Arbitrary point on BC and its relation with orthocenter

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Source: Balkan MO 2025 P2

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#1 h Apr 27, 2025, 11:47 AM • 5 Y

Y by farhad.fritl, Frd_19_Hsnzde, ehuseyinyigit, pomodor_ap, Nuran2010

In an acute-angled triangle $ABC$ , $H$ be the orthocenter of it and $D$ be any point on the side $BC$ . The points $E, F$ are on the segments $AB, AC$ , respectively, such that the points $A, B, D, F$ and $A, C, D, E$ are cyclic. The segments $BF$ and $CE$ intersect at $P.$ $L$ is a point on $HA$ such that $LC$ is tangent to the circumcircle of triangle $PBC$ at $C.$ $BH$ and $CP$ intersect at $X$ . Prove that the points $D, X,$ and $L$ lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus

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#2 h Apr 27, 2025, 11:49 AM

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nice problem

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#3 h Apr 27, 2025, 11:50 AM • 4 Y

Y by alexanderhamilton124, Nuran2010, Amkan2022, ihatemath123

MuradSafarli wrote:

nice problem

gurt:yo

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#4 h Apr 27, 2025, 11:52 AM

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Sadigly wrote:

MuradSafarli wrote:

nice problem

gurt:yo

yo:what?

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#5 h Apr 27, 2025, 12:02 PM • 1 Y

Y by GeorgeRP

Trig setup

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#6 h Apr 27, 2025, 12:14 PM • 1 Y

Y by khina

Feels a bit troll, solved it in around 5 minutes.
Simple angle chase gives that $BCPH, AEPF, BDPE, CDPF$ are all cyclic. Let $A'$ be reflection of $A$ in $D$ . Then $A'$ is obviously on $(BCPH)$ . Also $\angle BPD=\angle BED=\angle BCA=180^\circ-\angle BHA=\angle BHA'=\angle BPA'$ so $P, D, A'$ are collinear. Now Pascal on $CCPA'HB$ solves.

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#7 h Apr 27, 2025, 1:46 PM

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Diagrams

Solution

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#8 h Apr 27, 2025, 2:05 PM

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Solution: We first claim the following:

This claim wrote:

The reflection of $P$ over $BC$ is the second intersection of $AD$ and $(ABC)$ .

Proof. Let $P'$ be the second intersection of $AD$ and $(ABC)$ . Then, since $\begin{align*} \measuredangle PBC &= \measuredangle FBD \\ &= \measuredangle FAD \\ &= \measuredangle CAP' \\ &= \measuredangle CBP' \\ &= -\measuredangle P'BC \end{align*}$ then $BP$ and $BP'$ are reflections over $BC$ . Note that since $\measuredangle AEP = \measuredangle ADC = \measuredangle ADB = \measuredangle AFP$ , then $AEPF$ is cyclic, implying that $\measuredangle BPC = \measuredangle FPE = -\measuredangle BAC = \measuredangle BP'C$ , so $P$ and $P'$ are indeed reflections over $BC$ .

Now we reflect everything except $A$ over $BC$ , without overlaying the new diagram with the old one. We can also do barycentrics on $\triangle ABC$ now.
Let $a$ , $b$ , $c$ , $A$ , $B$ , $C$ , $S_A$ , $S_B$ , $S_C$ denote $BC$ , $CA$ , $AB$ , $(1, 0, 0)$ , $(0, 1, 0)$ , $(0, 0, 1)$ , $\frac{-a^2+b^2+c^2}{2}$ , $\frac{a^2-b^2+c^2}{2}$ , $\frac{a^2+b^2-c^2}{2}$ respectively. (I know that's a lot but they're just common notation anyway)
We first calculate $H$ . Let $H=(t:S_C:S_B)$ . Then, $\begin{align*} -a^2S_BS_C - b^2S_Ct - c^2tS_B &= 0 \\ \iff t &= -\frac{a^2S_BS_C}{b^2S_B+c^2S_C} \end{align*}$ so $H=(-a^2S_BS_C: S_C(b^2S_B+c^2S_C): S_B(b^2S_B+c^2S_C))$ . Let $P = (x:y:z)$ . Then obviously $-a^2yz-b^2zx-c^2xy=0$ . We can also calculate $X=(-a^2S_BS_Cx : -a^2S_BS_Cy : S_Bx(b^2S_B+c^2S_C))$ , $D=(0:y:z)$ and $L=(-a^2S_C:b^2S_C:b^2S_B)$ . Finally, $\begin{align*} \begin{vmatrix} 0&y&z\\ -a^2S_C&b^2S_C&b^2S_B\\ -a^2S_BS_Cx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\ \end{vmatrix} &= -a^2S_C \begin{vmatrix} 0&y&z\\ 1&b^2S_C&b^2S_B\\ S_Bx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\ \end{vmatrix} \\ &=-a^2S_BS_C \begin{vmatrix} 0&y&z\\ 1&b^2S_C&b^2S_B\\ x&-a^2S_Cy&x(b^2S_B+c^2S_C)\\ \end{vmatrix} \\ &= -a^2S_BS_C((-a^2S_Cyz - xy(b^2S_B+c^2S_C))+(b^2S_Bxy-b^2S_Cxz)) \\ &= -a^2S_BS_C(-a^2S_Cyz-b^2S_Czx-c^2S_Cxy) \\ &= 0 \end{align*}$ so we're done.

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#9 h Apr 27, 2025, 2:50 PM

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Here is non-projective synthetic solution.
Since $\angle BEC=\angle ADB=\angle AFB$ , $AEPF$ and $BHPC$ are cyclic and $\angle FPC=\angle BAC=\angle FDC$ , so $CFPD$ is cyclic. Now, we claim that $L$ lies on the radical axis of $(BHD)$ and $(CDPF)$ , which clearly finishes the problem as this radical axis is $XD$ due to $XH \cdot XB=XP\cdot XC$ . Let $AH \cap (BHD)=Q$ and $LC \cap (CPD)=R$ . Observe that $\angle LCB=\alpha$ and $\angle DRC=\angle DFC=\beta$ , so $\angle RDC=\gamma=\angle BHQ=\angle BDQ$ , so $Q, D, R$ are collinear. Then $\angle HQR=\angle HQD=\angle HBC=\angle RCH$ , i.e. $HQCR$ is cyclic, i.e. $LH \cdot LQ=LR \cdot LC$ and thus $L$ lies on the radical axis.

This post has been edited 2 times. Last edited by VicKmath7, Apr 27, 2025, 3:38 PM

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#10 h Apr 27, 2025, 3:12 PM • 2 Y

Y by vi144, Ciobi_

Note that saying points $D,X,L$ lie on the same line is equivalent to saying $BH,CE,DL$ are concurrent lines.

It is then natural to apply Desargues's Theorem on $\triangle LHC$ and $\triangle DBE$ .

Let $A'$ and $C'$ be the feet of the heights from $A$ and $C$ respectively.
Since we need to prove that $LH\cap DB, HC\cap BE, LC\cap DE$ are collinear,
and since (by Reim's Theorem) $A'C'\parallel DE$ ,
then it would be sufficient to prove that $LC$ is parallel to these two lines as well.

As noted before, by rather straightforward angle chasing, $P$ lies on the circle $(BHC)$ .

Hence $\angle LCH= \angle HBC=\angle HAC$ . And since $\angle HCB= \angle HAB$ , we get $\angle LCB= \angle A=\angle EDB$ and the conclusion follows.

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#11 h Apr 27, 2025, 3:13 PM

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Complex bash with $\triangle ABC$ inscribed in the unit circle, and let $AD$ meet the unit circle again at $U$ , so that
$|a|=|b|=|c|=|u|=1$ $h = a+b+c$ $d = \frac{au(b+c) - bc(a+u)}{au-bc}$ Let lines $BF,CE$ intersect the unit circle again at $V,W$ respectively. Now since $A,B,D,F$ are concyclic, we have
$\frac{(a-d)(b-f)}{(a-f)(b-d)} \in \mathbb{R} \implies \frac{(a-u)(b-v)}{(a-c)(b-c)} = \frac{c^2(a-u)(b-v)}{uv(a-c)(b-c)} \implies c^2 = uv$ So
$v = \frac{c^2}u$ and similarly
$w = \frac{b^2}u$ Then
$\begin{align*} p &= \frac{bv(c+w) - cw(b+v)}{bv-cw} \\ &= \frac{\frac{bc^2}u\left(c+\frac{b^2}u\right) - \frac{b^2c}u\left(b+\frac{c^2}u\right)}{\frac{bc^2}u-\frac{b^2c}u} \\ &= \frac{bc^3u+b^3c^2-b^3cu-b^2c^3}{bc^2u-b^2cu} \\ &= \frac{c^2u+b^2c-b^2u-bc^2}{cu-bu} \\ &= \frac{bu+cu-bc}u \end{align*}$ (So $P$ is the reflection of $U$ over line $BC$ .) Now since $L$ lies on line $HA$ , we have
$\overline{\ell} = \frac{a\ell + bc - a^2}{abc}$ And since $LC$ is tangent to the circumcircle of $\triangle PBC$ , we have
$\frac{(c-\ell)(b-p)}{(b-c)(c-p)} \in \mathbb{R}$ $\frac{c(c-\ell)(b-u)}{b(b-c)(c-u)} = \frac{\frac1c\left(\frac1c-\frac{a\ell + bc - a^2}{abc}\right)\left(\frac1b-\frac1u\right)}{\frac1b\left(\frac1b-\frac1c\right)\left(\frac1c-\frac1u\right)} = -\frac{(a^2+ab-a\ell-bc)(b-u)}{a(b-c)(c-u)}$ $ac(c-\ell) = -b(a^2+ab-a\ell-bc) \implies \ell = \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)}$ Now we find the coordinate of $X$ . Since $X$ lies on line $BH$ , we have
$\overline{x} = \frac{bx+ac-b^2}{abc}$ Since $X$ lies on line $CP$ , we have
$\frac{c-x}{c-p} \in \mathbb{R}$ $\frac{u(c-x)}{b(c-u)} = \frac{\frac1u\left(\frac1c - \frac{bx+ac-b^2}{abc}\right)}{\frac1b\left(\frac1c-\frac1u\right)} = -\frac{ab-ac+b^2-bx}{a(c-u)}$ $au(c-x) = -b(ab-ac+b^2-bx) \implies x = \frac{ab^2-abc+acu+b^3}{au+b^2}$ Now we find the vectors
$\begin{align*} d-\ell &= \frac{au(b+c) - bc(a+u)}{au-bc} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\ &= \frac{a(b+c)(abu+acu-abc-bcu) - (au-bc)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au-bc)} \\ &= \frac{-a^3bu-a^2bc^2+2a^2bcu+ab^3c+abc^3-abc^2u-b^3c^2}{a(b+c)(au-bc)} \\ &= -\frac{b(a-c)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au-bc)} \end{align*}$ and
$\begin{align*} x-\ell &= \frac{ab^2-abc+acu+b^3}{au+b^2} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\ &= \frac{a(b+c)(ab^2-abc+acu+b^3) - (au+b^2)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au+b^2)} \\ &= \frac{-a^3bu-a^2b^2u-a^2bc^2+a^2bcu+ab^3c-ab^2c^2+ab^2cu+b^4c}{a(b+c)(au+b^2)} \\ &= -\frac{b(a+b)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au+b^2)} \end{align*}$ Then
$\frac{d-\ell}{x-\ell} = \frac{(a-c)(au+b^2)}{(a+b)(au-bc)}$ which is real. $\blacksquare$

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#12 h Apr 27, 2025, 3:35 PM • 1 Y

Y by egxa

Let $A'$ be the reflection of $A$ over $BC$ . $D$ is the miquel of $AEPF$ . Since $A,E,F,P$ are concyclic, $P\in (BHCA')$ . Also $\measuredangle (DE,AH)=\measuredangle (AH,DF)$ hence projecting DDIT at $AEPF$ from $D$ , there exists an involution $(A,DP\cap AH),(DE\cap AH,DF\cap AH),(AH\cap BC,AH\cap BC)$ . This must be reflection over $BC\cap AH$ thus, $D,P,A'$ are collinear. Pascal at $BHA'PCC$ gives the result as desired. $\blacksquare$

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#13 h Apr 27, 2025, 4:05 PM

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Proposed by Theoklitos Parayiou, Cyprus

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#14 h Apr 27, 2025, 4:12 PM

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Orestis_Lignos wrote:

Proposed by Theoklitos Parayiou, Cyprus

is that the same guy proposed the 2020 JBMO-P2?

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Marios

#15 h Apr 27, 2025, 4:21 PM

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giangtruong13 wrote:

Orestis_Lignos wrote:

Proposed by Theoklitos Parayiou, Cyprus

is that the same guy proposed the 2020 JBMO-P2?

Yes, It is the same person. He proposed a handful of other geometry problems for Balkan olympiads as well.

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#16 h Apr 27, 2025, 5:44 PM

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WLOG, solve the analogous problem where $L$ is the intersection of the tangent at $B$ and line $AH$ , so $X = BF \cap CH$ . Let $H_A = AH \cap BC$ . Let $x = \angle CBF = \angle DAC$ . By Menelaus applied on the transversal $L,X,D$ on triangle $HH_AC$ it is enough to prove: $\frac{LH}{LH_A} \cdot\frac{H_AD}{DC} \cdot \frac{CX}{XH} = 1$ Using the Ratio Lemma on triangle $BH_AH$ we get: $\frac{LH}{LH_A} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)}$ Using the Ratio Lemma on triangle $BCH$ we get: $\frac{CX}{CH} = \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)}$ Using the Ratio Lemma on triangle $H_AAC$ we get: $\frac{H_AD}{DC} = \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)}$ Multiplying the three expressions, we get:
$\frac{LH}{LH_A} \cdot \frac{H_AD}{DC} \cdot \frac{CX}{XH} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)} \cdot \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \cdot \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} = 1$

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#17 h Apr 28, 2025, 12:21 AM

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From miquelpoint we have $BEPD, DPFC$ cyclic and additional trivially from the angles we have $BHPC$ cyclic, let $A'$ reflection of $A$ over $BC$ then from the angles from miquel config at $P$ we trivially have $D,P,A'$ colinear and thus a pascal at $(BHC)$ finishes.

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#19 h Apr 28, 2025, 7:18 AM

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This is a quick one I found when I was testing the problem.Same as previous solutions we prove that $H_CH_A\parallel ED\parallel l$ , where $l$ is the tangent to $(BHPC)$ at $C$ . Now define $L$ as the intersection of $AH$ and $DX$ . Apply Desargues's theorem on triangles $\triangle BH_AH_C$ and $\triangle XLC$ . Since $LH_A, CH_C$ and $BX$ are concurrent at $H$ , we get the intersection of $BH_C$ and $CX$ which is $E$ , the intersection of $BH_A$ and $LX$ which is $D$ and the intersection of $LC$ and $H_AH_C$ are collinear. However, since $DE \parallel H_AH_C$ , $LC$ is also parallel to these lines, therefore it coincides with the tangent and we are done. Not surprisingly Desargues's theorem on triangle $\triangle BDE$ and $\triangle LXC$ also works.

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Rayvhs

#20 h Apr 28, 2025, 3:53 PM

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Let $DP\cap AH=Q$ .

From ABDF and ACDE being cyclic, we get that BEPDand CDPF are cyclic as well.
Thus, we have
$\angle BDP = \angle AEC = \angle ADC = \angle CDQ.$ Also, since $AQ\perp BD$ , $\bigtriangleup ADQ$ is isosceles.
Therefore, Q is the symmetric point of A wrt BC.
Apply Pascal to BHQPCC and we're done.

This post has been edited 1 time. Last edited by Rayvhs, Apr 28, 2025, 3:55 PM

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#21 h Apr 28, 2025, 4:30 PM

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Let $A’$ be the image of $A$ under reflection across $BC$ .

By cyclic chasing, we obtain that $P$ lies on $(BHC)$ and $B,E,P,D$ are concyclic,

implying $P,D,A’$ are collinear. Note that $A’$ also lies on $(BHC)$ .

Applying Pascal on $PCCBHA’$ , we are done.

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#22 h Apr 28, 2025, 6:51 PM

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MMP solution:

Claim: $B$ , $H$ , $P$ , $C$ are cyclic.
Proof: $180 - \angle BPC = \angle PBC + \angle PCB = \angle FAD + \angle EAD = \angle BAC = 180 - \angle BHC \square$

Let $AH$ intersect $(BHPC)$ at $K \neq H$ .

Claim: $P$ , $D$ , $K$ collinear.
Proof: Reflect $P$ over $BC$ , name the point $P'$ . Notice $P'$ lies on $(ABC)$ .
$\angle DAC = \angle DBP = \angle DBP' = \angle P'AC$
Thus $P'$ , $D$ , $A$ collinear, so $P$ , $D$ , $K$ collinear $\square$

Now notice that if we let $\deg(D) = 1$ , then $P = KD \cap (BHC)$ , thus $\deg(P) = 2$ , and then by Zach's, $\deg(CP) = 1$ and $\deg(X) = 1$ . Notice $L$ is fixed since $P$ is on $(BHC)$ . Thus we need to check $1 + 1 + 0 + 1 = 3$ cases. Pick $D$ on $AH \cap BC$ , $B$ and $C$ .

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Mapism

#23 h Apr 28, 2025, 8:18 PM

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We switch the points $E$ and $F$ . Notice $D$ is the miquel point of complete quadrilateral $FPEA$ . This yields,
$BEPD,\ CFPD\ \text{cyclic}\ ,\ \ D\in BC \implies FPEA\ \text{cyclic}$ $180-\angle BHC=\angle BAC=\angle FAC=\angle FDB=\angle FPB=180-\angle BPC \implies BHPC\ \text{cyclic}$ Let $A'$ be the reflection of $A$ across $BC$ , it is well known that $A'\in BHPC$ . Pascal's theorem on cyclic quadrilateral $CCPA'HB$ gives
$D,X,L\ \text{collinear} \iff P,D,A' \ \text{collinear}\iff \angle PDC=\angle BDA'$ $\angle BDA'=\angle BDA=\angle BEA=180-\angle PEC=\angle PDC$ thus we're done $\Box$

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#24 h Apr 29, 2025, 7:15 AM

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2 days and P1 and P4 are still nowhere to be found

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EVKV

#25 h Apr 29, 2025, 1:34 PM • 1 Y

Y by Rotten_

@above they are on AoPS forums but not in contest collections for some reason

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SomeonesPenguin

#26 h Apr 29, 2025, 3:53 PM

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A quick angle chase yields $\angle BPC=180^\circ-\angle A=\angle BHC$ which means that $BHPC$ is cyclic. Therefore, $L$ is the fixed point on $HA$ such that $\angle BCL=\angle A$ . Now, the function $f:BC\mapsto BH$ defined by $D\mapsto E\mapsto X$ is projective since $E=D\infty_{\ell_b}\cap AB$ , where $\ell_B$ is the tangent line to $(ABC)$ through $B$ and $X=BH\cap CE$ . Notice that $f(B)=B$ so by prism lemma (or Steiner conic) the line $DX$ passes through a fixed point as $D$ moves along $BC$ . When $D$ is the foot of the $A$ -altitude, $DX$ becomes $HA$ and when $D=C$ we get $\angle BCX=\angle A$ , hence the fixed point is indeed $L$ .

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#27 h May 1, 2025, 4:50 AM

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MMP lets gooo
Consider a variable point $X$ on $BH$ .
We let $LX$ intersect $BC$ at $D’$
Notice that deg $E =1$ and deg $D’ =1$ .
Perform an inversion at $A$ radius $AC$ that map $E$ to $E_1$ and $D’$ to $D_1$ . Hence, the condition $E_1, D_1, C$ are collinear has deg $2$ . And since the inversion is projective, the condition $E, D’, A, C$ concyclic has a deg $2$ .
Hence, its suffice to check $3$ cases.
Consider $X = B, H, BE \cap LC$
Case $1$ : $X=B$
This one is trivial.
Case $2$ : $X=H$ .
This one is normal orthocenter config.
Case $2$ : $X= BE \cap LC = Y$
Let $LC \cap AB = Z$ , we need to show that $(AZC)$ is tangent to $BC$
This could be done by angle chasing, hence we are done.

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#28 h May 1, 2025, 11:29 AM

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Solution. We start with a few basic observations.
Claim. $B,H,C,P$ are concyclic.
Proof. This is true since
$\angle{BPC}=180-\angle{PBC}-\angle{PCB}=180-\angle{FBD}-\angle{ECD}=180-\angle{DAB}-\angle{DAC}=180-A=\angle{BHC}$ holds. $\blacksquare$

Claim. $P,E,B,D$ and $P,F,C,D$ are concyclic.
Proof. The following angle equalities hold.
$\angle{PBD}=\angle{FBD}=\angle{FAD}=\angle{CAD}=\angle{CED}=\angle{PED}$ One can similiarly prove that $\angle{PCD}=\angle{PFD}$ , which implies the claim. $\blacksquare$

Define $A'$ as the reflection of $A$ w.r.t $BC$ .
Claim. $\overline{P-D-A'}$ are collinear.
Proof. It is well known that $(BHC)$ is the reflection of $(ABC)$ w.r.t $BC$ . From this we obtain $A'\in(BHC)$ . The aforementioned claim implies that
$\angle{BPD}=C=\angle{A'CB} \text{ and } \angle{CPD}=B=\angle{A'BC}$ hold. This is enough to prove the claim. $\blacksquare$

Pascal in $(CCPA'HB)$ implies $\overline{CC\cap A'H-CP\cap HB-PA'\cap BC}\Rightarrow \overline{L-X-D}$ are collinear. $\blacksquare$

This post has been edited 2 times. Last edited by Yagiz_Gundogan, May 1, 2025, 12:54 PM
Reason: grammar

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#29 h May 1, 2025, 12:13 PM

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$\angle BPC=180^\circ-\angle PBC-\angle PCB=180^\circ-\angle FBD-\angle ECD=180-\angle FAD-\angle EAD=180-\angle BAC=\angle BHC\Longrightarrow$ $BCPH$ inscribed $(XH\cdot XB=XP\cdot XC)$ . Let $AH\cup (BDH)=G$ , $LC\cup (CDP)=I$ and $GD\cup AC=J\Longrightarrow \angle GAJ=\angle HAC=\angle HBC=\angle HBD=\angle HGD=\angle AGJ=\alpha$ $\angle BHG=\angle HAB+\angle HBA=\angle HCB+\angle HCA=\angle ACB=90^\circ-\alpha;$ $\angle DJC=\angle AGJ+\angle GAJ=2\alpha \Longrightarrow 90^\circ-\alpha=\angle CDJ=\angle BDG \Longrightarrow G,D,J$ lie on the same straight line. This means that $\angle ICH=\angle LCH=\angle LCP+\angle PCH=\angle CPB+\angle PBH=\angle CBH=\angle DBH=\angle DGH=\angle IGH\Longrightarrow CIHG$ , inscribed in $LH\cdot LG=LI\cdot LC\Longrightarrow D,L,X$ , lies on the radical axis of the circumscribed circles $(BDH)$ and $(CDP)$ .

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#30 h May 1, 2025, 3:38 PM

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After some angle chasing, let $CE \cap AH$ at $R$ and connect $CH$ then do ratio lemma with menelaus.

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optimusprime154

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optimusprime154

#31 h Yesterday at 6:22 PM

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let $deg(D)=1$ move on BC, similar to other solutions we know $BHPC$ cyclic so $deg(P)=2$ if we let $D=C$ we get $P=C$ so $deg(PC)=1$ since $BH$ is fixed then $deg(X) = 1$ $L$ is a fixed point so it suffices to check 3 values of $D$ we check $D=B$ , $D=C$ and $D=V$ where $V$ is the foot from $A$ to $BC$ all are trivial.

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#32 h 3 hours ago

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AMAZING PROBLEM ,
Let, $AH \cap (BPC) =L'$
$P$ is clearly the Miquel of triangle $ABC$ , easily proven by angle chasing into those two cyclic quads.
CLAIM:
$\angle DGB= \angle C$

PROOF:
$\angle PDF =\angle PCF$ ( $X$ being Miquel point). Hence we have $\angle DGB =\angle GDF+\angle GFD=\angle ECD +\angle FCG= \angle C$

CLAIM:
$B,H,P,C$ are concylic points.

PROOF:
$\angle BGD +\angle CGD =\angle BED +\angle CFD =\angle C +\angle B =180^{0}-\angle A =\angle BHC$
Hence we are done.

Now notice that , $\angle BHL' =\angle BGL'=\angle C$ ,
$\implies$ $\boxed{G,D,L}$ is collinear.

Now we are easily done by pascals theorem on $B,H,L',P,C,C$

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This post has been edited 2 times. Last edited by Mathgloggers, 3 hours ago
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#33 h 40 minutes ago

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Fun geometry.

Call the angles of the triangle $\alpha, \beta, \gamma$ and sides $a,b,c$ and $A'$ the foot of the perpendicular from $A$ to $BC$ . Angle chasing gives us $x\equiv \measuredangle DAC=\measuredangle DEC=\measuredangle FBC=\measuredangle PCL=\measuredangle ECL$ and therefore $ED\parallel LC$ . It suffices to prove $LC/ED=CX/EX$ . We also have $\measuredangle ECB=\alpha-x$ and so $\measuredangle A'CL=\alpha$ . We also have $\measuredangle BEC=\gamma+x$ , $\measuredangle EBX=90^{o}-\alpha$ , $\measuredangle CBX=90^{o}-\gamma$ , $\measuredangle BED=\gamma$ and $\measuredangle EDB=\alpha$ .

Now, $LC=\frac{A'C}{\cos \alpha}=b\frac{\cos \gamma}{\cos \alpha}$ . We have $\triangle EDB\sim \triangle CAB$ and therefore $ED=b\frac{EB}{CB}=b\frac{\sin(\alpha-x)}{\sin(\gamma+x)}$ by applying the Sine theorem to $\triangle ECB$ . Applying Sine theorems to $\triangle XEB$ and $\triangle XCB$ we obtain $EX=BX\frac{\cos\alpha}{\sin(\gamma+x)}$ and $CX=BX\frac{\cos\gamma}{\sin(\alpha-x)}$ . Combining all the results gives us: $\frac{LC}{ED}=\frac{\cos(\gamma)\sin(\gamma+x)}{\cos(\alpha)\sin(\alpha-x)}=\frac{CX}{EX}.$ Q.E.D.

This post has been edited 1 time. Last edited by NZP_IMOCOMP4, 36 minutes ago

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