4 lines concurrent

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Source: Homework

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16 posts

#1 h Apr 9, 2025, 11:51 AM • 1 Y

Y by PikaPika999

Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$ . $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$ .

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34 posts

#2 h Apr 10, 2025, 2:40 AM • 1 Y

Y by PikaPika999

............

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#3 h Apr 10, 2025, 3:08 AM • 1 Y

Y by PikaPika999

First three are trivial by pascal, but AD looks a bit hard / '

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149 posts

#4 h Apr 10, 2025, 5:40 AM • 1 Y

Y by PikaPika999

Redefine
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$ . $A'$ is antipode of $A$ . $O'$ is circumcenter of $(BOC)$ . Point $E,F$ satisfied $OE // A'F // AB , OF // A'E // AC$ then prove $OH,O'A',BE,CF$ concurrent .

Let $B',C'$ be antipode of $B,C$ respectively.
MMP: Fix $(O),B,C$ . Move $A$ on $(O)$ deg 2.
Since $A',E,C'$ collinear and $A',F,B'$ collinear
By $\angle C'EO = \angle BAC = \angle BC'C = \angle C'BO$ so $C',B,O,E$ concyclic.
implies that $E$ deg 2. Also $F$ deg 2.
So line $BE,CF$ deg 1.
$H=$ reflection of $A \infty_{\perp BC} \cap (O)$ across $BC$
Since $H,A'$ deg 2. implies that line $O'A'.OH$ deg 2.
We want to prove $BE,CF,OH$ concurrent first and $BE,CF,O'A'$ concurrent.
but both have deg 1+1+2+1 = 5 .

choose $A= B,C,B',C'$ and midpoint of arc $BC$
the rest of problem is easy. $\square$

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49 posts

#5 h Apr 10, 2025, 3:35 PM • 1 Y

Y by PikaPika999

My solution may not be elegant but here's some DDIT spam for you guys to enjoy :blush:

:blush:

.
Let $X = KE \cap LF$ . I will show that $X, A, D$ and $X, O, H$ are collinear.

Part 1:
I first claim that $\angle \infty_{CH}XL = \angle \infty_{BH}XK$ . This follows from
$\angle \infty_{CH}XL = \angle FLH = \angle FHL = 90^\circ - \angle BHL = 90^\circ - \angle CHK = \angle EHK = \angle XKH = \angle \infty_{BH}XK.$ Then, applying DDIT on $X \cup LDKH$ we see that $(XK, XL), (XH, XD), (X\infty_{BH}, X\infty_{CH})$ are reciprocal pairs under some involution on $\mathcal{P}_X$ . This involution must be a reflection in the angle bisector of $\angle KXL$ . Hence $XH, XD$ are isogonal in $\angle KXL$ .

Next, since $AK=AL$ (well-known), $LF=FH=AE$ , $AF=EH=EK$ , we yield $\triangle LAF \cong \triangle AKE$ .
I claim that $X\infty_{AB}, X\infty_{AC}$ are isogonal in $\angle XKL$ . This is because $\angle \infty_{AB}XF = 180^\circ - \angle AFL = 180^\circ - \angle AEK = \angle AEX = \angle \infty_{AC}XE.$ Applying DDIT on $X \cup AEHF$ would then give $XA, XH$ are isogonal in $\angle EXF$ . Since $XH, XD$ , $XH, XA$ are isogonal in $\angle KXL = \angle EXF$ we conclude that $XA \equiv XD$ , or $X \in AD$ .

Part 2:
I first prove that $X\infty_{CH}, OL, AK$ concur at a point $S$ on $(XLK)$ . For this, let $S = X\infty_{CH} \cap OL$ , where from a short angle chase we get $\angle XSL = \angle OLH = B-A = (90^\circ - A) - (90^\circ - B) = \angle AKL - \angle LCB$ $= \angle AKL - \angle LAF = \angle AKL - \angle AKE = \angle EKL = \angle XKL$ Hence $XSKL$ are cyclic, and from
$\angle SKX = \angle SLX = \angle FLO = \angle ALO - \angle ALF = A - \angle EAK = A - (90^\circ - C)$ which equals
$= A+C-90^\circ = 90^\circ - B = \angle LAB = \angle AKE = \angle AKX$ $\implies S \in AK$ . The claim is proven.

From DDIT in $X \cup AKOL$ we have the reciprocal pairs $(XA, XO), (XK, XL), (XS, XT)$ where $T = AL \cap KO \cap X \infty_{BH}$ (similarly).
since $(XS, XT) = (X\infty_{CH}, X\infty_{BH})$ and we have established $(XA, XH), (X\infty_{CH}, X\infty_{BH})$ are isogonal in $\angle KXL$ we get $XH \equiv XO \implies X \in OH$ . The problem is solved. $\blacksquare$

Attachments:

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#6 h May 1, 2025, 5:18 PM • 1 Y

Y by Amkan2022

We need two well-known lemmas about the isogonal line:
1. Given a $\triangle ABC$ and a point $P$ satisfied $\angle ABP=\angle ACP$ . Let $Q$ be the reflection of $P$ in the midpoint of $BC$ . Then $AP$ and $AQ$ are isogonals wrt $\angle BAC$ .
2. Given a trapezoid $ABCD$ $(AB\parallel CD)$ is inscribed $(O)$ . Let $E,F$ be the intersections of $BC$ and $AD$ ; $AC$ and $BD$ . Let $S$ be a abitary point on $(O)$ . Then $SE,SG$ are isogonals wrt $\angle ASB$ .

Back to the problem: Let $X,Y$ be the intersections of $OL$ and $AK$ ; $AL$ and $OK$ . By symmetric, we have $AK=AH=AL$ so $AO$ is the perpendicular bisector of $KL$ then $XKLY$ is a trapezoid. Let $LK$ intersects $KE$ at $Z$ .

We have
$\angle FZK=\angle ZEH-\angle ZFC=180^\circ-\angle KEH-\angle HFC=180^\circ-3\angle BAC$ and
$\angle LYK=\angle AOK-\angle OAL=2(90^\circ-\angle OAL)-\angle OAL=180^\circ-3\angle OAL=180^\circ-3\angle BAC=\angle FZK\,(2)$ since
$\angle OAL=\dfrac{1}{2}\angle KAL=\dfrac{1}{2}(\angle KAB+\angle BAH+\angle HAC+\angle CAL)=\angle BAC$ $(2)$ leads to $Z$ lies on the circumcircle of $XKLY$ . From the lemma 2, we obtain that $ZA,ZO$ are isogonals wrt $\angle LZK$ $(1)$ . We also have $\angle ZFE=\angle LFE=\angle KEH=180^\circ-\angle ZEH$ and $\angle HEZ=\angle HKE=\angle FLH=180^\circ-\angle HLZ$ , using lemma 1 and get $(XD,XH)$ and $(XH,XA)$ are two isogonal pairs wrt $\angle LZK\equiv\angle FZE$ , so $A,D,Z$ are collinear. Combine with $(1)$ and we conclude $Z,O,H$ are collinear or $AD,LF,KE,OH$ are concurrent at $Z$ .

https://i.postimg.cc/hJQdTNdW/image.png

This post has been edited 2 times. Last edited by tomsuhapbia, May 1, 2025, 5:25 PM

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#7 h May 2, 2025, 2:37 AM

Y by

I apologize for the complex bash, but I couldn't find another way.

Let $B'$ and $C'$ be the antipodes of $B$ and $C$ .
Since $L$ is the reflection of $H$ over $AB$ , we have $\angle FLB = \angle FHB = 90^\circ$ (since $FH \parallel AC \perp BH$ ).
Analogously, $\angle EKC = 90^\circ$ .
$\Rightarrow B', F, L$ are collinear and $C', E, K$ are collinear.
By Pascal's Theorem on
$\binom{B, L, C'}{C, K, B'}$ we get that $KE, LF$ , and $OH$ are concurrent.
Now it suffices to prove that $AD, KE$ , and $LF$ are concurrent.

We use complex numbers, where $(ABC)$ is the unit circle and $a = 1$ .
Let
$k = -\frac{c}{b}, \quad l = -\frac{b}{c}, \quad c' = -c, \quad b' = -b, \quad h = b + c + 1, \quad o = 0$ $d + b + c + 1 = d + h = l + k = -\frac{b}{c} - \frac{c}{b}$ $\Rightarrow d = -\frac{b^2 + c^2 + b^2c + bc^2 + bc}{bc}$
Let $X = KC' \cap LB'$ .
We have:
$\frac{x + c}{\overline{x + c}} = \frac{c - \frac{c}{b}}{\overline{c - \frac{c}{b}}} = -\frac{c^2}{b} \Rightarrow \overline{x} = -\frac{xb + bc + c}{c^2}$ Analogously,
$\overline{x} = -\frac{xc + bc + b}{b^2}$ Equating both expressions:
$(b - c)(b^2c + bc^2 + bc - x(b^2 + bc + c^2)) = 0 \Rightarrow x = -\frac{bc(b + c + 1)}{b^2 + bc + c^2}$
Points $A, D, X$ are collinear if and only if
$\frac{d - 1}{x - 1} \in \mathbb{R}$ Substituting:
$\frac{(b^2 + c^2 + b^2c + bc^2 + 2bc)/bc}{(b^2c + bc^2 + 2bc + b^2 + c^2)/(b^2 + bc + c^2)} = \frac{b^2 + bc + c^2}{bc}\in \mathbb{R}_\blacksquare$

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733 posts

#8 h May 2, 2025, 9:31 AM

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Let $B',C'$ be the antipodes of $B,C$ on $(ABC)$ . Also let $LC'\cap KB'=W,KC'\cap LB'=P$ .
Claim: $B',F,L$ and $C',E,K$ are collinear.
Proof: Pascal at $KB'LCAB$ yields $AC_{\infty},B'L\cap AB,H$ are collinear thus, $B'L\cap AB=F$ . Similarily $C',E,K$ are collinear.
Claim: $P$ lies on $OH$ .
Proof: Pascal at $BKC'CLB'$ gives $H,P,O$ are collinear.
Claim: $A,D,P$ are collinear.
Proof: Notice that $W,L,H,K$ lie on the circle with diameter $WH$ and since $AH=AK=AL$ , $A$ must be the circumcenter of $(KLHW)$ . Hence $W,A,H$ are collinear.
DDIT at $DLHK$ implies $(\overline{AD},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. DDIT at $B'KC'L$ gives $(\overline{AP},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. Combining these implies $AD\equiv AP$ hence $A,D,P$ are collinear as desired. $\blacksquare$

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