Line Perpendicular to Euler Line

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#1 h Jun 29, 2017, 2:55 AM • 15 Y

Y by adamov1, Davi-8191, doxuanlong15052000, tenplusten, Mathuzb, nguyendangkhoa17112003, rashah76, mathematicsy, megarnie, samrocksnature, HWenslawski, son7, Adventure10, Mango247, duhGOAT

Let $ABC$ be a triangle with circumcircle $\Gamma$ , circumcenter $O$ , and orthocenter $H$ . Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$ . Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$ , respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$ , respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$ . Let $Q$ be the intersection point, other than $A$ , of $\Gamma$ with the circumcircle of $\triangle AEF$ . Let $R$ be the intersection of lines $AQ$ and $EF$ . Prove that $PR\perp OH$ .

Proposed by Ray Li

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v_Enhance

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v_Enhance

#2 h Jun 29, 2017, 2:56 AM • 23 Y

Y by tiwarianurag021999, huricane, Vrangr, jam10307, brokendiamond, brainiacmaniac31, abbosjon2002, green_leaf, amar_04, rashah76, brendan_cape, v4913, vvluo, Lamboreghini, megarnie, math31415926535, AllanTian, samrocksnature, son7, Kameawtamprooz, Adventure10, Mango247, cursed_tangent1434

First solution (power of a point) Let $\gamma$ denote the nine-point circle of $ABC$ .

$[asy] pair A = dir(125); pair B = dir(210); pair C = dir(330); pair M = midpoint(A--B); pair N = midpoint(A--C); pair O = origin; pair H = A+B+C; draw(A--B--C--cycle, blue); pair E = foot(B, A, C); pair F = foot(C, A, B); draw(B--E, lightblue); draw(C--F, lightblue); pair R = extension(E, F, B, C); pair Q = -A+2*foot(O, A, R); filldraw(unitcircle, invisible, blue); draw(A--R--E, heavygreen); pair P = extension(M, N, A, A+dir(90)*A); draw(A--P--N, red); filldraw(circumcircle(A, M, N), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavygreen); filldraw(circumcircle(M, N, E), invisible, heavycyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(145)); dot("$N$", N, dir(20)); dot("$O$", O, dir(315)); dot("$H$", H, dir(H)); dot("$E$", E, dir(40)); dot("$F$", F, dir(F)); dot("$R$", R, dir(R)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); /* TSQ Source: A = dir 125 B = dir 210 C = dir 330 M = midpoint A--B R145 N = midpoint A--C R20 O = origin R315 H = A+B+C A--B--C--cycle blue E = foot B A C R40 F = foot C A B B--E lightblue C--F lightblue R = extension E F B C Q = -A+2*foot O A R unitcircle 0.1 lightcyan / blue A--R--E heavygreen P = extension M N A A+dir(90)*A A--P--N red circumcircle A M N 0.1 lightred / red circumcircle A E F 0.1 lightgreen / heavygreen circumcircle M N E 0.1 lightcyan / heavycyan */ [/asy]$

Note that

$PA^2 = PM \cdot PN$ , so $P$ lies on the radical axis of $\Gamma$ and $\gamma$ .
$RA \cdot RQ = RE \cdot RF$ , so $R$ lies on the radical axis of $\Gamma$ and $\gamma$ .

Thus $\overline{PR}$ is the radical axis of $\Gamma$ and $\gamma$ , which is evidently perpendicular to $\overline{OH}$ .

Remark: In fact, by power of a point one may also observe that $R$ lies on $\overline{BC}$ , since it is on the radical axis of $(AQFHE)$ , $(BFEC)$ , $(ABC)$ . Ironically, this fact is not used in the solution.

Second solution (barycentric coordinates) Again note first $R \in \overline{BC}$ (although this can be avoided too). We compute the points in much the same way as before. Since $\overline{AP} \cap \overline{BC} = (0 : b^2 : -c^2)$ we have $P = \left( b^2-c^2 : b^2 : -c^2 \right)$ (since $x=y+z$ is the equation of line $\overline{MN}$ ). Now in Conway notation we have $R = \overline{EF} \cap \overline{BC} = (0 : S_C : -S_B) =\left( 0 : a^2+b^2-c^2 : -a^2+b^2-c^2 \right).$ Hence $\overrightarrow{PR} = \frac{1}{2(b^2-c^2)} \left( b^2-c^2 , c^2-a^2 , a^2-b^2 \right).$ On the other hand, we have $\overrightarrow{OH} = \vec A + \vec B + \vec C$ . So it suffices to check that $\sum_{\text{cyc}} a^2\left( (a^2-b^2) + (c^2-a^2) \right) = 0$ which is immediate.

Third solution (complex numbers) Let $ABC$ be the unit circle. We first compute $P$ as the midpoint of $A$ and $\overline{AA} \cap \overline{BC}$ : $\begin{align*} p &= \frac{1}{2} \left( a + \frac{a^2(b+c)-bc\cdot2a}{a^2-bc} \right) \\ &= \frac{a(a^2-bc)+a^2(b+c)-2abc}{2(a^2-bc)}. \end{align*}$ Using the remark above, $R$ is the inverse of $D$ with respect to the circle with diameter $\overline{BC}$ , which has radius $\left\lvert \frac{1}{2}(b-c) \right\rvert$ . Thus $\begin{align*} r - \frac{b+c}{2} &= \frac{\frac14(b-c)\left( \frac1b-\frac1c \right)} {\overline{\frac{1}{2}\left( a-\frac{bc}{a} \right)}} \\ r &= \frac{b+c}{2} + \frac{-\frac{1}{2} \frac{(b-c)^2}{bc}}{\frac1a-\frac{a}{bc}} \\ &= \frac{b+c}{2} + \frac{a(b-c)^2}{2(a^2-bc)} \\ &= \frac{a(b-c)^2+(b+c)(a^2-bc)}{2(a^2-bc)}. \end{align*}$ Expanding and subtracting gives $p-r = \frac{a^3-abc-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} = \frac{(a+b+c)(a-b)(a-c)}{2(a^2-bc)}$ which is visibly self-conjugate once the factor of $a+b+c$ is deleted.

(Actually, one can guess this factorization ahead of time by noting that if $A=B$ , then $P=B=R$ , so $a-b$ must be a factor; analogously $a-c$ must be as well.)

This post has been edited 3 times. Last edited by v_Enhance, Dec 23, 2019, 12:40 AM

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WizardMath

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WizardMath

#4 h Jun 29, 2017, 12:55 PM • 4 Y

Y by myang, samrocksnature, son7, Adventure10

My solution:

It is well known that $R$ is on the radical axis of the circumcircle and the npc of $ABC$ . (Proof: radical axis on the these circles and the circle with $BC$ as a diameter.). Now doing the same trick with the circle with $AO$ as diameter and the npc and circumcircle of $ABC$ does the job as circle with diameter $AO$ is tangent to the circumcircle of $ABC$ by a simple homothety at $A$ or using collinearity of their centers.

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BobsonJoe

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BobsonJoe

#6 h Jun 29, 2017, 8:35 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

see here
https://artofproblemsolving.com/community/c312162h1469210

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EulerMacaroni

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EulerMacaroni

#8 h Jun 29, 2017, 10:33 PM • 6 Y

Y by anantmudgal09, Wizard_32, samrocksnature, son7, Adventure10, Mango247

What a beautiful problem! I add my sol

Let $X\equiv MN\cap EF$ , $Y\equiv ME\cap NF$ ; then $AX$ is the polar of $Y$ with respect to the nine-point circle by Brokard's theorem. In addition, we know by Pappus that $Y\in OH$ , so $AX$ is perpendicular to the Euler line. Then since $R\in BC$ we have $\overline{PX}$ bisects $\overline{AR}$ and from $AP\parallel RX$ we know quadrilateral $APRX$ is a parallelogram, as desired.

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PepsiCola

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PepsiCola

#9 h Jul 3, 2017, 12:17 PM • 4 Y

Y by samrocksnature, SerdarBozdag, Adventure10, Mango247

Draw the $9$ -point circle, $(N)$ . Radical Axis Theorem on $(ABC), (AMN), (N)$ implies $P$ lies on the radical axis of $(ABC)$ and $(N)$ . Radical Axis Theorem on $(AEFH), (ABC), (N)$ implies $R$ lies on the radical axis of $(ABC)$ and $(N)$ .
Then the radical axis of $(ABC), (N)$ is just $PR$ , so $PR$ is perpendicular to the line connecting their centers, which is just the Euler Line! Done!

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droid347

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droid347

#13 h Jul 8, 2017, 4:12 AM • 2 Y

Y by samrocksnature, Adventure10

Let $X$ be the midpoint of $BC$ and $Y=MN\cap AR$ . $Q$ is the Miquel point of cyclic $BFEC$ so $R\in BC$ and $XQ\perp QR$ . Let $L$ be the antipode of $A$ ; by the tangent $\angle XLO=\angle PAY$ and we can chase that $\angle XOL=\angle YPA=|B-C|$ so $\triangle XOL\sim \triangle YPA$ . However, $HL=2XL$ and $RA=2YA$ so $\triangle HOL\sim \triangle RPA$ and thus $XQ\perp QR\implies PR\perp OH$ .

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enhanced

515 posts

enhanced

#18 h Dec 12, 2017, 3:39 PM • 6 Y

Y by One2three, Vrangr, brokendiamond, amar_04, samrocksnature, Adventure10

We will prove a more general claim

In a $\Delta ABC$ with circumcenter $O$ and circumcircle $\Gamma$ and Let $K$ be a point on the line $AO$ and $K^*$ denote the isogonal conjugate of $K$ with respect to $\Delta ABC$ and let $O_1$ be the midpoint of $KK^*$ and $M,N$ are projection of $P$ in $AB,AC$ and $E,F$ are projection of $K^*$ in $AB,AC$ let tangent at $A$ to $\Gamma$ meet $MN$ at $P$ and $(AEF)\cap (ABC)=A,Q$ and $EF\cap AQ=R$ prove that $PR\perp OO_1$

PROOF
By six point circle theorem we see that $(EFMN)$ is cyclic with center $O_1$ and denote it's circumcircle by $\gamma$
Since $K \in AO$ it is easy to see that $MN\parallel BC$ so $(ABC),(AMNK)$ are tangent to each other,on the other hand
By power of a point
$(RQ)(RA)=(RE)(RF)$
$PA^2=(PM)(PN)$
so $PR$ is the radical axis of $\Gamma,\gamma$ ,so $PR\perp OO_1$
and we are done $\blacksquare$
When, $K\equiv O$ ,we get the problem

This post has been edited 9 times. Last edited by enhanced, Dec 14, 2017, 3:05 AM

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PROF65

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PROF65

#19 h Dec 12, 2017, 4:28 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

enhanced wrote:

...
$(PM)(PN)=PA^2$

this result doesn't occur unless $NM\parallel BC$

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Magikarp1

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Magikarp1

#20 h Dec 12, 2017, 10:07 PM • 5 Y

Y by wu2481632, amar_04, benk25, samrocksnature, Adventure10

$M$ and $N$ are the midpoints of $AB$ and $AC$ so $NM\parallel BC$

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One2three

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One2three

#21 h Dec 17, 2017, 4:28 AM • 2 Y

Y by samrocksnature, Adventure10

@PROF65 I don't see any flaw in enhanced's proof,

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tworigami

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tworigami

#22 h Mar 2, 2018, 6:22 AM • 2 Y

Y by samrocksnature, Adventure10

Since there was some controversy, here is a little bit more explanation: (Basically v_enhance's solution)

We have $AMON$ cyclic and internally tangent to $\Gamma$ as they share a diameter. Hence, we have from power of a point,

$PA^2 = PM \cdot PN$ .

Additionally, let $\gamma$ denote the nine-point circle of $\triangle{ABC}$ . One key lemma which was left out as it is very well known is that the nine-point circle of a triangle is the triangle passing through the three midpoints, the altitudes, and the midpoints of $AH, BH, CH$ . It is easy to see from a homothety centered at $H$ with dilation factor $2$ that the center $N$ of the nine point circle is the midpoint of $OH$ , hence $O, N, H$ are collinear. Also, it is ovious that $AEHF$ is cyclic, so from power of a point, we also have

$RA \cdot RQ = RE \cdot RF$ .

The key observation is that $Pow_{\gamma}P = Pow_{\Gamma}P$ , and similarly, $Pow_{\gamma}R = Pow_{\Gamma}R$ . Thus, both points lie on the radical axis of circles $\Gamma, \gamma$ , and as two points uniquely determine a line, $PR$ is the radical axis of $\Gamma$ and $\gamma$ . The definition of the radical axis is that it is perpendicular to the line connecting the centers of the circles, so we see that $\overline{OH} \perp \overline{PR}$ and from the previous collinearity with $H$ , the desired result follows. $\blacksquare$

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math_pi_rate

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math_pi_rate

#23 h Nov 9, 2018, 5:39 PM • 2 Y

Y by samrocksnature, Adventure10

Nice and easy. Here's my solution: Let $AP \cap BC=T$ , and let $X$ be the point such that $ATRX$ is a parallelogram. Also let $K$ be the midpoint of $RX$ . From here, we get that $X$ lies on $EF$ , and so $K$ is the point where $EF$ and $MN$ meet. But, as $P$ is the midpoint of $AT$ , we get that $APRK$ is also a parallelogram, or equivalently that $AK \parallel PR$ . Restating the problem in terms of $\triangle DEF$ , where $D$ is the foot of $A$ -altitude, we are left with the following problem:-

New problem wrote:

Let $O$ be the circumcenter, $\omega$ be the circumcircle, $I$ be the incenter, $I_A$ be the $A$ -excenter of $\triangle ABC$ . Let $BI_A \cap \omega=M$ and $CI_A \cap \omega=N$ . Finally let $BC \cap MN=K$ . Then show that $I_AK$ is perpendicular to line $OI$ .

Let $BN \cap CM=L$ . Note that, by Brokard's Theorem, $I_AK$ is the polar of $L$ wrt $\omega$ , and so $I_AK$ is perpendicular to $OL$ . So it suffices to show that $O,I,L$ are collinear. Let $BI \cap \omega=U$ and $CI \cap \omega=V$ . Then by Pascal's Theorem on $UMCVNB$ , we get the desired result. $\blacksquare$

This post has been edited 1 time. Last edited by math_pi_rate, Nov 9, 2018, 5:39 PM

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RC.

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RC.

#24 h Jan 14, 2019, 7:53 AM • 5 Y

Y by AlastorMoody, MarkBcc168, samrocksnature, Adventure10, Mango247

Let $A'$ denote the antipode of $A$ in $\Gamma$ and $AP \cap BC= X$ . Then, $\Delta AHA' \sim \Delta ARX \Rightarrow \angle AHO= \angle PRX \therefore PR\perp OH.$

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jayme

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jayme

#25 h Jan 15, 2019, 7:01 AM • 2 Y

Y by samrocksnature, Adventure10

Dear RC.
very nice and elegant proof...

Sincerely
Jean-Louis

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ike.chen

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ike.chen

#51 h Dec 14, 2021, 6:59 PM • 1 Y

Y by Jalil_Huseynov

Remarks: This problem took me way too long (around three hours). Amusingly enough, I finally found this solution while watching a ballet performance

.

Let $\omega$ denote the Nine-Point Circle of $ABC$ and $N_9$ be the Nine-Point center of $ABC$ . It's clear that $Pow_{\Gamma}(P) = PA^2 = Pow_{(AMN)}(P) = PM \cdot PN = Pow_{\omega}(P)$ and $Pow_{\Gamma}(R) = RA \cdot RQ = Pow_{(AEF)}(R) = RE \cdot RF = Pow_{\omega}(R)$ so $PR$ is the Radical Axis of $\Gamma$ and $\omega$ . Because $H, N_9, O$ are collinear on the Euler Line, the desired result follows immediately from properties of Radical Axes. $\blacksquare$

This post has been edited 1 time. Last edited by ike.chen, Aug 21, 2022, 9:46 AM
Reason: Format

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Tafi_ak

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Tafi_ak

#52 h Dec 16, 2021, 9:04 PM • 1 Y

Y by Jalil_Huseynov

We Claim that $PR$ is the radical axis of $(ABC), (EFM)$ . For the proof, notice that $P$ is the radical center of $(AMN), (ABC), (EFM)$ . Therefore the radical axis of $(ABC), (EFM)$ passes through $P$ . Similarly $R$ is the radical center of $(BCF), (ABC), (EFM)$ . So the radical axis of $(ABC), (EFM)$ passes through $R$ . $\blacksquare$

It is very well known that the center of $(ABC)$ and $(EFM)$ are collinear with $H$ . Hence done.

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lazizbek42

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lazizbek42

#53 h Jan 4, 2022, 8:25 AM

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$K$ is midpoint $BC$ . $D$ feet of the altitudes from $A$
$K,H,Q$ collinear.
$AQEF$ , $EFDK$ , $AQDK$ cyclic.
by radical axis $AQ,EF,DK$ concurrent.
$R,B,C$ collinear.
Carnot theorem.
$HP^2-OP^2=HR^2-OR^2$ $( AP^2 +AH^2 - 2AP*AH*cos(\angle PAH))-(PA^2+OA^2)=RD^2+HD^2-RK^2-OK^2$ This expression can be clearly proved because it will all be expressions related to $ABC$ .

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Mogmog8

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Mogmog8

#54 h Feb 17, 2022, 2:44 AM • 2 Y

Y by centslordm, megarnie

By Radical Axis on $(AEF),(BCEF),$ and $\Gamma,$ we see $R$ lies on $\overline{BC}.$ Then, the power of $R$ with respect to the Nine-Point circle of $\triangle ABC,$ $(N_9),$ is $RF\cdot RE=RB\cdot RC=\textrm{pow}_{\Gamma}(R)$ as $\triangle RFB\sim\triangle RCE.$ Also, $\angle PAN=180-\angle CBA=\angle AMP$ so $\triangle AMP\sim\triangle NAP$ and $\textrm{pow}_{(N_9)}=PM\cdot PN=PA^2=\textrm{pow}_{\Gamma}.$ Hence, $\overline{PR}$ is the radical axis of $(N_9)$ and $\Gamma.$ $\square$

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JAnatolGT_00

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JAnatolGT_00

#55 h Feb 17, 2022, 4:32 PM

Y by

$R$ is radical center of $\Gamma, (AFHE), (BCEF), (MFNE)$ and therefore $R=BC\cap EF.$
$P$ is radical center of $\Gamma, (AMN), (MFNE).$ Hence $PR$ is radical axis of $\Gamma, (MFNE).$
But $OH$ passes through centers of these circles, so we are done.

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MathLuis

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MathLuis

#56 h Feb 17, 2022, 5:03 PM

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Another speedrun les goooo.
By radax on $(AQEF), (BFEC), (ABC)$ we have that $R$ lies on $BC$ now let $D$ the projection from $A$ to $BC$ , then its known that $DFMEN$ is cyclic becuase they lie on the nine point circle and since the nine point center lies on $OH$ is enough to show that $PR$ is the radical axis of the nine-point circle and $(ABC)$ . Now by PoP
$RE \cdot RF=RB \cdot RC \implies R \; \text{lying inside the radax of} \; (ABC), (DFMEN)$ Now by angle chasing knowing that $D$ and $A$ are symetric w.r.t. $MN$ we have that
$\angle RAM=\angle ACB=\angle ANM=\angle DNM \implies PD \; \text{tangent to} \; (DFMEN) \implies PD^2=PN \cdot PM$ Wait, rememebr when i said $D$ and $A$ are symetric w.r.t. $MN$ , hence $PD=PA$ meaning that $PA^2=PM \cdot PN$ hence $P$ lies on the radax of $(ABC), (DFMEN)$
Thus, we are done :blush:

This post has been edited 1 time. Last edited by MathLuis, Feb 17, 2022, 5:04 PM

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pi271828

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pi271828

#58 h Mar 15, 2023, 5:07 PM

Y by

Let $\omega$ denote the nine-point circle. Notice that since the center of $\omega$ is the centroid $G$ , and $O$ , $H$ , $G$ are collinear, it suffices to prove that $PR$ is the radical axis of $\omega$ and $\Gamma$ . Let $\gamma$ denote $(OMAN)$ . Note that $\mathrm{Pow}(P, \omega) = PM \cdot PN = \mathrm{Pow}(P, \gamma)$ . Now, since $A$ is the antipode of $O$ across $\gamma$ , we have that $PA$ is tangent to $\gamma$ and therefore $\mathrm{Pow}(P, \gamma) = \mathrm{Pow}(P, \Gamma)$ . It suffices to prove that $R$ lies on the radical axis. Note that $BFEC$ is cyclic. Therefore by radical axis theorem, $AQ, EF, BC$ all concur at point $R$ . Because $R$ is the radical center, we are done.

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IAmTheHazard

5000 posts

IAmTheHazard

#59 h Jul 3, 2023, 2:11 PM • 1 Y

Y by centslordm

Basically was told the fact that nuked the problem but I think typing this up will help me remember it:

$P$ lies on the orthic axis of $\triangle ABC$ , since $PA^2=PM\cdot PN$ . More concretely, $\overline{PR}$ is the radical axis of $(ABC)$ and the 9-point circle, since $RE\cdot RF=RB\cdot RC$ (where $R$ lies on $\overline{BC}$ because of radical center) and $PA^2=PM\cdot PN$ , so $\overline{PR}$ is perpendicular to $\overline{ON_9}$ which is just the Euler line. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Jul 3, 2023, 2:12 PM

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IAmTheHazard

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IAmTheHazard

#60 h Jul 3, 2023, 2:22 PM • 1 Y

Y by centslordm

Here's a complex solution (some calculations omitted, but I did do them):

First note that by radical center on $(AEF)$ , $(BCEF)$ , and $(ABC)$ , we find that $R$ is also the intersection of $\overline{AQ}$ and $\overline{BC}$ . Since it is well-known that $Q$ , the $A$ -antipode, and the midpoint of $\overline{BC}$ are collinear, we set up the equation
$\frac{q+a}{q-\frac{b+c}{2}}=\frac{\frac{1}{q}+\frac{1}{a}}{\frac{1}{q}-\frac{\frac{1}{b}+\frac{1}{c}}{2}}.$ This can be neatly cross-multiplied, and since $q=-a$ is a factor we divide out $q+a$ and obtain a linear equation in $q$ , which we can solve to yield
$q=\frac{2a+b+c}{2+\frac{a}{b}+\frac{a}{c}}.$ Now let $P'=\overline{AA} \cap \overline{BC}$ . By complex intersection,
$p'=\frac{a^2(b+c)-2abc}{a^2-bc},$ and since $P$ is the midpoint of $\overline{AP}$ for homothety reasons we find that
$p=\frac{a^3+a^2b+a^2c-3abc}{2(a^2-bc)}.$ We then calculate $R$ , which is surprisingly neat since a lot of $2+\tfrac{a}{b}+\tfrac{a}{c}$ terms die:
$r=\frac{a^2b+a^2c+ab^2+ac^2-2abc-b^2c-bc^2}{2(a^2-bc)},$ so
$p-r=\frac{a^3-ab^2-ac^2-abc+b^2c+bc^2}{2(a^2-bc)}.$ We would like to show that this is a pure imaginary multiple of $a+b+c$ , so using the Euclidean algorithm we try to divide out $a+b+c$ from the numerator and find that it actually equals $(a+b+c)(a-b)(a-c)$ , so we want to show that
$\frac{(a-b)(a-c)}{a^2-bc}=-\frac{(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}-\frac{1}{c})}{\frac{1}{a^2}-\frac{1}{bc}},$ but this is obvious after multiplying the numerator and denominator of the RHS by $a^2bc$ , so we're done. $\blacksquare$

This post has been edited 2 times. Last edited by IAmTheHazard, Aug 29, 2023, 12:08 AM

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trinhquockhanh

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trinhquockhanh

#61 h Aug 22, 2023, 1:09 PM

Y by

https://i.ibb.co/hZvgFkX/Screenshot-2023-08-22-200814.png

It's easy to prove that $R\in BC,$ also note that $PR$ is the radical axis of $(O)$ and the Euler circle of $\triangle ABC$

$\Rightarrow PR\perp OH.$

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Pyramix

419 posts

Pyramix

#62 h Apr 9, 2024, 7:19 PM

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Let $S$ be the mid-point of $\overline BC$ , $D$ be the feet of altitudes from $A$ in $\triangle ABC$ . Then, $M,N,D,E,F,S,$ are cyclic as they form the nine-point circle.

Note that $R$ is the radical center of circles $(AQEF)$ , $(MNDS)$ , $(ABC)$ . So, $R,B,C$ are collinear. Since $(BCEF)$ is cyclic, we have $RB\cdot RC=RE\cdot RF$ . Since $(EFDS)$ is cyclic, we have $RE\cdot RF=RD\cdot RS$ . Hence, we have $RD\cdot RS=RB\cdot RC$ , which means $Pow_{(ABC)}(R)=Pow_{(MNDS)}(R)$ and hence $R$ lies on the radical axis of the circumcircle and the nine-point circle.

Similarly, since $\overline{PA}$ is tangent to $(ABC)$ which is also tangent to $(AMN)$ because $(AMN)$ is just $(ABC)$ with homothety at $A$ with factor $\frac12$ , we have $PA^2=PM\cdot PN$ which means $Pow_{(ABC)}(P)=Pow_{(MNDS)}(P)$ and hence $P$ lies on the radical axis of the circumcircle and the nine-point circle. So, $PR$ is the radical axis, which means $PR\perp ON_9$ and hence $PR\perp OH$ . $\blacksquare$

Remark. If $U,S$ are the intersections of the $A-$ median and the nine-point circle, then $\overline{PU},\overline{PD}$ are the tangents from $P$ to nine-point circle. So, $DU$ is the pole of $P$ in nine-point circle. Also, the mid-point of $\overline{AH}$ , say $L$ , lies on the pole of $R$ .

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ihatemath123

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ihatemath123

#63 h Apr 21, 2024, 8:29 PM • 1 Y

Y by GeoKing

Applying the radical axis theorem on $(ABC)$ , $(AMN)$ and $(MNEF)$ , it suffices to show that $R$ has equal power WRT $(ABC)$ and $(MNEF)$ . This is true because $RQ \cdot RA = RF \cdot RE$ .

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OronSH

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OronSH

#64 h Jun 18, 2024, 3:11 PM • 1 Y

Y by GeoKing

Since $(AMN)$ is tangent to $(ABC)$ by homothety, $PA^2=PB\cdot PC,$ so $P$ lies on the radical axis of the circumcircle and the nine point circle. Since $(AEFQ)$ is cyclic, $RQ\cdot RA=RE\cdot RF,$ so $R$ lies on the radical axis of the circumcircle and the nine point circle. Thus $PR$ is perpendicular to the line through their centers which is $OH.$

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Eka01

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Eka01

#65 h Aug 20, 2024, 1:41 PM • 1 Y

Y by AaruPhyMath

It is well known that $R$ lies on the radical axis of the nine point circle and the circumcircle, also called the orthic axis.
By radax on $(ABC),(AMN),(MNEF)$ , we see that $P$ also lies on this radical axis and the line $OH$ is the line joining the centers of the $NPC$ and $(ABC)$ so it trivially follows that $PR \perp OH$ .

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ohiorizzler1434

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ohiorizzler1434

#67 h Yesterday at 12:07 AM

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Bro what the sigma? This is triv! This is trivialised by radical axis because P and R have equal power wrt the NPC and the circumcircle!

This post has been edited 2 times. Last edited by Luis González, Yesterday at 1:22 PM

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