# ELMO 2019/G3: Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.

Let $$\triangle ABC$$ be an acute triangle with incenter $$I$$ and circumcenter $$O$$. The incircle touches sides $$BC,CA,$$ and $$AB$$ at $$D,E,$$ and $$F$$ respectively, and $$A'$$ is the reflection of $$A$$ over $$O$$. The circumcircles of $$ABC$$ and $$A'EF$$ meet at $$G$$, and the circumcircles of $$AMG$$ and $$A'EF$$ meet at a point $$H\neq G$$, where $$M$$ is the midpoint of $$EF$$. Prove that if $$GH$$ and $$EF$$ meet at $$T$$, then $$DT\perp EF$$.

My Progress: After seeing this problem, the first thing that struck my mind was sharky devil lemma (not a very known lemma )

Here's the lemma: In triangle $$ABC$$, let $$DEF$$ be the contact triangle, and let $$(M)$$ be the midpoint of the arc $$(BC)$$ not containing $$(A)$$ in $$(ABC)$$. Suppose ray $$MD$$ meets $$(ABC)$$ again at $$R$$. If $$I$$ is the incenter of $$(ABC)$$ and ray $$RI$$ intersects $$(ABC)$$ again at $$A'$$, then $$A'$$ is the antipode of $$A$$. If $$P=RA'\cap EF$$, then $$DP\perp EF$$. Anyways, here's the problem's diagram: Here $$J$$ is defined as $$(ABC)\cap (AEF) .$$

Now, if I am able to show that $$JITA'$$ are collinear, then I am done.

Moreover, I got that $$T$$ is the radical centre of $$(AEF),(GHA')$$ and $$(AHG)$$. Here, I defined $$K$$ as $$AT\cap (AEF)$$.

Now,I thought of using Phantom points . So I defined $$T'= \overline{JIA'}\cap EF$$ .

We want to show that $$T'=T$$ . To show that $$T'=T$$ , we can also show $$G,T',H$$.

Now, note that $$AM\perp EF$$.

Let $$AJ\cap EF=L$$.

So, by radical axis lemma on $$(AEF),(ABC) ,(GH'EF)$$ , we get $$AJ,EF,GA'$$ concur at $$L$$ .

Also we have $$T'KMI$$ and $$AJT'M$$ cyclic .

Again by radical axis lemma on $$(AEF),(AJT'M),(T'MKI)$$ , we get $$AJ,TM(EF),KI$$ concur at $$L$$.

Note that $$\angle AGA'=90=\angle AMF$$ . Since $$LFE$$ and $$LGA$$ are collinear , we get $$(AHMGLK)$$ concyclic.

Also note that $$T'$$ is the orthocentre of $$\Delta ALI$$. This is what I got till now. Now after showing that J,I,T are collinear , by applying sharky devil lemma , we will be done . I know that this problem has a 1 para solution( by @Anand ), but can someone provide a non-projective solution ?

• not related to the question but do you know some link where they prove the sharky devil lemma? thanks Aug 27, 2020 at 4:21
• @hellofriends maybe this might help artofproblemsolving.com/community/c776104h1954097 Aug 27, 2020 at 4:32
• I saw the configuration in the Sharky-Devil lemma after a long time, saw it first in a proof of $IO^2=R^2-2Rr$. Aug 28, 2020 at 3:22

Note that $$\odot(AEF)$$ has $$AI$$ as diameter. Also, as $$T$$ is the radical center of $$\{\odot(AEF),\odot (GHA'),\odot (AGH)\}$$ and thus, $$AK\perp LI$$ and $$LT\perp AI\implies T$$ is orthocenter of $$\triangle ALI$$ and thus, $$IT\perp AL$$. Let, $$IT\cap AL=J'$$ and thus, $$J'\in\odot(AI)\implies J'=J\implies$$ by sharky devil lemma, we get, $$J-I-T$$ collinear completing the proof.
• is $K$ defined to be in line $AT$ or $AT'$? Because it seems to me when you prove $T'$ is orthocenter you already have that $T=T'$ Aug 29, 2020 at 9:03
• @hellofriends The above proof does not require the definition of $T'$ Aug 29, 2020 at 9:08