on the incircle of $\triangle ABC$ . Further,
$\text{Pow}_{(CEI_a)}(D)=DP \cdot DC = 2DB \cdot DC = DQ \cdot DB = \text{Pow}_{(BFI_a)}(D)$ which indicates that $D$ lies on the radical axis of circles $(BFI_a)$ and $(CEI_a)$ and hence points $N$ , $D$ and $I_a$ are collinear. We now simply show that this point $N$ is the desired point of tangency.

Claim : Circle $(BNC)$ is tangent to the incircle at $N$ .

Proof : Let $M$ denote the midpoint of segment $DI_a$ (overrides a point definition in the problem). It is clear that $M$ lies on the perpendicular bisector of segment $BC$ since it is well known that the midpoint of $BC$ is the midpoint of the segment joining the $A-$ intouch and $A-$ extouch points. Further,
$DB \cdot DC = \frac{DP \cdot DC}{2} = \frac{DN\cdot DI_a}{2}= DM \cdot DN$ which implies that $M$ lies on $(NBC)$ . Thus, $M$ must be the minor $BC-$ arc midpoint in $(NBC)$ . By Shooting Lemma it now suffices to show that $MD \cdot MN = MB^2$ which is exactly what we shall do next.

We now define the function, $f(\bullet)=\text{Pow}_{(I_aBF)}(\bullet)-\text{Pow}_{(B)}(\bullet)$ which by the Linearity of Power of Point, we know is linear. Let $X_a$ denote the $A-$ extouch point and $M_a$ the midpoint of segment $BC$ in $\triangle ABC$ . Thus,
$\begin{align*} f(M)&= \frac{f(D)+f(I_a)}{2}\\ &= \frac{-2DB\cdot DC - DB^2 + 0 - I_aB^2}{2}\\ &= \frac{-2DB \cdot DC-DB^2 - I_aX_a^2 - BX_a^2}{2}\\ &= \frac{-2DB\cdot DC - DB^2 - 4MM_a^2-BX_a^2}{2}\\ &= \frac{-2(s-b)(s-c)-(s-b)^2 - 4MM_a^2 - (s-c)^2}{2}\\ &= \frac{-((s-b)+(s-c))^2-4MM_a^2}{2}\\ &= \frac{-a^2 - 4MM_a^2}{2}\\ &= \frac{-4(BM_a^2+MM_a^2)}{2}\\ &= \frac{-4BM^2}{2}\\ &= -2BM^2 \end{align*}$ Thus,
$-MN \cdot MI_a-MB^2=-2MB^2$ so
$MB^2 = MN\cdot MI_a = MN \cdot MD$ as desired.

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ihategeo_1969

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ihategeo_1969

#125 h Monday at 11:08 AM • 1 Y

Y by cursed_tangent1434

Why don't most solutions do this?

Let $I_A$ be the $A$ -Excenter and by Midpoint of Altitude lemma, we have $N=\overline{KI_A} \cap$ incircle.

Now invert about the incircle and so we solve this instead

Quote:

Let $\triangle ABC$ be a triangle with circumcenter $O$ and $\triangle MNP$ as medial triangle. If $T=\overline{MO} \cap \overline{NP}$ then if $Y=(ATO) \cap (ABC)$ then prove that $(PNY)$ is tangent to $(ABC)$ .

We will prove $Y$ is $A$ -Why point. Now we will phantom point this and use two well known facts about $Y_A$ (the Why point).

$Y_A=\overline{DGA'} \cap (ABC)$ where $G$ is centroid.
$(Y_APN)$ is tangent to $(ABC)$ .

So we just need to prove $Y_A \in (ATO)$ . Firstly by $-\frac 12$ homothety at $G$ we get $T \in \overline{DGA'}$ and so $\measuredangle AY_AT=\measuredangle AY_AA'=\measuredangle ODA=\measuredangle AOM=\measuredangle AOT$ And done.

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