perpendicularity wanted, intersections of tangents with perp. bisectors related

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perpendicularity wanted, intersections of tangents with perp. bisectors related

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#1 h May 5, 2019, 4:00 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$ . Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$ . Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$ . Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$ . Prove that $AK$ and $LM$ are perpendicular.

by Michael Sarantis, Greece

This post has been edited 1 time. Last edited by parmenides51, May 5, 2019, 6:32 PM
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#2 h May 5, 2019, 9:18 PM • 3 Y

Y by HAFER, Adventure10, Mango247

Let $D, E, F$ be the feet of the altitudes from $A, B, C$ respectively and denote by $T$ the intersection of the tangents to $\Gamma$ at $B$ and $C$ . Let $O'$ be the center of $\odot(BHC)$ . Define $P = DE \cap HC$ and $Q = DF \cap HB$ .

First note that a homothety at $A$ takes the nine-point circle to $\odot(BHC)$ , so $A, K, O'$ collinear. Next, note that $P$ is the radical center of $\odot(BHC), \odot(HDCE)$ , and the nine-point circle, and similar observations with $Q$ imply that $PQ$ is the radical axis of $\odot(K)$ and $\odot(BHC)$ , i.e. $PQ \perp O'K$ . So it remains to show that $LM \parallel PQ$ .

It is easy to see that $PD \parallel MT$ and $HD \parallel OT$ and $HP \parallel OM$ . So $\triangle HDP \sim OTM$ . Similarly $\triangle HDQ \sim \triangle OTL$ . Thus (by similar quadrilaterals, or some applications of Desargues) $\triangle HPQ \sim \triangle OML$ . But two pairs of sides are parallel, so the third pair of sides is parallel, i.e. $LM \parallel PQ$ as desired.

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#3 h Aug 16, 2019, 12:57 PM • 5 Y

Y by teomihai, Ifhml, DrYouKnowWho, Neo-Pythagorean, Adventure10

parmenides51 wrote:

Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$ . Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$ . Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$ . Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$ . Prove that $AK$ and $LM$ are perpendicular.

by Michael Sarantis, Greece

Let $A’$ be a point such that $ABCA’$ is a parallelogram. Let $K_c, K_b$ be points on $\Gamma$ such that $K_cC \parallel AB$ , $K_bB \parallel AC$ . Let $P$ be the reflection of $O$ in $BC$ . Note that $A, K, P$ are collinear. Also, $AP \parallel OA’$ as $A, P, O, A’$ form a parallelogram. So, $AK \parallel PA’$ .
It is easy to see that $BK_b, CK_c$ are the polars of $L, M$ respectively. Since $OP \perp LM$ , we have $AK \perp LM$ . $\blacksquare$

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#4 h Aug 16, 2019, 4:27 PM • 6 Y

Y by RudraRockstar, Ifhml, primesarespecial, Jasurbek, Iora, Adventure10

Complex coordinates: $|a|=|b|=|c|=1$ . Then we have $h=a+b+c\wedge k=\frac{a+b+c}{2}$ $OM\perp AB\iff \frac{m}{a-b}=-\overline{\left(\frac{m}{a-b}\right)}\iff \overline{m}=\frac{m}{ab}$ $CM\perp OC\iff \frac{c-m}{c}=-\overline{\left(\frac{c-m}{c}\right)}$ Two equations give $m=\frac{2abc}{ab+c^2}$ .
In the same way we obtain $l=\frac{2abc}{ac+b^2}$ Finally
$\frac{l-m}{k-a}=\frac{4abc\cdot (ab+c^2-ac-b^2)}{(ab+c^2)(ac+b^2)(b+c-a)}=\frac{4abc\cdot (c-b)}{(ab+c^2)(ac+b^2)}$ Clearly $\frac{l-m}{k-a}=-\overline{\left(\frac{l-m}{k-a}\right)}\iff LM\perp AK$

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#5 h Feb 10, 2022, 12:09 PM • 2 Y

Y by Om245, Funcshun840

$[asy] size(250); pair P=dir(105),Q=dir(-120),R=dir(-60),O=incenter(P,Q,R),A=foot(O,Q,R),B=foot(O,R,P),C=foot(O,P,Q),K=circumcenter( 1/2*(B+C),1/2*(C+A),1/2*(A+B)),L=extension(P,B,Q,O),M=extension(P,C,R,O),IP=2*dir(-90)-O,T=circumcenter(P,Q,R),IR=extension(Q,IP,R,O),IQ=extension(R,IP,P,IR); draw(circumcircle(A,B,C),blue); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$K$",K,dir(230)); dot("$M$",M,dir(180)); dot("$L$",L,dir(90)); dot("$I_P$",IP,dir(IP)); draw(P--Q--R--P,red); draw(M--L,brown); draw(IP--T^^A--K,green); draw(Q--L^^R--M,purple); [/asy]$
Let $PQR$ be the tangential triangle of $ABC$ . Note $\overline{QL},\overline{RM}$ are angle bisectors. Let $I_P,I_Q,I_R$ be the excenters of $\triangle PQR$ . By Theorem 2.7.2 in Muricaaaaaaa (on Page 50), we know line joining $I_P$ and nine-point center of $\triangle I_PI_QI_R$ is perpendicular to $\overline{LM}$ . Since $\triangle ABC, \triangle I_PI_QI_R$ are homothetic, thus $\overline{LM} \perp \overline{AK}$ . $\blacksquare$

This post has been edited 1 time. Last edited by guptaamitu1, Feb 10, 2022, 12:11 PM
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#6 h Feb 26, 2022, 8:53 AM

Y by

Let $X = ML \cap AK$
just draw Tangent from X to nine-point circle with angle bisector & harmonic

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onyqz

#7 h Oct 3, 2024, 1:49 PM

Y by

nice bashing exercise
solution

This post has been edited 2 times. Last edited by onyqz, Oct 3, 2024, 8:05 PM

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#8 h Mar 31, 2025, 8:24 AM

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Let $B'$ be the reflection of $B$ in perpendicular bisector of $AC$ and $C'$ the reflection of $C$ in the perpendicular bisector of $AB$ . By applying Pascal twice it follows that line $LM$ is the same line as the line through $BC\cap B'C'$ and $BC'\cap B'C$ . By Brocard this line is perpendicular to the line through $O$ and $BB'\cap CC'$ . Let $D$ be the intersection of $BB'$ and $CC'$ . Then by parallel lines we see that $ABDC$ is a parallelogram. So $D$ is the reflection of $A$ in midpoint of $BC$ . It is well-known that the reflection of $O$ in $BC$ is on line $AN$ . Call that reflection $O'$ . We want to show $AN\parallel OD$ , or $AO'\parallel OD$ but this is obvious because $AODO'$ is a parallelogram so we are done.

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