# 2012 EGMO P7: Show that the lines $KH$, $EM$ and $BC$ are concurrent [ Proof Verification needed]

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.)

Surely a very hard problem ! It took me 6 hrs to solve without any hints ! But this question is very diagram dependent, so if possible can someone verify my proof ? Thanks in advance.

Plus you can send your solution too, It helps me a lot.

My Proof: Now, let $$X$$ be the reflection of $$H$$ over side $$BC$$. It is well known that $$ABCX$$ is cyclic.

Claim: $$BMHC$$ and $$LBHA$$ are cyclic quads.

Proof: Note that $$ABCK$$ is cyclic ( It is given) . So $$\angle BMC=\angle BKC= \angle BXC= \angle BHC$$ and hence $$BMHC$$ is cyclic. Similarly, we can prove it for $$LBHA$$.

Let $$MX\cap HK=Y$$ . Note that by angle chase, we have $$Y \in BC$$.

So it is enough to show that $$E,M,X$$ are collinear .

Now, since $$BC$$ is the perpendicular bisector of $$MK$$ and $$AB$$ is the perpendicular bisector of $$LK$$, note that $$B$$ is the circumcentre of $$\Delta KLM$$ .

Define $$I=MK\cap BC$$ and $$G= LK\cap AB$$. Note that $$BIGK$$ is cyclic.

Now, we move to our next claim. ( Note: the Proof might look simple but it took me 4 hrs)

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

Proof: Since $$BIGK$$ is cyclic, we get $$\angle ABC=\angle GKI=\frac {1}{2} \angle LBM \implies \angle BML=90-\angle ABC$$ .

So it is enough to show that $$\angle HMB= 90+\angle ABC$$ or $$\angle HCB=90-\angle ABC$$ (which is true by angle chase , $$HC \perp AB$$)

Now,the main proof .

Claim: $$E,M,X$$ are collinear

Proof: Note that by using the cyclic quads $$(BMHC)$$, $$(LBHA)$$, $$(ABCEX)$$ and $$(LBME)$$, we note that

$$\angle BEM=\angle MLB=\angle BLH=\angle BAH=\angle BAX= \angle BEH =\angle BEX$$ .

Hence $$\angle BEM=\angle BEX$$. Hence $$EMX$$ are collinear.

And we are done!

• Hey! I haven't read your proof but why solving EGMO p7's . Note that INMO<<<<<EGMO – Raheel Aug 1 '20 at 11:53
• I read it up to just before the second claim --- why are you defining $I=MK\cap BK$? Isn't it obvious $I=K$ then (and "BIGK" concyclic is trivial)? By the way, you are missing the word "of " in "BC is the perpendicular bisector MK". – user10354138 Aug 1 '20 at 12:20
• @user10354138 that is a typo .. I edited it , it should be $I=MK\cap BC$ – Sunaina Pati Aug 1 '20 at 12:54
• @Raheel , INMO has become really hard.. – Sunaina Pati Aug 1 '20 at 12:58
• Hmm... that follows from the well-known result the reflection of the circumcircle in a side passes through the orthocenter, but OK it isn't too much of a problem. The final displayed formula has a lot of typos in there. It should be (omitting the angle signs) BEM=BLM=BLH=BAH=BAX=BEX. – user10354138 Aug 1 '20 at 13:13

This proof relies on a following lemma (which is easy to prove):

Lemma: Reflection of $$H$$ across side of a triangle lies on circumcircle of that triangle.

Let $$H'$$ and $$H''$$ be respectively reflections of $$H$$ across $$BC$$ and $$AB$$. If we prove $$E,M,H'$$ are collinear we are done since $$H'M$$ and $$HK$$ meets in side $$BC$$.

Let $$H'M$$ and $$H''L$$ meet at point $$F$$. If we prove that $$F$$ lies on both circles we are done.

Let $$\angle H'HC = x$$, $$\angle H'HK = y$$ and $$\angle MKB =z$$.

• Circle $$ABC$$:

Clearly $$\angle HCB = 90-x$$ and so $$\angle BCH' = 90-x$$. Also $$\angle HH'F = y$$ and $$\angle H''HK = 180-x-y$$ and thus $$\angle FH''H = 180-x-y$$. Since the sum of all angles in quadrilateral is $$360$$ we have (look at $$H''HKF$$) $$\angle H''FK = 2x$$ and thus $$F$$ is on circle $$ABC$$ (since $$\angle H''CH' +\angle H''FH' =180$$).

• Circle $$MBL$$:

Since reflection preserves angles we have $$\angle H'MB = y+z$$ and $$\angle BLF = y+z$$ and we are done.

• Great solution! I didn't think of reflecting H over AB . I will remember this. – Sunaina Pati Aug 1 '20 at 16:57