Polish MO Finals 2018, Problem 5

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Polish MO Finals 2018, Problem 5

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j___d

#1 h Apr 19, 2018, 12:26 PM • 7 Y

Y by Mathuzb, khoonyu, mathematicsy, itslumi, tiendung2006, Adventure10, Mango247

An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$ , respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$ . The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$ . Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$ .

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fastlikearabbit

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fastlikearabbit

#2 h Apr 19, 2018, 12:52 PM • 4 Y

Y by microsoft_office_word, guptaamitu1, Adventure10, Mango247

j___d wrote:

An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$ , respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$ . The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$ . Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$ .

Let $H_A$ be the A-HM point. As the polar of $M$ -the midpoint of $BC$ passes through $Q$ , by La Hire’s theorem we obtain that $QA=QH_A$ .
It’s well known that $PH_A$ is tangent to $\odot(BH_AC)$ * . So $PH_A^2=PB\cdot PC=PA^2$ . So $PQ$ is the perpendicular bisector of $AH_A$ .

* look at Example 2 from https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2017-02/article_1_a_special_point_on_the_median.pdf

This post has been edited 2 times. Last edited by fastlikearabbit, Apr 19, 2018, 1:06 PM

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#3 h Apr 19, 2018, 12:53 PM • 1 Y

Y by Adventure10

Nice problem!

Let $T$ be the reflection of $H$ w.r.t. $A$ , $M$ be the midpoint of $BC$ and $A'$ be the antipode of $A$ w.r.t. $\odot(ABC)$ .

Since $\triangle ABC\cup P\cup A'\sim\triangle AEF\cup Q\cup H$ , we see that $\frac{AP}{AQ} = \frac{AA'}{AH}= \frac{AT}{AA'}$ . Combining with $\angle TAQ=\angle PAA'=90^{\circ}$ gives $\triangle ATA'\sim\triangle AQP$ . So $\angle(TA', PQ) = \angle(AT, AQ)=90^{\circ}$ .

Clearly $BHCA'$ is parallelogram so $M$ is the midpoint of $HA'$ . So $TA'\parallel AM$ . Combining with the previous paragraph gives $PQ\perp AM$ as desired.

This post has been edited 1 time. Last edited by MarkBcc168, Apr 6, 2020, 10:07 AM

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#4 h Apr 19, 2018, 1:10 PM • 3 Y

Y by microsoft_office_word, Adventure10, Mango247

Let $H_A$ be the A-Humpty point, clearly $(AH_AEF)=-1$ and because $QA$ is tangent to $\odot(AEF)$ and $Q$ lies on $EF$ we have that $QA=QH_A$ . Because $P$ is the center of the A-Apollonius circle, which passes through both $A$ and $H_A$ we have that $PQ$ is perpendicular to $\overline{AH_A}$ , which is precisely the A-median of $\triangle{ABC}$ .

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GGPiku

#5 h Apr 20, 2018, 1:49 PM • 12 Y

Y by Anar24, mijail, platypus43, tapir1729, ApraTrip, Maths_is_my_life, PRMOisTheHardestExam, guptaamitu1, packlj, Adventure10, Mango247, endless_abyss

Is this correct?
Take the circles $C(A,0)$ and $C(BCEF)$ . We have $PA^2=PB\times PC$ and $QA^2=QE\times QF$ hence $PQ$ is the radical axis the 2 circles with centers $A$ and $M$ , hence we are done.

This post has been edited 3 times. Last edited by GGPiku, Apr 20, 2018, 1:55 PM

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cirey

#7 h Apr 21, 2018, 2:05 PM • 3 Y

Y by Omeredip, Adventure10, Mango247

Here is a bashy solution, in which I will prove that $\overrightarrow{AM}\bullet\overrightarrow{PQ}=0$ , $M$ being the midpoint of $BC$ .

http://i64.tinypic.com/25qvkuq.png

Note that this is equivalent to $(\overrightarrow{AB}+\overrightarrow{AC})\bullet(\overrightarrow{PA}+\overrightarrow{AQ})=0$ .

Now, observe these four equalities:

$\overrightarrow{AB}\bullet\overrightarrow{PA}=|AB|\cdot|PA|\cdot\cos{(180-\hat{C})}$

$\overrightarrow{AB}\bullet\overrightarrow{AQ}=|AB|\cdot|AQ|\cdot\cos{(180-\hat{B})}$

$\overrightarrow{AC}\bullet\overrightarrow{PA}=|AC|\cdot|PA|\cdot\cos{\hat{B}}$

$\overrightarrow{AC}\bullet\overrightarrow{AQ}=|AC|\cdot|AQ|\cdot\cos{\hat{C}}$

Adding these with using $|PA|=|AQ|\cdot\cos{\hat{A}}$ leaves to prove that $\frac{|AC|}{|AB|}=\frac{\cos{\hat{B}}+\cos{\hat{A}}\cdot\cos{\hat{C}}}{\cos{\hat{C}}+\cos{\hat{A}}\cdot\cos{\hat{B}}}$ ,

which is equal to $\frac{\cos{\hat{B}}+\cos{(\hat{A}-\hat{C})}}{\cos{\hat{C}}+\cos{(\hat{A}-\hat{B})}}=\frac{\sin{\hat{B}}\cdot\sin{\hat{A}}}{\sin\hat{C}\cdot\sin\hat{A}}$ , the conclusion follows.

This post has been edited 1 time. Last edited by cirey, Apr 22, 2018, 8:38 AM

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#8 h Apr 21, 2018, 8:17 PM • 3 Y

Y by MatBoy-123, Adventure10, Mango247

j___d wrote:

An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$ , respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$ . The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$ . Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$ .

Note that $Q$ is the radical center of $\odot(A), \odot(AEF), \odot(BC)$ hence $Q$ lies on the radical axis of $\odot(A), \odot(BC)$ . Also $P$ is the radical center of $\odot(A), \odot(ABC), \odot(BC)$ hence $P$ also lies on the radical axis of $\odot(A), \odot(BC)$ . Now the $A$ -median passes through the center of $\odot(BC)$ , the result follows.

This post has been edited 1 time. Last edited by anantmudgal09, Apr 21, 2018, 8:17 PM

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IstekOlympiadTeam

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IstekOlympiadTeam

#9 h Apr 22, 2018, 5:34 PM • 3 Y

Y by bluedragon17, Adventure10, Mango247

Let $Z=EF\cap BC,K=EF\cap AH, T=AH\cap BC$ .
It is well known fact that $ZH\perp AM$ $\implies$ it is enough to show that $ZH\parallel PQ$ .

Let $F=ZH\cap AQ$ . Let's show that $AQ=QF$ $\iff$ $\frac{AQ}{AF}=\frac{1}{2}$ .
Using parellelism we have that $AQ=\frac{|AK||ZT|}{KT} \ ,\ AF=\frac{|AH||ZT|}{HT}$ $\implies$ $\boxed{\frac{AQ}{AF}=\frac{|AK||KT|}{|HT||AH|}}$ It is easy to show that last relatioon is equal to $\frac{1}{2}$ (Sine's Law).

Finally we have that $AQ=QF=PZ$ ( $APZQ$ is alrealy a parallelogram since $AP\parallel EF$ ) $\implies$ $PQFZ$ is a parallelogram $\implies$ $ZH\parallel PQ$ as desired $\blacksquare$

This post has been edited 1 time. Last edited by IstekOlympiadTeam, Apr 22, 2018, 5:35 PM

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#10 h Apr 23, 2018, 8:03 AM • 1 Y

Y by Adventure10

This is a very nice algebra problem.(Codyj,USAMO 2018,P5)
We can find easily with barycentric coordinate
$P=(0,\frac{-2b^2}{2(c^2-b^2)},\frac{2c^2}{2(c^2-b^2)}),$ and
$Q=(\frac{2(c^2-b^2)}{2(c^2-b^2)},\frac{-a^2+b^2+c^2}{2(c^2-b^2)},\frac{-(-a^2+b^2+c^2)}{2(c^2-b^2)}).$
The rest is very easy.

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#11 h Jun 28, 2018, 8:07 AM • 4 Y

Y by lahmacun, srijonrick, Adventure10, Mango247

Let $T$ be the midpoint of $AH$ , where $H$ is the orthocenter of $\triangle ABC$ . Also, Let $M_A$ be the midpoint of $BC$ .

Note that $EF$ is the radical axis of $\odot (AEHF)$ and $\odot (BFEC) \Rightarrow TM_A$ is the perpendicular bisector of $EF$ .

Also, $\angle PAB = \angle ACB \Rightarrow \angle PAF = \angle AFE \Rightarrow PA \parallel EF$

But, $TM_A$ is perpendicular to $EF \Rightarrow TM_A \perp PA \Rightarrow T$ is the orthocenter of $\triangle AM_AP \Rightarrow PT \perp AM_A$

Now, Let $H_A$ be the A-Humpty point and $PT \cap AM_A = X \Rightarrow TX \perp AH_A \Rightarrow X$ is the midpoint of $AH_A$ .

Also, $M_AE$ and $M_AF$ are tangents to $\odot (AEHH_AF) \Rightarrow AEH_AF$ is a harmonic quadrilateral.

$\Rightarrow QH_A$ is also tangent to $\odot (AEH_AF) \Rightarrow Q, X, T, P$ are collinear $\Rightarrow PQ \perp AM_A \text{ } \blacksquare$

This post has been edited 2 times. Last edited by math_pi_rate, Oct 29, 2018, 1:12 PM

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soryn

#12 h Jun 28, 2018, 8:09 AM • 2 Y

Y by Adventure10, Mango247

Nice problem from Poland..Nice solutions..

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#13 h Oct 29, 2018, 1:05 PM • 2 Y

Y by Adventure10, Mango247

Sorry for double posting, but here's a bit different ending: Let $H_A$ be the $A$ -Humpty point, and let $EF \cap BC=X$ . From the various properties proved here, we have that $X$ lies on $HH_A$ , and that $HH_A$ is perpendicular to the $A$ -median. Thus it suffices to show that $PQ$ is parallel to $XH$ . Let $XH \cap AQ=T$ . Now, as both $EF$ and $AP$ are antiparallel to $BC$ , we get that $EF$ is parallel to $PQ$ . But, as $PX$ is parallel to $AQ$ , we have that $APXQ$ is a parallelogram, and so $AQ=PX$ . Also, as $AEH_AF$ is a harmonic quadrilateral, and cause $AQ$ is tangent to $\odot (AEHH_AF)$ at $A$ , we must have that $QH_A$ is tangent to $\odot (AEHH_AF)$ at $H_A$ , which gives $QH_A=QA$ . But, $TH_A$ is perpendicular to $AH_A$ , which means that $Q$ is the circumcenter of $\triangle AH_AT$ , or equivalently $QT=QA=PX$ . As $PX \parallel QT$ , quadrilateral $PXTQ$ is a parallelogram $\Rightarrow PQ \parallel TX$ . Hence, done. $\blacksquare$

This post has been edited 1 time. Last edited by math_pi_rate, Oct 29, 2018, 1:10 PM

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#14 h Feb 23, 2019, 3:53 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be midpoint of $BC$ then we have: $PA^2$ $-$ $QA^2$ = $\overline{PB}$ . $\overline{PC}$ $-$ $\overline{QE}$ . $\overline{QF}$ = $P_{P / (BCEF)}$ $-$ $P_{Q / (BCEF)}$ = $PM^2$ $-$ $QM^2$ or $PQ$ $\perp$ $AM$

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Optx

#15 h May 9, 2019, 3:16 PM • 2 Y

Y by Adventure10, Mango247

Let T denote midpoint of AH where H is orthocenter of ABC. It easy to prove that T is orthocenter of APM since AT is perpendicular to BC and MT is perpendicular to AP. Therefore, PT is perpendicular to AM. Now, since Q lies on EF which is polar of M, then AM is polar of Q so QT is perpendicular to AM and we are done.

This post has been edited 1 time. Last edited by Optx, May 9, 2019, 3:16 PM
Reason: Mistake

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#16 h Aug 1, 2019, 4:25 PM • 3 Y

Y by Myriam2003, Adventure10, Mango247

let $T=EF \cap BC$ , $N=EF \cap AD$ , $X=AQ \cap TH$
from brocard on $EFBC$ we have that $AM$ is perpendicular to $TH$
so it suffices to show that $QP \parallel TX$
we have $AP \parallel TQ \implies AQTP$ is parallelogram so it suffices to show that $Q$ is the midpoint of $AX$
but we have $(A,H;N,D) =_T (A,X:Q,P_\infty)=-1$ and we done

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9767 posts

jayme

#17 h May 30, 2020, 8:56 AM • 1 Y

Y by Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%205.pdf p. 48-50.

Sincerely
Jean-Louis

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#18 h Aug 8, 2020, 5:51 PM

Y by

Let $H$ be the orthocenter of $\triangle ABC$ , and let $AH$ intersect $BC$ at $D$ . Let $X_A$ be the $A$ -humpty point of $\triangle ABC$ . Let $L$ be the midpoint of $AH$ , and let $\omega$ be the circle with diameter $AH$ , let $K$ be the foot of the perpendicular from $H$ to $AP$ . Let $J$ be the midpoint of $EF$ , which lies on line $LM$ by inversion around $\omega$ .

Claim: $QA = QX_A$ .

Proof: Note that $-1 =(AX_A;EF)$ by Three Tangents Lemma, and since $QA$ is a tangent to $\omega$ , $QX_A$ must also be, as desired.

Claim: $PA = PX_A$ .

Proof: First, remark that $PKHD$ is cyclic, so by radical axes on $(PKHD)$ and $(DHX_AM)$ we find that $PKX_AM$ is cyclic. We claim that $KLX_AM$ is also cyclic, which would imply the $L$ is the arc midpoint of $\triangle PKX_A$ , and thus $\angle APL = \angle LPX_A$ , which would finish the claim.

To this end, we invert around $\omega$ . We wish to show that $K,J,X_A$ are collinear. Redefine $J$ to be the intersection of $KX_A$ and $EF$ . Remark that $AK \parallel EF$ , so $-1 = (EF;AX_A) \stackrel{K}{=}(EF;\infty_{EF}J)$ , so $J$ is the midpoint of $EF$ , as desired.

Finally, consider the circles centered around $P$ and $Q$ passing through $A$ and $X_A$ . Their radical axis is $AM$ , thus $PQ \perp AM$ , as desired.

Remarks: After reading the solutions I realized how much I overcomplicated this problem. However, I think most steps in this proof are really natural if you've seen HM point before; most points that showed up in my diagram are standard pieces of the orthocenter puzzle; the only strange one was $P$ and $K$ , which was dealt with by only considering the property that $AP \parallel EF$ instead of trying to use the tangent.

This post has been edited 1 time. Last edited by mathlogician, Aug 8, 2020, 5:53 PM

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#19 h Oct 17, 2020, 11:00 PM

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Let the midpoint of $BC$ be $M$ .

Invert about the circle centered at $A$ with radius $\sqrt{AE \cdot AC}=\sqrt{AF \cdot AB}$ . Then $E$ and $F$ go to $B$ and $C$ , so $(AEF)$ gets sent to line $BC$ and $(ABC)$ gets sent to line $EF$ .

$P$ goes to the other intersect of line $AP$ and $(AEF)$ and $Q$ goes to the second intersection of $(ABC)$ and line $QA$ . Call these two points $P'$ and $Q'$ , respectively.

We want to show that lines $\overline{AM} \perp \overline{PQ}$ . It suffices to show that their images under the inversion are orthagonal. Since $\overline{AM}$ goes through the center of inversion ( $A$ ), it gets sent to itself. $\overline{PQ}$ gets sent to $(P'AQ')$ . For a circle and a line to be orthogonal, the line must go through the circle's center. So to finish, it suffices to show that $MP'=MA=MQ'$ .

The angle between $\overline{AH}$ and $\overline{BC}$ maps to the angle between $\overline{AH}$ and $\overline{Q'A}$ so $\angle Q'AH=90^\circ$ and $QA \parallel BC$ . Hence $ABCQ'$ is a trapezoid. $ABCQ'$ is cyclic and consequently an isocoleces trapezoid so, combined with the fact that $M$ is a midpoint, $\overline{MA}=\overline{MQ'}$ . Similarly, $\overline{PA} \parallel \overline{EF}$ so $P'AEF$ is a trapezoid. $P'AEF$ is also cyclic and consequently an isocoleces trapezoid. Now, we prove the following

Claim. $M$ lies on the perpendicular bisector of $EF$ .
Proof. Notice that $\measuredangle BFC = \measuredangle BEC$ so $BFEC$ is cyclic. It is well known that the circumcenter of a right triangle is the midpoint of the hypotenuse. The circumcenter of both $\triangle BFC$ and $\triangle BEC$ is $M$ . Hence, it must also be the circumcenter of $(BFEC)$ . The circumcenter of a cyclic quadrilateral is the intersection of the perpendicular bisectors of its sides. So the perpendicular bisector of $EF$ meets $BC$ at $M$ . $\square$

Hence $M$ is on the perpendicular bisector of $AP'$ and $MA=MP'$ . Thus, $MP'=MA=MQ'$ and the problem is solved. $\blacksquare$

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#20 h May 13, 2021, 7:28 PM

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j___d wrote:

An acute triangle $ABC$ in which $AB<AC$ is given. Points $E$ and $F$ are feet of its heights from $B$ and $C$ , respectively. The line tangent in point $A$ to the circle escribed on $ABC$ crosses $BC$ at $P$ . The line parallel to $BC$ that goes through point $A$ crosses $EF$ at $Q$ . Prove $PQ$ is perpendicular to the median from $A$ of triangle $ABC$ .

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First of all note that $(GD,BC)=-1$ since $BEFC$ is a complete quadrilateral. also $\angle PAF=\angle ACB=\angle AFE$ so $AP||QG$ , now from $(GD,BC)=-1$ we understand that $(AE,Ft)=-1$ which gives that $QA=QT$ on the other hand $T$ is A-humpty point so $PA=PT$ so $PQ$ is the perpendicular bisecot of $AT$ and we are done.(You can read more about A-Humpty point here)

This post has been edited 1 time. Last edited by iman007, May 13, 2021, 7:29 PM
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RM1729

#21 h Nov 9, 2021, 6:21 PM

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An excellent problem! Let $H$ denote the orthocentre of $ABC$ . Let $\omega_1$ denote the circumcircle of $ABC$

First, note that $AEHf$ and $EFCB$ are cyclic quadrilaterals. Denote by $\omega_2$ the circle through $AEFH$ (diameter $AH$ ) and $\omega_3$ the circle through $EFCB$ (diameter $BC$ and centre $M_{BC}$ ).

Now observe that the parallel through $A$ to $BC$ is simply the tangent at $A$ to $\omega_2$

We now define a circle of radius zero at point $A$ as $\omega_4$ and consider the radical centre of $\omega_1$ , $\omega_3$ , $\omega_4$

Note that the radical axes are $BC$ , tangent at $A$ and the radical axis of the zero radius circle at $A$ and the circle through $EFCB$

Since $BC \cap$ tangent at $A = P$ , $P$ lies on the 3rd radical axis too.

We now consider the radical centre of $\omega_2$ , $\omega_3$ and $\omega_4$

Note that the radical axes are $EF$ , parallel at $A$ and the same radical axis of the zero radius circle at $A$ and the circle through $EFCB$

Since $EF \cap$ parallel at $A = Q$ , $Q$ lies on the same radical axis too

Thus line $PQ$ is just the radical axis of the radius $0$ circle at $A$ and the circle through $EFCB$ . Since the line joining the centres is perpendicular to the radical axis, $AM \perp PQ$

This post has been edited 1 time. Last edited by RM1729, Nov 9, 2021, 6:22 PM

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DottedCaculator

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DottedCaculator

#22 h Dec 11, 2021, 3:55 AM • 2 Y

Y by MiraclesINmaths, Elnuramrv

Solution

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#23 h Dec 13, 2021, 6:58 PM

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Let $\omega_a$ denote the point circle $A$ and $M$ be the midpoint of $BC$ . It's clear that $Pow_{\omega_a}(P) = PA^2 = Pow_{(ABC)}(P) = PB \cdot PC = Pow_{(BC)}(P)$ and the Three Tangents Lemma gives $Pow_{\omega_a}(Q) = QA^2 = Pow_{(AEF)}(Q) = QE \cdot QF = Pow_{(BC)}(Q)$ so $PQ$ is the Radical Axis of $\omega_a$ and $(BC)$ . But $(BC)$ is centered at $M$ , so we're done. $\blacksquare$

Remark: Utilizing the $A$ -Humpty point is probably more motivated than this solution.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#24 h Dec 16, 2021, 7:20 AM

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In order to prove PQ ⊥ AM we need to prove |PA^2 - PM^2| = |QA^2 - QM^2| or |PA^2 - QA^2| = |PM^2 - QM^2|
|PA^2 - QA^2| = PB.PC - QE.QF = power of P w.r.t BFEC - power of Q w.r.t BFEC
Let M be midpoint of BC. M is center of BFEC so :
power of P w.r.t BFEC - power of Q w.r.t BFEC = |PM^2 - QM^2|
so at the end we have |PA^2 - QA^2| = P w.r.t BFEC - power of Q w.r.t BFEC = |PM^2 - QM^2| and this will prove our problem so we're Done.

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#25 h Jan 22, 2022, 5:05 AM • 2 Y

Y by centslordm, megarnie

Let $M$ be the midpoint of $\overline{BC},$ $\omega_1=(M,\overline{MB}),$ and $\omega_2=(A,0)$ with $(O,r)$ denoting the circle with center $O$ and radius $r.$ Notice $\angle PAB=\angle ACB$ so $\triangle ACP\sim\triangle BAP.$ Hence, $\text{pow}_{\omega_2}{P}=AP^2=PB\cdot PC=\text{pow}_{\omega_1}{P}.$ Similarly, $Q$ lies on the radical axis of $\omega_1$ and $\omega_2$ by the three tangents lemma. Radical Axis finishes. $\square$

Remarks: (Motivation) My solution was motivated by the three tangents lemma, which had occurred to me after seeing $A,E,F,$ and $\overline{AQ}\parallel\overline{BC}.$ Since $M$ was the center of $(BCEF),$ radical axis was natural and the solution followed.

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#27 h Jun 3, 2022, 10:05 AM

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Special Thanks to guptaamitu1 for referring and motivating me to solve this problem

.

consider a circle of radius zero

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#28 h Jun 3, 2022, 5:29 PM • 1 Y

Y by Mango247

From $\angle BAP=\angle BCA=\angle EFA =\angle EAQ$ we deduce $|QA|^2=|QE|\cdot |QF|,|PA|^2=|PB|\cdot |PC|,$ hence $PQ$ is the radical axis of $A,\odot (BCEF),$ the conclusion follows.

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#29 h Jun 11, 2022, 4:08 AM

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Sol:- Note that $P$ is the radical center of $(A, 0), (BC), (ABC)$ and $Q$ is the radical center of $(A,0), (BC), (AED)$ .Hence the radax of $(A, 0)$ and $(BC)$ is $A$ - median.

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#30 h Sep 11, 2022, 6:22 AM • 1 Y

Y by Mango247

We claim $P$ and $Q$ lie on the radax of $(A)$ and $(BFEC)$ (which is sufficient). By PoP on $(ABC)$ we have $PA^2 = PB \cdot PC$ . It is well known that $AQ$ is tangent to $A$ . By PoP $AQ^2 = QE \cdot QF$ . Proving our result.

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#31 h Apr 20, 2023, 10:54 PM

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Lol I complicated the solution so much although iÍ still think it is really nice.

Let $X$ be the intersection of $AM$ with the nine-point circle $\gamma$ . Also, let $Y$ be the intersection of $AH$ with $\gamma$ . It is well-known that $MY$ is a diameter of $\gamma$ , so $\angle YXM=90^\circ$ . It now suffices to show that $P,Q\in XY$ .

It is well-known that $(AEHF)$ is tangent to the line through $A$ to $BC$ . To see that, simply note $\angle QAE=\angle ACB=180^\circ-\angle BFE=\angle EFA$ . It is also easy to see that $\triangle AXY$ and $\triangle AMD$ are homothetic at $A$ , so $(AXY)$ is tangent to $QA$ as well. Applying the radical axis theorem to $\gamma$ , $(AEHF)$ and $(AXY)$ shows that $Q\in XY$ .

An angle chase shows that
$\angle PAB=\angle ACB=\angle EFA=\angle AEH$ meaning that $AP\parallel EF$ . As $MY\perp EF$ , we also have $MY\perp AP$ . Therefore, letting $Z=MY\cap AP$ , we see that $Z$ lies on $(AYX)$ . Therefore, $AZDM$ is cyclic, and apllying the radical axis theorem on $(AZDM)$ , $\gamma$ and $(AYX)$ we see that $P\in YX$ , so we are done.

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#32 h Feb 4, 2024, 6:16 PM

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Let $M$ be the midpoint of $BC$ .
We claim that $PQ$ is the radical axis of the circle with radius $0$ centered at $A$ , and $(BFEC)$ .
Since by Power of a Point on $(ABC)$ , we have $PA^2 = PB \cdot PC$ , $P$ has equal power to both $(A)$ and $(BFEC)$ . Then we have $\angle QAE = \angle C = \angle AFE$ , and $\angle Q = \angle Q$ , $\triangle QAF \sim \triangle QEA$ . From this, we have $\frac{QA}{QE} = \frac{QF}{QA} \implies QA^2 = QE \cdot QF$ so $Q$ also lies on the radical axis of $(A)$ and $(BFEC)$ . We finish by noting that $A$ and $M$ are the respective centers of $(A)$ and $(BFEC)$ , so $AM \perp PQ$ .

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#33 h Mar 25, 2024, 4:52 AM

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oops radius 0 is so smart i didn't think of that

here is a probably isomorphic solution

We first eliminate $P$ . The conclusion is equivalent to $PM^2-PA^2=QM^2-QA^2.$ However, note that $PM^2-PA^2=PM^2-PB\cdot PC=PM^2-(PM^2-MB^2)=MB^2=\frac{a^2}{4}.$ Thus, it remains to show that $QM^2-QA^2=\frac{a^2}{4}.$
Note that $\angle QAE=\angle QFA=\gamma$ since $AQ\parallel BC$ and $BFEC$ is cyclic. Thus, $QA^2=QE\cdot QF.$ Hence, $QE\cdot QF=Pow_{(BFEC)}(Q)=QM^2-\frac{a^2}{4},$ as desired.

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ismayilzadei1387

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ismayilzadei1387

#34 h Mar 26, 2024, 2:52 PM

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Decent problem indeed.
$EF \cap BC$ at $X_{A}$
$X_{A}H \cap AQ$ at $R$
$EF \cap AH$ at $L$
We have $X_{A}H$ is prependicular to $AM$ that motivates us to prove $PQ$ is parallel to $X_{A}H$
We also have $AQ$ is parallel to $BC$ and $PX_{A}$ and $AP$ is parallel to $X_{A}Q$ so that $APX_{A}Q$ is parallelogram
It remains to prove $Q$ is midpoint of $AR$
From easy harmonic bundle $(A,H;L,AH \cap BC) \equiv (A,R;Q,V_\infty)=-1$ .
which leads conclusion.

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#35 h Mar 12, 2025, 11:11 PM

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Idk isn't the question trivial? PQ is the radical axis between (A) (radius 0) and the circle of diameter BC?

I think post #23 said it but it's quite confused ?

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#36 h Mar 13, 2025, 2:28 AM • 1 Y

Y by peace09

theres no way this worked

We claim that $N$ , the midpoint of $AH$ , is exactly the orthocenter of triangle $PAM$ . Indeed, it lies on $AH$ , and also $MNAO$ is a parallelogram so $MN\perp AP$ . Now if $PN\cap AM=X$ and $R$ is the $A$ -Humpty point, then $X$ is the midpoint of $AR$ . Therefore, it suffices to prove that $QX$ is also perpendicular to $AM$ .

Note that $Q$ is the radical center of $(BFEC)$ , $(AFE)$ , and the point circle at $A$ . Therefore, the foot of the perpendicular from $Q$ to $AM$ should have equal power to the point circle at $A$ and $(BFEC)$ . Call that point $Y$ . Then
$AY^2=YM^2-MC^2\Longrightarrow (YM-YA)(YM+YA)=MC^2\Longrightarrow YM-YA=RM.$ Therefore, $Y$ is the midpoint of $AR$ , so $Y=X$ , as desired. $\blacksquare$

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#37 h Mar 15, 2025, 12:46 AM

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Let $\omega$ be the circle at $A$ with radius $0$ and let $M$ be the midpoint of $\overline{BC}$ . It is obvious that $M$ is the center of $(BFEC)$ .

Note that $PA^2 = PB \cdot PC$ from Power of a Point on $(ABC)$ , which means $P$ lies on the radical axis of $\omega$ and $(BFEC)$ . Moreover, some angle chasing yields

$\angle QAE = \angle ECB = \angle EFA,$
so $\triangle QAE \sim \triangle QFA$ . This implies $QA^2 = QE \cdot QF$ , so $Q$ also lies on the radical axis of $\omega$ and $(BFEC)$ . Hence, the radical axis is $\overline{PQ}$ , and it is perpendicular to the line connecting the centers of the circles, which is $\overline{AM}$ . $\blacksquare$

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