$KH$, $EM$ and $BC$ are concurrent

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$KH$, $EM$ and $BC$ are concurrent

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Source: 2012 European Girls’ Mathematical Olympiad P7

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yunxiu

#1 h Apr 13, 2012, 12:34 PM • 4 Y

Y by Tumon2001, iman007, NVA9205, Adventure10

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$ . Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$ . Let $L$ be the reflection of $K$ in the line $AB$ , and let $M$ be the reflection of $K$ in the line $BC$ . Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$ .
Show that the lines $KH$ , $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Luxembourg (Pierre Haas)

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teps

#2 h Apr 13, 2012, 1:09 PM • 2 Y

Y by Adventure10, Mango247

You did not define $K$ well.

[WakeUp: this has been edited]

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WakeUp

#3 h Apr 13, 2012, 1:15 PM • 4 Y

Y by vsathiam, Adventure10, Mango247, ehuseyinyigit

Let $KL,KM$ meet $BC,BA$ at $L',M'$ respectively.

Let us prove that $L'$ and $M'$ lie on the circle $(BLM)$ .

Now $BK=BM$ since $M$ is the reflection of $K$ through $BC$ but $L$ is the reflection of $K$ through $AB$ so clearly we have $BK=BL$ too. Thus $BK=BL=BM$ and so $B$ is the circumcentre of $\triangle KLM$ . Then an angle chase suffices. If $K$ lies between the antipode of $A$ on $(ABC)$ and $C$ for example (as in the diagram), then:

$B$ is the circumcentre of $\triangle MLK$ and so $\angle MBK=2\angle MLK$ but $\angle MBK=2\angle MBC$ and thus $\angle MLK=\angle MBC=\angle MBL'$ . So $L'$ lies on $(BKL)$ . Also, if $\angle M'ML=\alpha$ then $\angle LMK=180^{\circ}-\alpha$ so $\angle LBK=2\alpha$ since the chord $LK$ subtends a angle of $\alpha$ using directed angles mod 180. But then $2\alpha =\angle LBK=2\angle LBA$ and hence $\alpha =\angle LBA=\angle LBM'$ and so it follows $M'$ lies on $(BKL)$ too.

Now, suppose $EM$ and $AH$ meet at $A'$ . Note $AH$ and $MK$ are both perpendicular to $BC$ , and so $MK||AH$ . Therefore $\angle AA'M=\angle A'MK=\angle M'ME=\angle EBM'=\angle EBA$ . Therefore $\angle AA'E=\angle ABE$ , so $A'$ lies on the circumcircle of $\triangle ABC$ . It is well known that if $AH$ meets $(ABC)$ at $P$ then $P$ is the reflection of $H$ through $BC$ (it's easy to prove, note that $\angle BHC=180^{\circ}-A$ by an easy angle chase, but also $\angle BPC=180^{\circ}-A$ so by symmetry it follows $BC$ is the perpendicular bisector of $HP$ ).

Therefore, $BC$ is both the perpendicular bisector of $MK$ and $HA'$ . So $HA'KM$ is an isosceles trapezoid. Therefore, its diagonals meet on the perpendicular bisector, i.e. $BC$ .

The angles here assume that $K$ lies on the arc $A_1C$ of $(ABC)$ not containing $A$ , where $A_1$ is the antipode of $A$ on $(ABC)$ . In this case, $HM$ is the smaller side of the isosceles trapezoid $HA'KM$ and so $HM,KA'$ and $BC$ concur at a point on $BC$ closer to $H$ than $M$ . If $K=A_1$ then $HA'KM$ is a rectangle, and if $A_1$ is on the arc $(BA_1)$ not containing $A$ then $HA'KM$ is an isosceles trapezoid with $HA'$ the larger parallel side.

Remark: You can also prove $L,M,H$ are collinear.

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yunxiu

#4 h Apr 13, 2012, 2:26 PM • 2 Y

Y by jlammy, Adventure10

Let ${c_2}$ is the reflection of $\Gamma$ in $AB$ , ${c_3}$ is the reflection of $\Gamma$ in $BC$ , then the orthocentre $H$ is on the ${c_2}$ and ${c_3}$ .
Let $AH \cap \Gamma = P$ , then $P$ is the reflection of $H$ in the line $BC$ , thus $D = HK \cap PM$ is on the line $BC$ , suppose $PM \cap \Gamma = E'$ .
By symmetry we have $\angle BMH = \angle PKB = \angle PAB = \angle BLH = 90^\circ - \angle ABC$ .
Because $\angle LBM = \angle LBA + \angle ABM = \angle KBA + \angle ABM = 2\angle ABC$ , $\angle LBM + \angle BMH + \angle BLH = 180^\circ$ , so $L,H,M$ are collinear .
Then have $\angle ME'B = \angle PE'B = \angle PAB = \angle BLM$ , so $EMBL$ is cyclic, hence $E = E'$ , $HK \cap EM \cap BC = D$ .

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#5 h Apr 13, 2012, 9:57 PM • 3 Y

Y by ahaanomegas, Adventure10, Mango247

yunxiu wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$ . Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$ . Let $L$ be the reflection of $K$ in the line $AB$ , and let $M$ be the reflection of $K$ in the line $BC$ . Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$ .
Show that the lines $KH$ , $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Solution using complex numbers:

Let $\Gamma$ be the unit circle, let $BC\cap HK=\{F\}$ and $BC\cap ME=\{D\}$ . It is enough to show that $D\equiv F$ .
We can easily get that:
$m=\frac{k(b+c)-bc}{k};\bar{m}=\frac{b+c-k}{bc};l=\frac{k(a+b)-ab}{k};\bar{l}=\frac{a+b-k}{ab}(1)$
Because $\angle MEB=\angle MLB$ we have $\frac{e-m}{\bar{e}-\bar{m}}:(-be)=\frac{l-m}{\bar{l}-\bar{m}}:\frac{l-b}{\bar{l}-\bar{b}}$ (2) so together with (1) we get:
$e=\frac{k(ab+bc+ca)-abc}{k(a+b+c-k)}=\frac{abc(\bar{h}-\bar{k})}{h-k}$
In above equation I used $h=a+b+c$ . From (2) we get $\frac{e-m}{\bar{e}-\bar{m}}=\frac{bce}{a}=\frac{b^2c^2(\bar{h}-\bar{k})}{h-k}$ (3).
Because $D\in\widehat{BC}$ we have $\bar{d}=\frac{b+c-d}{bc}$ (4) and because $E,M$ and $D$ are collinear we have $\frac{e-m}{\bar{e}-\bar{m}}=\frac{m-d}{\bar{m}-\bar{d}}$ but together with (3),(4) we get $\frac{b^2c^2(\bar{h}-\bar{k})}{h-k}=\frac{m-d}{\bar{m}-\frac{b+c-d}{bc}}$ which is with (1) equivalent with:
$d=\frac{(h-k)(k(b+c)-bc)+kbc(k\bar{h}-1)}{bc(\bar{h}k-1)+(h-k)k}$
On the other side $K,F,H$ are collinear and $F\in\widehat{BC}$ so $\frac{f-k}{\bar{f}-\bar{k}}=\frac{h-k}{\bar{h}-\bar{k}}$ and $\bar{f}=\frac{b+c-f}{bc}$ , therefore:
$f=\frac{(b+c)(h-k)k+bc(k^2\bar{h}-h)}{bc(\bar{h}k-1)+(h-k)k}$
We see that $d=f$ so $D\equiv F.\square$

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#6 h Dec 14, 2013, 10:39 PM • 9 Y

Y by ahaanomegas, vsathiam, pika7_7, BobaFett101, Brian_Xu, HamstPan38825, Dimanas23, Adventure10, Mango247

$[asy]size(8cm); pointpen=black; pathpen=black; pair A = Drawing("A", dir(50), dir(50)); pair B = Drawing("B", dir(209), dir(180)); pair C = Drawing("C", dir(331), dir(0)); Drawing(A--B--C--cycle); pair H = orthocenter(A,B,C); pair Ka = foot(A,B,C); pair Kc = foot(C,A,B); Drawing("H", H, dir(45)); pair Ha = Drawing("H_A", 2*Ka-H, dir(-70)); pair Hc = Drawing("H_C", 2*Kc-H, dir(135)); pair K = Drawing("K", dir(-64), dir(-90)); pair M = Drawing("M", reflect(B,C) * K, dir(180)); pair L = Drawing("L", reflect(B,A) * K, dir(90)); pair E = Drawing("E", 2*foot(origin,M,Ha)-Ha, dir(70)); Drawing(A--Ha--B--Hc--C); Drawing(L--Hc); Drawing(Hc--E, dotted); draw(unitcircle); draw(Ha--E); draw(B--H--K); draw(L--K--M,dashed); [/asy]$

In what follows, all angles are directed. Let $H_A$ and $H_C$ be the reflections of $H$ across $\overline{BC}$ and $\overline{BA}$ , which lie on $\Gamma$ . Let $E'$ be the second intersection of line $H_AM$ with $\Gamma$ . By construction, lines $E'M$ and $HK$ concur on $\overline{BC}$ .
First, we claim that $L$ , $H_C$ , and $E'$ are collinear. Notice that $\measuredangle LH_CB = -\measuredangle KHB = \measuredangle MH_AB$ by reflections, and that $\measuredangle MH_AB = \measuredangle E'H_AB = \measuredangle E'H_CB$ as desired. Now, $\measuredangle LE'M = \measuredangle H_CE'H_A = \measuredangle H_CBH_A = 2\measuredangle ABC$ and $\measuredangle LBM = \measuredangle LBK + \measuredangle KBM = 2\measuredangle ABK + 2\measuredangle KBC = 2\measuredangle ABC$ so $B$ , $L$ , $E'$ , $M$ are concyclic. Hence $E=E'$ and we are done.

This post has been edited 1 time. Last edited by v_Enhance, Jul 20, 2017, 1:03 AM
Reason: \overline

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jlammy

#7 h Jul 17, 2014, 12:59 PM • 1 Y

Y by Adventure10

My solution is essentially the same as yunxiu's, but it is less diagram dependent, and perhaps has technical advantages through the use of directed angles.

One possible diagram

Another possible diagram

Let $H'$ be the reflection of $H$ in $BC$ , which lies on $\Gamma$ . Let $\Gamma_A$ be the reflection of $\Gamma$ in $BC$ ; analogously define $\Gamma_C$ . Define $X \equiv HK \cap H'M$ ; which lies on $BC$ , the axis of reflection. Finally, let $E' \equiv H'M \cap \Gamma$ .

Using the reflections, we have $\measuredangle HMB = \measuredangle BKH' = \measuredangle BAH' = 90^{\circ} - \measuredangle CBA$ ; similarly, $\measuredangle BLH = 90^{\circ} - \measuredangle CBA$ . Now $\begin{align*}\measuredangle MBL = \measuredangle ABL + \measuredangle MBA &= \measuredangle KBA + \measuredangle MBA \\ &= (\measuredangle MBA + \measuredangle CBM + \measuredangle KBC) + \measuredangle MBA \\ &= 2\measuredangle CBA. \end{align*}$ $\therefore \measuredangle MBL + \measuredangle HMB + \measuredangle BLH = 180^{\circ}$ , so $L, M, H$ are collinear.

$\therefore \measuredangle BE'M = \measuredangle BE'H' = \measuredangle BAH' = \measuredangle BLM$ , so $E'MBL$ is a cyclic quadrilateral, and $E'$ lies on the circumcircle of $\triangle BLM$ and also on $\Gamma$ ; it follows that $E' \equiv E$ .

$\therefore KH \cap EM \cap BC \equiv X$ , so these lines are concurrent, as required.

This post has been edited 3 times. Last edited by jlammy, Jul 17, 2014, 5:39 PM

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jayme

#8 h Jul 17, 2014, 1:26 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
1. according to Carnot, the symmetric U od H wrt BC is on gamma
2. HUKM is a isoceles trapeze with BC as axis of symmetry
And we are done…
Sincerely
Jean-Louis

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Eray

#9 h Oct 23, 2014, 1:49 PM • 2 Y

Y by Adventure10, Mango247

@jayme, aren't you suppose to prove E,M,U are collinear?
Sorry if I'm missing something..

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#10 h Jun 13, 2015, 12:23 AM • 2 Y

Y by Adventure10, Adi1005247

My proof is the same as that of v_Enhance so i wouldnot post it. But one observation: $L,M,H$ are collinear.

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#11 h Dec 23, 2016, 2:41 AM • 1 Y

Y by Adventure10

It's pretty evident that L,M,H are collinear by homothety about K, but is there a way to use that to prove the problem statement? I thought this fact seemed interesting and nontrivial, which suggeted that it might be useful for solve the problem. However, I couldn't actually put it into use.

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HadjBrahim-Abderrahim

#13 h Mar 27, 2017, 3:04 PM • 1 Y

Y by Adventure10

yunxiu wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$ . Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$ . Let $L$ be the reflection of $K$ in the line $AB$ , and let $M$ be the reflection of $K$ in the line $BC$ . Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$ .
Show that the lines $KH$ , $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Luxembourg (Pierre Haas)

Here is my solution. First, we present two simple lemmas,
Lemma $(1).$ Let $A,B,C,D$ be four distinct points such that, the quadrilateral $ABCD$ is convex. If there exists a line $\ell$ such that: the reflection of the points $A,D$ in $\ell$ are the points $B,C,$ respectively. Then, the intersection of $AC$ and $BD$ lies on $\ell.$
Proof. Let $X$ be the intersection of $AC$ and $\ell,$ and let $D_1$ be the intersection of $BX$ and $CD.$ We deduce from thales' theorem that $XC=XD_1,$ but $XC=XD$ and the points $C,D,D_1$ are collinear, hence $D\equiv D_1$ and this complete the proof.
Lemma $(2).$ the reflections of the orthocenter of any triangle $XYZ$ about its sides lie on its circumcircle.
Proof. Just angle chasing.
Let $H_1$ be the reflection of $H$ in the line $AB,$ from $(1)$ we deduce that the points of intersection of the lines $HL$ and $KH_1$ lies on $AB,$ and from $(2)$ we deduce that $H_1$ lies on the circle $(\Gamma),$ hence, we get $\angle KLH=\angle KH_1H=\angle KH_1C.$ Because $M,L$ are the reflection of $K$ in the lines $BC,AB$ respectively, we have $BM=BK=BL$ and $B$ is the circumcircle of the triangle $LMK,$ hence, we have $\angle KLM=\frac{1}{2}\angle KBM=\angle KBC=\angle KH_1C.$ We deduce that the points $L,H,M$ are collinear. Now, because $\begin{align*}\angle BMC=\angle CKB=180^\circ-\angle BAC=180^\circ-\left[ (90^\circ-\angle ACB)+(90^\circ-\angle CBA)\right]=\angle BHC,\end{align*}$ we deduce that the quadrilateral $BHMC$ is cyclic. Therefore, $\begin{align*}\angle BEM=\angle BLM=\angle LMB=\angle HMB=\angle HCB=\angle BAH. \end{align*}$ Beacuse the triangle $ABC$ is acute-angled, then $H$ lies inside it, and since $\angle BAH=\angle BEM$ we deduce that the intersection of $AH$ and $EM$ is a point $H_2$ lies on the circle $(\Gamma).$ From $(2)$ we deduce that $H_2$ is the reflection of $H$ in the line $AB$ and since, $M$ is the reflection of $K$ in this line we deduce from $(1)$ that the point of intersection of $MH_2$ and $HK$ lies on the line $AB.$ We conclude that the lines $KH$ , $EM$ and $BC$ are concurrent.

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#14 h Apr 27, 2017, 5:17 AM • 2 Y

Y by Adventure10, Mango247

Solution:

Claim: $L$ , $H$ and $M$ are collinear.

Proof of the claim: Let $LM\cap AH = I$ and let $H'$ be the reflection of $H$ in $BC$ (which lies on $\Gamma$ ). Observe that $K$ is the center of the spiral similarity that takes $LM$ to $AC$ . So, $\measuredangle MKL = \measuredangle CKA$ . Since, $B$ is the circumcenter of $\Delta MKL$ , so, $\measuredangle BLM = 90 - \measuredangle MKL = 90 - \measuredangle CKA = \measuredangle BAH$ . Thus, $B$ , $L$ , $A$ and $I$ are concyclic.
Also, $\measuredangle AHB = \measuredangle BCA= \measuredangle ALB$ . So, $B$ , $L$ , $A$ and $H$ are also concyclic.
Hence, $H = I$ .

Back to the problem:
Since, $HMKH'$ is an isosceles trapezoid, so, $BC$ , $HK$ and $H'M$ are concurrent. Let $H'M\cap \Gamma = N$ . It would be enough to prove that $L$ , $B$ , $M$ and $N$ are concyclic, which follows directly, as $\measuredangle BNH' = \measuredangle BAH' = \measuredangle BLH$ .

P.S.- All angles above are directed modulo $180$ °.

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maxo

#15 h Jul 18, 2017, 10:58 PM • 2 Y

Y by Adventure10, Mango247

v_Enhance wrote:

$[asy]size(8cm); pointpen=black; pathpen=black; pair A = Drawing("A", dir(50), dir(50)); pair B = Drawing("B", dir(209), dir(180)); pair C = Drawing("C", dir(331), dir(0)); Drawing(A--B--C--cycle); pair H = orthocenter(A,B,C); pair Ka = foot(A,B,C); pair Kc = foot(C,A,B); Drawing("H", H, dir(45)); pair Ha = Drawing("H_A", 2*Ka-H, dir(-70)); pair Hc = Drawing("H_C", 2*Kc-H, dir(135)); pair K = Drawing("K", dir(-64), dir(-90)); pair M = Drawing("M", reflect(B,C) * K, dir(180)); pair L = Drawing("L", reflect(B,A) * K, dir(90)); pair E = Drawing("E", 2*foot(origin,M,Ha)-Ha, dir(70)); Drawing(A--Ha--B--Hc--C); Drawing(L--Hc); Drawing(Hc--E, dotted); draw(unitcircle); draw(Ha--E); draw(B--H--K); draw(L--K--M,dashed); [/asy]$

In what follows, all angles are directed. Let $H_A$ and $H_C$ be the reflections of $H$ across $\ol{BC}$ and $\ol{BA}$, which lie on $\Gamma$ . Let $E'$ be the second intersection of line $H_AM$ with $\Gamma$ . By construction, lines $E'M$ and $HK$ concur on $\ol{BC}$.
First, we claim that $L$ , $H_C$ , and $E'$ are collinear. Notice that $\measuredangle LH_CB = -\measuredangle KHB = \measuredangle MH_AB$ by reflections, and that $\measuredangle MH_AB = \measuredangle E'H_AB = \measuredangle E'H_CB$ as desired. Now, $\measuredangle LE'M = \measuredangle H_CE'H_A = \measuredangle H_CBH_A = 2\measuredangle ABC$ and $\measuredangle LBM = \measuredangle LBK + \measuredangle KBM = 2\measuredangle ABK + 2\measuredangle KBC = 2\measuredangle ABC$ so $B$ , $L$ , $E'$ , $M$ are concyclic. Hence $E=E'$ and we are done.

why does showing those 4 four points being concyclic imply that $E=E'$ , and that the other three points are collinear?

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maxo

#16 h Jul 19, 2017, 10:22 PM • 1 Y

Y by Adventure10

bump? /8char

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#17 h Jul 19, 2017, 10:46 PM • 2 Y

Y by Adventure10, Mango247

By definition $E'$ lies on $(ABC)$ and once we show those points are concyclic then we know $E'$ lies on $(BLM)$ , thus $E=E'$ , as $E$ is the points on both $(BLM)$ and $(ABC)$ .

This post has been edited 3 times. Last edited by claserken, Jul 19, 2017, 10:49 PM

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maxo

#18 h Jul 20, 2017, 12:55 AM • 2 Y

Y by Adventure10, Mango247

claserken wrote:

By definition $E'$ lies on $(ABC)$ and once we show those points are concyclic then we know $E'$ lies on $(BLM)$ , thus $E=E'$ , as $E$ is the points on both $(BLM)$ and $(ABC)$ .

ahh okay thanks, that makes sense. having trouble wrapping my head around on
how to be rigorous for some reason lol

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#19 h Oct 30, 2017, 3:16 AM • 2 Y

Y by Adventure10, Mango247

Tumon2001 wrote:

Solution:

Claim: $L$ , $H$ and $M$ are collinear.

Proof of the claim: Let $LM\cap AH = I$ and let $H'$ be the reflection of $H$ in $BC$ (which lies on $\Gamma$ ). Observe that $K$ is the center of the spiral similarity that takes $LM$ to $AC$ . So, $\measuredangle MKL = \measuredangle CKA$ . Since, $B$ is the circumcenter of $\Delta MKL$ , so, $\measuredangle BLM = 90 - \measuredangle MKL = 90 - \measuredangle CKA = \measuredangle BAH$ . Thus, $B$ , $L$ , $A$ and $I$ are concyclic.
Also, $\measuredangle AHB = \measuredangle BCA= \measuredangle ALB$ . So, $B$ , $L$ , $A$ and $H$ are also concyclic.
Hence, $H = I$ .

Back to the problem:
Since, $HMKH'$ is an isosceles trapezoid, so, $BC$ , $HK$ and $H'M$ are concurrent. Let $H'M\cap \Gamma = N$ . It would be enough to prove that $L$ , $B$ , $M$ and $N$ are concyclic, which follows directly, as $\measuredangle BNH' = \measuredangle BAH' = \measuredangle BLH$ .

P.S.- All angles above are directed modulo $180$ °.

How to prove $K$ is the center of a spiral similarity that takes $LM$ to $AC$ ?

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niyu

#20 h Apr 22, 2018, 7:59 PM • 3 Y

Y by BobaFett101, Adventure10, Mango247

Let $H'$ be the reflection of $H$ wrt $\overline{BC}$ , $P = \overline{KL} \cap \overline{AB}$ , and $Q = \overline{MK} \cap \overline{BC}$ .

Since $\angle BPK = \angle BQK = 90^\circ$ (this follows from reflections), $BPQK$ is cyclic. Then,
$\begin{eqnarray*} \angle LKM = \angle PKQ = \angle PBQ = B \end{eqnarray*}$ (where $B = \angle ABC$ ). It follows from reflections that $BK = BM = BL$ . Therefore, $B$ is the circumcenter
of $\triangle LMK$ . Hence,
$\begin{eqnarray*} \angle LBM = 2\angle LKM = 2B. \end{eqnarray*}$ Since $LBM$ is isosceles,
$\begin{eqnarray*} \angle BLM & = & \frac{180^\circ - \angle LBM}{2} \\ & = & \frac{180^\circ - 2B}{2} \\ & = & 90^\circ - B. \end{eqnarray*}$ Therefore,
$\begin{eqnarray*} \angle BEM = \angle BLM = 90^\circ - B. \end{eqnarray*}$
It is well known that $H'$ lies on $(ABC)$ . Noting that $\overline{AH'} \perp \overline{BC}$ , we have
$\begin{eqnarray*} \angle BEH' = \angle BAH' = 90^\circ - B. \end{eqnarray*}$
Therefore,
$\begin{eqnarray*} \angle BEM = \angle BEH', \end{eqnarray*}$ which is enough to imply that $E, M, H'$ are collinear. Hence, $\overline{EM} = \overline{MH'}$ is the reflection of $\overline{HK}$ wrt $\overline{BC}$ . This implies that $\overline{HK} \cap \overline{BC} = \overline{EM} \cap \overline{BC}$ . Thus, $\overline{HK}, \overline{EM},$ and $\overline{BC}$ are concurrent, as desired. $\Box$

This post has been edited 2 times. Last edited by niyu, Aug 8, 2019, 9:21 PM

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Mr_ONE

#21 h May 15, 2018, 11:40 AM • 2 Y

Y by Adventure10, Mango247

Let $F, N$ be the reflections of $H$ in $BC, AB$ respectively, it's well known that both $F,N$ lie on $\Gamma$ .
By their definition we have: $FM,BC,HK$ are concurrent in a point $X$ and $LN,AB,HK$ are also concurrent in a point $Y$ .
Let $E'\equiv (FM)\cap (LN)$ ; we claim that $E=E'$ which is equivalent to $E\in (MX)$ which is what we want to prove.
$\textbf{Claim\ 1.}\ E'\in \Gamma.$
$\textbf{Proof.}\ \angle NE'F=\angle YE'X=180-\angle E'YX-\angle E'XY=180-(180-2\angle XYB)-(180-2\angle YXB)$
$=180-2\angle YBX=180-2\angle YBH -2\angle HBX=180-\angle NBF$ .
$\textbf{Claim\ 2.}\ E'\in (LBM).$
$\textbf{Proof.}\ \angle LBM=\angle LBA+ \angle ABM=\angle KBA+ \angle ABM$
$=\angle KBC+\angle CBA +\angle CBA -\angle CBM=2\angle CBA=\angle NBF=180-\angle NE'F.$
By Claim 1 and 2 we deduce that $E'=E$ as wanted.

This post has been edited 1 time. Last edited by Mr_ONE, May 15, 2018, 11:42 AM

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#22 h Aug 25, 2018, 4:09 PM • 2 Y

Y by Pluto04, Adventure10

anantmudgal09 wrote:

My proof is the same as that of v_Enhance so i wouldnot post it. But one observation: $L,M,H$ are collinear.

vsathiam wrote:

It's pretty evident that L,M,H are collinear by homothety about K, but is there a way to use that to prove the problem statement? I thought this fact seemed interesting and nontrivial, which suggeted that it might be useful for solve the problem. However, I couldn't actually put it into use.

Here is an outline for a solution that uses this fact. Let $N$ be the reflection of $K$ over $AC$ , so $L,H,M,N$ are collinear. Let $X$ , $Y$ , and $Z$ be the feet of the perpendiculars from $K$ to $AB$ , $AC$ , and $BC$ , respectively. Let $KL$ and $KN$ meet $BC$ at $R$ and $S$ , respectively. Define $P=HK \cap BC$ . We wish to show that $E, M, P$ are collinear.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.221172766627305, xmax = 9.021251475796944, ymin = -5.927701692247144, ymax = 7.209370326643048; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-3.62,0.33), 4.8845880071916), linewidth(1) + wrwrwr); draw(circle((-10.834010068032772,5.993741903880838), 8.682595696196357), linewidth(1) + wrwrwr); draw(circle((-0.750226398326001,1.446911282614519), 3.8459692000793124), linewidth(1) + wrwrwr); draw((-6.185603077820917,-0.4826952164090483)--(-4.38781227002869,-4.493864044310681), linewidth(1) + wrwrwr); draw((-4.38781227002869,-4.493864044310681)--(2.832183361700406,2.846132279893439), linewidth(1) + wrwrwr); draw((-11.96255112469488,-2.6151989779602896)--(2.832183361700406,2.846132279893439), linewidth(1) + linetype("4 4") + wrwrwr); draw((-8.175181697361785,-3.554531511135485)--(-0.7778144541641416,-0.8238658822086206), linewidth(1) + linetype("4 4") + wrwrwr); draw((-6.18,4.49)--(-8.175181697361785,-3.554531511135485), linewidth(1) + wrwrwr); draw((-6.18,4.49)--(0.5816300849103664,-2.161085030579269), linewidth(1) + wrwrwr); draw((-13.859135172380286,-2.144813626841671)--(0.5816300849103664,-2.161085030579269), linewidth(1) + wrwrwr); draw((-2.20778230183693,5.005985583061074)--(-5.4363942488225545,-2.154304108523395), linewidth(1) + linetype("4 4") + wrwrwr); draw((-13.859135172380286,-2.144813626841671)--(-4.38781227002869,-4.493864044310681), linewidth(1) + wrwrwr); /* dots and labels */ dot((-6.18,4.49),dotstyle); label("$A$", (-6.515938606847688,4.650810704447064), NE * labelscalefactor); dot((-7.827233162731285,-2.151610185829781),dotstyle); label("$B$", (-8.39983116883116,-2.070952380952381), NE * labelscalefactor); dot((0.5816300849103664,-2.161085030579269),dotstyle); label("$C$", (0.7253049980322824,-2.464498229043683), NE * labelscalefactor); dot((-6.185603077820917,-0.4826952164090483),linewidth(3pt) + dotstyle); label("$H$", (-6.515938606847688,-0.2606414797323897), NE * labelscalefactor); dot((-4.38781227002869,-4.493864044310681),dotstyle); label("$K$", (-4.390791027154653,-4.942998819362454), NE * labelscalefactor); dot((-8.175181697361785,-3.554531511135485),linewidth(3pt) + dotstyle); label("$X$", (-8.404958677685942,-4.040377804014154), NE * labelscalefactor); dot((-13.859135172380286,-2.144813626841671),linewidth(3pt) + dotstyle); label("$R$", (-14.150728059818961,-2.653014561196379), NE * labelscalefactor); dot((-4.385177461520879,-2.1554885868011855),linewidth(3pt) + dotstyle); label("$Z$", (-4.427823691460045,-2.0237268791814245), NE * labelscalefactor); dot((-0.7778144541641416,-0.8238658822086206),linewidth(3pt) + dotstyle); label("$Y$", (-0.8803620621802328,-0.6699291617473442), NE * labelscalefactor); dot((-2.0902102447239272,-2.1580744842577064),linewidth(3pt) + dotstyle); label("$S$", (-2.1239669421487495,-2.608949232585596), NE * labelscalefactor); dot((-11.96255112469488,-2.6151989779602896),linewidth(3pt) + dotstyle); label("$L$", (-12.072805981896883,-3.1312042502951586), NE * labelscalefactor); dot((2.832183361700406,2.846132279893439),linewidth(3pt) + dotstyle); label("$N$", (2.8976780794962735,2.96643447461629), NE * labelscalefactor); dot((-4.382542653013067,0.18288687070831036),linewidth(3pt) + dotstyle); label("$M$", (-4.1389216843762195,-0.07173947264856458), NE * labelscalefactor); dot((-2.20778230183693,5.005985583061074),linewidth(3pt) + dotstyle); label("$E$", (-2.1397087760724016,5.1388075560802795), NE * labelscalefactor); dot((-5.4363942488225545,-2.154304108523395),linewidth(3pt) + dotstyle); label("$P$", (-5.913105076741431,-2.5857890594254227), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
1. Show that $R \in (BLM)$ and $S \in (CNM)$ . This is angle chasing using the fact that $XY \parallel LN$ and cyclic quads $KZYC$ and $KZBX$ .

2. Let $E'$ be the second point of intersection of $(BLRM)$ and $(CNSM)$ . Show that $E'=E$ , or equivalently that $E' \in \Gamma$ . This is also angle chasing using cyclic quads and the fact that $MSKR$ is a kite.

3. Note that since $BH \parallel KS$ and $CH \parallel KR$ , we have similar figures $BHCP \sim SKRP$ . From this it follows that $\frac{PB}{PC}=\frac{PS}{PR} \implies PB \cdot PR=PC \cdot PS$ Thus $P$ lies on the radical axis of $(BME)$ and $(CME)$ , so $E,M,P$ are collinear as desired. $\blacksquare$

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572 posts

anser

#23 h Aug 8, 2019, 6:06 PM • 3 Y

Y by Adventure10, Mango247, sargamsujit

Let $H'$ be the reflection of $H$ over $BC$ . It is enough to show that $E, M, H'$ are collinear, since $HK$ and $MH'$ intersect on $BC$ . Since
$\begin{align*} \angle LBM &= \angle LBA + \angle ABM\\ &= \angle ABK + \angle B - \angle CBK\\ &= 2\angle B \end{align*}$ and $BL = BK = BM$ , $EB$ bisects $\angle LEM$ , and in particular, $\angle BEM = 90 - \angle B$ . Also, $\angle BEH' = \angle BAH' = 90 - \angle B$ , so $E, M, H'$ are collinear, as desired.

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270 posts

#24 h Sep 22, 2020, 3:15 PM

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yunxiu wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$ . Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$ . Let $L$ be the reflection of $K$ in the line $AB$ , and let $M$ be the reflection of $K$ in the line $BC$ . Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$ .
Show that the lines $KH$ , $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Luxembourg (Pierre Haas)

A projective solution:
$1.$ Let the intersection of HK with BC be S.Now fix $A,B,C$ and move $K$ along $\Gamma$
$2.$ look at this projective map : $S \mapsto K \mapsto L \mapsto M \mapsto E$
$3.$ now we need to determine $3$ special cases
$(i)$ $K$ on $B$
$(ii)$ $K$ on $C$
$(iii)$ $K$ on $AH \cap \Gamma$
finished!

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1002 posts

Afo

#25 h Dec 12, 2020, 11:48 AM

Y by

I think it's different :oops:

:oops:

https://i.ibb.co/gSnPc6h/geo3.png

Solution.

Claim 1. $L,M,H$ is collinear.
Proof. Take a homothety at $K$ with factor of $\frac 1 2$ , then by Simson Line Bisection, the three points are collinear.

Let $H'$ be the reflection of $H$ over $BC$ , it's well-known that $H' \in (ABC)$ . Let $KH \cap BC = D$ . Let $H'D \cap (ABC)=E$ . It's enough to show that $LBME$ is cyclic or $\angle BLM = \angle BLH = \angle BEH' = \angle BAH$ which is equivalent to proving $ALBH$ cyclic. This is true since $\angle ALB = \angle AKB = \angle C$ and by easy angle chasing $\angle AHB = \pi-\angle C$ .

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797 posts

#26 h Feb 24, 2021, 7:42 PM

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EGMO 2012 P7 wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$ . Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$ . Let $L$ be the reflection of $K$ in the line $AB$ , and let $M$ be the reflection of $K$ in the line $BC$ . Let $E$ be the second point of intersection of $\Gamma$ with the circumcircle of triangle $BLM$ .
Show that the lines $KH$ , $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)

Luxembourg (Pierre Haas)

Collinearity of $L,H,M$ directly follows from the fact that $L$ and $M$ lie on the Steiner line and the Steiner line passes through $H$ . Reflect orthocenter $H$ about $BC$ to $H_{A}$ which lies on $(ABC)$ . It is clear that $HMKH_{A}$ is an isoceles trapezoid since $BC$ is a common perpendicular bisector. Then extend $MH_{A}$ to meet $(ABC)$ at $E_1$ . By simple angle chasing it can be proven that $E_1$ lies on $(BLM)$ and we conclude $E_1= E$ .

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1079 posts

#27 h Feb 24, 2021, 8:58 PM

Y by

Clearly not a unique solution, but maybe a different approach:

Claim. $L,M,H$ are collinear.
We have $h=a+b+c$ , $m=b+c-\frac{bc}{k}$ and $l=a+b-\frac{ab}{k}$ .
We have $\frac{h-l}{\overline{h}-\overline{l}}=\frac{a+b+c-a-b+\frac{ab}{k}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a}-\frac{1}{b}+\frac{k}{ab}}=\frac{abc}{k}$ and $\frac{h-m}{\overline{h}-\overline{m}}=\frac{a+b+c-b-c+\frac{bc}{k}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{b}-\frac{1}{c}+\frac{k}{bc}}=\frac{abc}{k}.$ Claim follows.

Let $H'$ be the reflection of $H$ over $BC$ .
Claim. $E, M, H'$ are collinear.
$\measuredangle BEM=\measuredangle BLM=\measuredangle LMB=\measuredangle HMB=\measuredangle BKH'=\measuredangle BAH'=\measuredangle BEH'$ Claim follows.

Now notice that since $MHH'K$ is an isosceles trapezoid, the perpendicular bisector of $MK$ , $BC$ passes through the intersection of $KH$ and $MH'\equiv EM$ .

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5001 posts

#28 h May 26, 2021, 10:41 PM

Y by

wrong…

This post has been edited 2 times. Last edited by IAmTheHazard, May 27, 2021, 2:53 AM
Reason: Makes no sense

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11290 posts

#29 h Aug 13, 2021, 12:48 AM

Y by

Let $H_a$ be the reflection of $H$ across $BC$ , and let $H_c$ be the reflection of $H$ across $AB$ . Note that $MH_a,KH,BC$ concur. Ignore the defintion of point $E$ given and let $E=H_aM\cap\Gamma$ . It suffices to show that $LBME$ is cyclic.

Claim 1: (well-known) $H_a$ lies on $\Gamma$ .
Note that $\angle H_aBC=\angle CBH=90-\angle ACB$ while $\angle H_aCB=\angle BCH=90-\angle ABC$ . Summing these, we have $180-\angle BH_aC=\angle BAC$ , so $ABH_aC$ is cyclic.

Claim 2: $L,E,H_c$ are collinear.
This is true since
$\measuredangle LH_cB=-\measuredangle KHB=\measuredangle MH_aB=\measuredangle EH_aB=\measuredangle EH_cB.$
Claim 3: $L,B,M,E$ are concyclic
We have
$\measuredangle LEM=\measuredangle LEH_a=\measuredangle H_cEH_a=\measuredangle H_cBH_a=2\measuredangle ABC$ and
$\measuredangle LBM=\measuredangle LBK+\measuredangle KBM=2\measuredangle ABK+2\measuredangle KBC=2\measuredangle ABC,$ so the result holds.

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5001 posts

#30 h Mar 17, 2022, 12:45 AM

Y by

ok I think it's right this time

Let $H'$ denote the reflection of $H$ over $\overline{BC}$ , $K'$ the midpoint of $\overline{MK}$ (which lies on $\overline{BC}$ ), and $L'$ the midpoint of $\overline{LK}$ (which lies on $\overline{AB}$ ). It suffices to show that $H', M, E$ are collinear, since $\overline{KH}$ and $\overline{MH'}$ intersect on $\overline{BC}$ by symmetry. Note that by reflections we have $BL=BM=BK$ , so $B$ is the center of $(LMK)$ . A homothety of scale factor $\tfrac{1}{2}$ at $K$ thus shows that $BL'M'K$ is cyclic. Further, from $BL=BM$ it follows that $\overline{EB}$ bisects $\angle LEM$ . Thus we have
$\begin{align*} \measuredangle BEM&=\measuredangle LEB\\ &=\measuredangle LMB\\ &=90^\circ-\frac{\measuredangle LBM}{2}\\ &=90^\circ-\measuredangle LKM\\ &=90^\circ-\measuredangle L'KM'\\ &=90^\circ-\measuredangle L'BM'\\ &=90^\circ-\measuredangle ABC\\ &=\measuredangle BAH'\\ &=\measuredangle BEH', \end{align*}$ where we are allowed to write $\tfrac{\measuredangle LBM}{2}$ because we treat it as an arc measure (modulo 360 degrees) rather than a conventional directed angle modulo 180 degrees. This implies that $E,M,H'$ are collinear, so we're done. $\blacksquare$

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1481 posts

CT17

#31 h Apr 4, 2022, 11:55 AM

Y by

Let $AH$ intersect $\Gamma$ again at $A'$ , and let $KL$ intersect $BC$ at $P$ . Moreover, let $KM$ and $KL$ intersect $BC$ and $BA$ at $X$ and $Y$ , respectively, so that $BKXY$ is cyclic.

Claim: $P$ lies on $(BLM)$ .

Proof: We have

$\measuredangle MBL = \measuredangle KBL - \measuredangle KBM = 2\measuredangle KBA - 2\measuredangle KBC = 2\measuredangle CBA = 2\measuredangle XBY = 2\measuredangle XKY = 2\measuredangle MKP = \measuredangle MPK = \measuredangle MPL$
as desired.

Claim: $A'$ , $M$ , and $E$ are collinear.

Proof: We have

$\measuredangle A'EP = \measuredangle A'EB - \measuredangle PEB = \measuredangle A'AB - \measuredangle PLB = (90^\circ - \measuredangle ABC) - (90^\circ - \measuredangle ABK) = \measuredangle CBK = \measuredangle MBP = \measuredangle MEP$
as desired.

Since $HA'KM$ is an isosceles trapezoid symmetric about $BC$ , $MA'$ and $HK$ concur on $BC$ , as desired.

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1079 posts

#32 h Apr 7, 2022, 6:05 PM

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EGMO predicted BLM?!

Let $H_A$ denote the reflection of $H$ over $BC$ . It is sufficient to show that $EM$ passes through $H_A$ .
Observe that $AHBL$ and $BHMC$ are cyclic as they are reflections of $(ABC)$ over $AB,BC$ , respectively.
Thus, $\measuredangle LHB=\measuredangle LAB=\measuredangle BAK=\measuredangle BCK=\measuredangle MCB=\measuredangle MHB,$ which proves Steiner's theorem, which states that $L,H,M$ are collinear. Finally, $\measuredangle BEM=\measuredangle BLM=\measuredangle BLH=\measuredangle BAH=\measuredangle BAH_A=\measuredangle BEH_A,$ which proves the desired. $\blacksquare$
$[asy]import olympiad;import geometry; size(10cm);defaultpen(fontsize(10pt)); pair A,B,C,H,K,L,M,Ha,E; A=dir(120);B=dir(205);C=dir(335);H=orthocenter(A,B,C);K=dir(265);L=2*foot(K,A,B)-K;M=2*foot(K,B,C)-K;Ha=2*foot(A,B,C)-H; E=2*foot((0,0),M,Ha)-Ha; draw(A--B--C--cycle);draw(circumcircle(A,B,C));draw(L--M);draw(K--H^^Ha--E,dotted); draw(circumcircle(A,B,H)^^circumcircle(H,B,C)^^circumcircle(B,L,M),dashed);draw(A--Ha); clip((1.1,-1.1)--(-2,-1.1)--(-2,1.1)--(1.1,1.1)--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$K$",K,dir(K)); dot("$H_A$",Ha,dir(Ha)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$E$",E,dir(E)); [/asy]$

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#33 h Apr 11, 2022, 5:04 PM

Y by

Well $H$ has no usage in our problem so lets make it useful. Let $S,P,Q$ be reflections of $H$ across $BC,AB,AC$ . Note that $HSKM$ is isosceles trapezoid so $S,KH,BE$ are concurrent so we need to prove $E,M,S$ are collinear. Let $SM$ meet $\Gamma$ at $E'$ we will prove $LE'MB$ is cyclic.
Claim : $L,P,E'$ are collinear.
Proof : Note that $LPHK$ is isosceles trapezoid so $\angle LPB = \angle BHK = \angle BHS + \angle SHK = \angle C + \angle E'SA = \angle E'CB \implies L,E',P$ are collinear.
Note that $B$ lies on perpendicular bisector of $HS,KM$ so $\angle BMS = \angle BKH = \angle BLP = \angle BLE'$ so $LE'MB$ is cyclic as wanted.
we're Done.

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322 posts

BVKRB-

#34 h Apr 15, 2022, 7:09 AM

Y by

Let $X$ be the reflection of $H$ over $BC$
Note that the problem basically asks us to show that $X-M-E$ and also observe that $B$ is the centre of $\odot(KLM)$
Also notice that $L-H-M$ because $LM$ is the steiner line of $K$
Now $\angle MEB = \angle MLB = 90^{\circ} - \angle LKM = 90^{\circ}-\angle ABC = \angle XAH = \angle XEB \implies KH \ \cap \ ME \in BC \ \ \blacksquare$

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1080 posts

#35 h Apr 26, 2022, 4:59 AM • 1 Y

Y by centslordm

Let $H_A$ be the reflection of $H$ in $\overline{BC}.$ Notice $H,M,$ and $K$ are collinear by the Simson Line Bisection Lemma. Also, $AHBL$ is cyclic as $\measuredangle AHB=\measuredangle BCA=\measuredangle BKA=\measuredangle ALB.$ Thus, $\measuredangle BEM=\measuredangle BLM=\measuredangle BAH=\measuredangle BCH_A=\measuredangle BEH_A$ and $H_A,M,$ and $E$ are collinear. Since $\overline{KH_A}$ is the reflection of $\overline{MH}$ in $\overline{BC},$ we are done. $\square$

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113 posts

#36 h Nov 5, 2022, 3:37 PM • 1 Y

Y by jelena_ivanchic

Solved with L567, mxlcv and jelena_ivanchic!

Add $H_A,H_C$ , the reflection of $H$ in $BC,BA$ respectively. Due to reflections we have $KH, MH_A, BC$ concurrent. Hence we want to prove $H_A-M-E$ .
We have $L-H-M$ by Steiner Line. As $ABH_CK$ cyclic, reflection in $AB$ implies $ABHL$ cyclic. We have been given that $BMEL$ cyclic.
Hence, $\angle MEB=\angle MLB=\angle HLB=\angle HAB=\angle H_AAB=\angle H_AEB$ . This implies $H_A-M-E$ , and we are done!

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320 posts

PNT

#37 h Mar 4, 2023, 11:38 PM

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Let $H’$ be the reflection of $H$ across $BC$ . $I=EM\cap HK$ We prove $B,I$ and $C$ are collinear or by letting $E’=MH’\cap \Gamma$ it suffices to prove $LE’MB$ cyclic (because $MH’$ , $HK$ and $BC$ are obviously concurrent).
Angle chasing the diagram we get $LAHB$ and $BHMC$ are both cyclic. Which implies $M,H$ and $L$ are indeed collinear. Now we’re ready to prove the main cyclicity ,
We have $\angle LME’=180^{\circ}-\angle HMH’$ and $\angle LBE’= \angle LBA+ \angle ABE= \angle ABK+ \angle HH’M$ They’re equal iff $\angle HMH + \angle ABK+ \angle HH’M =180 ^{\circ}$ but $\angle HMH+ \angle HH’M=180 ^{\circ}-\angle H’HM$ $=180 ^{\circ}-\angle H’HC-\angle CHM=180 ^{\circ}-\angle FHA-\angle KBC$ $90 ^{\circ} + \angle BAD-\angle KBC=180 ^{\circ}-\angle ABK.$ Where $F$ and $D$ are the foot of the altitudes.

Attachments:

This post has been edited 3 times. Last edited by PNT, Mar 4, 2023, 11:46 PM
Reason: F,D

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5419 posts

#38 h Aug 3, 2023, 4:49 PM

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First, note that $H,L,M$ are collinear $(\ast)$ , because
$\angle BHL+\angle BHM\stackrel{(\ast)}{=}\angle BH_BK+\angle BH_AK\stackrel{(ABC)}{=}180^\circ.$ Let $H_A,H_C$ be the reflections of $H$ across $\overline{BC},\overline{AB}$ respectively, so that clearly $HH_AKM$ and $HH_BLK$ are isosceles trapezoids $(\dagger)$ . Since $BC,HK,H_AM$ concur, it remains to prove that $H_A,M,E$ are collinear. Indeed, we claim that the intersection $E'$ of $H_AM$ and $(BLM)$ lies on $(ABC)$ , which suffices since such a point is unique. Write
$\angle BE'H_A=\angle BE'M\stackrel{(BE'LM)}{=}\angle BLM\stackrel{(\ast)}{=}\angle BLH\stackrel{(\dagger)}{=}\angle BKH_B=90^\circ-B=\angle BAH_A,$ and we are done. $\square$

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1946 posts

#39 h Mar 21, 2024, 3:40 PM

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First, extend $\overline{EM}$ , and have $(ABC) \cap \overline{EM}$ be $X$ . Note that $\overline{BL}=\overline{BK}=\overline{BM}$ , and that $BMHC$ is cyclic (by a reflection over $\overline{BC}$ .

Claim: $KH$ , $EM$ and $BC$ are concurrent iff $X$ is the reflection of $H$ over $\overline{BC}$ .

Proof: We note that $MHH'K$ is an isosceles trapezoid with axis of symmetry $BC$ , implying the claim.

Claim: $\triangle{BHC}$ is congruent to $\triangle{BXC}$ .

Proof: We note that $\angle{BCX}=\angle{BEX}=\angle{BLM}=\angle{BML}=\angle{BCH}$ . We also note that $\angle{BHC}=\pi-(\frac{\pi}{2}-\angle{C}+\frac{\pi}{2}-\angle{B})=\pi-\angle{A}=\angle{BXC}$ . Therefore, the 2 triangles are congruent, implying the conclusion.

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165 posts

#40 h Jun 25, 2024, 2:54 AM • 1 Y

Y by GeoKing

Let me provide a solution using Stiener line

Let $H_a$ be the reflection of $H$ in the line $BC$ (It’s known that $H_a \in \Gamma$ )
Since $H$ , $H_a$ and $M$ , $K$ are symmetric to $BC$
So $HK \cap MH_a \in BC$
Therefore, we only have to prove $E$ , $M$ , $H_a$ collinear

Let $N$ be the reflection of $K$ in the line $AC$
Then $LMH$ is the Stiener line of $K$ with respect to $\triangle ABC$
So $L$ , $H$ , $M$ , $N$ are collinear

Let $H_aM \cap \Gamma =E’$ , through easy angle chasing
$\measuredangle BE’M=\measuredangle BE’H_a =\measuredangle BKH_a=\measuredangle BMH=\measuredangle BML=\measuredangle BLM$ $\therefore \odot(BLME’) \implies E=E’ \implies H_a$ , $M$ , $E$ collinear $\implies$ $KH$ , $EM$ and $BC$ are concurrent

This post has been edited 1 time. Last edited by shanelin-sigma, Jun 25, 2024, 2:56 AM

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L13832

#41 h Jul 29, 2024, 6:19 AM • 1 Y

Y by CRT_07

Note that $E'M$ intersects at $H_A$ . By construction $E'M$ intersects $HK$ at $BC$ . Now we will prove $LH_CE$ is collinear. It can be easily seen that $\triangle H_CBL \sim \triangle HBK \sim \triangle H_ABM$ . We can prove $\measuredangle LH_CB=\measuredangle BHK=\measuredangle MH_AB$ and $\measuredangle EH_CB=\measuredangle E'AB =\measuredangle E'CB= \measuredangle E'KB=\measuredangle E'CB=\measuredangle MH_AB$ $\Rightarrow \measuredangle LH_CB=\measuredangle EH_CB$ , so $LH_CE$ is collinear. Now we prove $LE'MB$ is concyclic. $\measuredangle LE'M=\measuredangle H_CBH_A= \measuredangle ABC$ . $\measuredangle LBM = \measuredangle LBK =\measuredangle KBC=\measuredangle MBC$ . $\measuredangle CBD =\measuredangle ABC$ this gives use $\measuredangle H_ABD=\angle ABM$ $\Rightarrow LE'MB$ is concyclic.
$\therefore$ $E \equiv E'$ and we are done.

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1618 posts

#42 h Jul 29, 2024, 9:08 AM • 1 Y

Y by CRT_07

Redefine $E$ is Anti - Steiner point of $HK$ WRT $\triangle ABC$ . Then $EL, EM$ must pass through the intersections of $AH, CH$ with $(ABC),$ let them be $D, F$ . We have $\angle{LEM} = \angle{DEF} = \angle{DEB} + \angle{BEF} = \angle{DAB} + \angle{BCF} = 2\angle{OAC} = 180^{\circ} - \angle{AOC} = 180^{\circ} - 2\angle{ABC} = 180^{\circ} - 2\angle{LKM} = 180^{\circ} - \angle{LBM}$ . So $E \in (LBM)$

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cursed_tangent1434

616 posts

cursed_tangent1434

#43 h Oct 19, 2024, 12:36 AM

Y by

Unlike many of the solutions my proof uses the fact that $L$ , $M$ and $H$ are collinear, instead of obtaining it as a corollary. Let $H'$ be the reflection of $H$ across side $BC$ which is well known to lie on $(ABC)$ . First of all note that $\overline{ML}$ is the Steiner Line of $K$ with respect to $\triangle ABC$ which is well known to pass through its orthocenter $H$ . Thus, points $L$ , $M$ and $H$ are collinear. With this in hand we prove the following key claim.

Claim : Points $E$ , $M$ and $H'$ are collinear.
Proof : First of all, since $H'$ is the reflection of $H$ across $BC$ and $M$ is the reflection of $K$ across $BC$ , the quadrilateral $KH'HM$ is in fact an isosceles trapezoid, and hence cyclic. Simply note that,
$\measuredangle H'EB = \measuredangle H'KB = \measuredangle BMH = \measuredangle BML = \measuredangle MLB = \measuredangle MEB$ which indeed implies that points $E$ , $M$ and $H'$ are collinear.

Now we are done since $\overline{EM}$ and $\overline{KH}$ are the diagonals of isosceles trapezoid $KMHH'$ which intersect on the common perpendicular bisector of the parallel sides, $\overline{BC}$ .

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156 posts

Ritwin

#44 h Feb 20, 2025, 3:45 AM • 2 Y

Y by Curious_Droid, cursed_tangent1434

oh hey no one used pascal yet!

Let $AH$ and $CH$ meet $\Gamma$ again at $D$ and $F$ respectively. Also select $N \neq K$ on $\Gamma$ so that $BN = BK$ . By reflecting over $BC$ , it suffices to show that $D$ , $M$ , and $E$ are collinear. We will prove $E = DM \cap FL$ .

Claim: $N$ lies on $AL$ and $CM$ .
Proof. We have $N \in AL$ from $\measuredangle LAB = \measuredangle BAK = \measuredangle NAB$ , and $N \in CM$ follows similarly. $\square$

Claim: $H$ , $M$ , and $L$ are collinear.
Proof. Notice $LM$ is the Steiner line of $K$ .
Alternatively, we can angle chase $\measuredangle LHB = \measuredangle BFK = \measuredangle BDK = \measuredangle MHB$ from the reflections. $\square$

Now letting $E' = DM \cap FL$ , converse Pascal on $ANCFE'D$ implies $E' \in \Gamma$ . All that remains is showing $E' = E$ . To prove $E' \in (BLM)$ , it suffices to chase $\begin{align*} \measuredangle ME'L &= \measuredangle DE'F = \measuredangle DBF = 2 \measuredangle CBA, \\ \measuredangle LBM &= \measuredangle LBA + \measuredangle ABC + \measuredangle CBM \\ &= \measuredangle ABK + \measuredangle ABC + \measuredangle KBC \\ &= 2 \measuredangle ABC. \end{align*}$ Hence $\measuredangle ME'L = \measuredangle MBL$ , so we are done. $\blacksquare$

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35 posts

#45 h Feb 20, 2025, 5:17 AM • 1 Y

Y by Ritwin

Did I... make a mistake? A weirdly speedy solution?

Let $H'$ be the reflection of $H$ over $BC$ . It is obvious that $MH'$ and $HK$ concur on $BC$ . Now, $\angle MEB = \angle MLB = 90^\circ - \frac12 \angle LBM = 90^\circ - \frac12 (2\angle KBA - 2\angle KBC) = 90^\circ - \frac12 (2\angle B) = 90^\circ - \angle B = \angle H'AB = \angle H'EB \implies M, H', E$ collinear. Thus we are done.

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#46 h Apr 25, 2025, 10:00 AM

Y by

My proof is similar to most angle chasing solutions, but i defined O to be the circumcenter so that I get a similarity

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