# ELMO 2013/G7:Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear.

Let $$ABC$$ be a triangle inscribed in circle $$\omega$$, and let the medians from $$B$$ and $$C$$ intersect $$\omega$$ at $$D$$ and $$E$$ respectively. Let $$O_1$$ be the center of the circle through $$D$$ tangent to $$AC$$ at $$C$$, and let $$O_2$$ be the center of the circle through $$E$$ tangent to $$AB$$ at $$B$$. Prove that $$O_1$$, $$O_2$$, and the nine-point center of $$ABC$$ are collinear.

My Progress:

Here's the diagram Define : $$F,X,Y$$ as midpoints of $$BC,CA,AB$$ .

$$N_9$$ as the nine point center

$$O$$ as the circumcentre

$$H'$$ as the orthocentre

$$BX \cap (O_1) = L$$

$$CY \cap (O_2)= I$$

Claim: $$ABCL$$ is a parallelogram

Proof: Since $$AX=CX$$ by midpoint condition and $$BX=XL$$ by POP ( Taking power of $$X$$ wrt both circles $$XD.XL=CX^2= XD.XB$$ )

Similarly $$ABCI$$ is a parallelogram

Claim: $$ALI$$ are collinear

Proof: That follows from BC parallel condition

We also know that $$H, M,N_9,O$$ are collinear , $$M$$ is the centroid

Now what we noticed was that $$OO_2H'O_1$$ are parallelogram with $$N_9$$ as the intersection of the diagonals .

What I think is that showing $$OO_2H'O_1$$ a parallelogram is enough , since we know that $$N_9$$ is the midpoint of OH'

There's also nice dilations happening , like $$N_9$$ dilating $$O$$ to $$H'$$ and $$O_2$$ to $$O_1$$ with scale $$-1$$ ( observation )

dilation centred at $$X$$ and $$Y$$ with scale factor -1 too.

Moreover, we also know that $$OO_2 \perp BE$$ , so it's enough to show that $$O_1H' \perp BE$$ .

Also I want to find a pure synthetic method ( not using tring, cord , Bary, vector , etc ) but can include inversion or projective .

EDIT: Since @Anand told me to define $$IB\cap LC$$ , I defined $$IB\cap LC=J$$ Since $$IL || BC$$ , $$A$$ is midpoint of $$IA$$ and $$F$$ is midpoint of $$BC$$ , we get that $$JFA$$ is collinear and $$B,F,C$$ are midpoints of $$IJ,AJ,LC$$

So $$AJ,BL,CI$$ concur at $$K$$ , $$K$$ is the centroid and we get that $$K$$ is the centre of dilation with scale factor $$-2$$ .

also $$K$$ dilates $$O$$ to $$H'$$ too . That's how much I could proceed till now :(

• Is $OO_1H'O_2$ always a parallelogram? There is a configuration in your diagram when it is a concave quadrilateral. – Fawkes4494d3 Sep 7 '20 at 20:58
• Woah! Really nice observations!! Here's a hint: Aren't your observations begging you to define $IB\cap LC$? – Anand Sep 8 '20 at 5:41
• @Anand actually no , Should I define IB\cap LC ..? – Sunaina Pati Sep 8 '20 at 8:13
• @Anand oh, H' is the centre of that circle , so everything is dilated with centre as G ..Nice! but how is this going to help us ? – Sunaina Pati Sep 8 '20 at 8:52
• Hello Sunaina! It's been over two weeks since i haven't seen one of your informative posts, hope you're doing good. – Baba Yaga Sep 23 '20 at 7:55 Here is my way to prove it. My apology if there is some point notation that is different that in your original problem as I only denote the point that is stated in your problem, not your figure. Moreover, there are idea of the problem that is obvious by angles transformation, hence I won't go detail into it.

Now let $$G$$ is the centroid of $$\triangle ABC$$, $$F$$ is midpoint of $$BC$$;$$K, L$$ is the intersection point of $$BX$$ with $$O_2$$ and $$CY$$ with $$O_1$$ respectively. Then denote $$J, I$$ are intersections point of $$BX, CY$$ with 9-point circle, respectively. And $$T,Z$$ are intersection points of $$CY$$ with $$O_2$$, $$BX$$ with $$O_1$$. Hence we will have: $$EYJB$$ are cyclic ($$\angle BEC = \angle XFY$$) which infer that $$BE\parallel XI$$. Similarly, we obtain $$DXIC, DZLC$$ are cyclic and $$CD \parallel YJ$$. Now since $$BE\parallel XI$$ and $$DXIC$$ are cyclic, we obtain that $$BE\parallel ZL$$. By the same way, $$CD \parallel TK$$.

Up to now, notice that $$TZLK$$ are cyclic (angle transformation), it follows that if $$O_3$$ is the center of $$(TZLK)$$ then $$O_3O_2 \perp TK$$ or $$O_3O_2 \perp CD$$. More insightful, $$(TZLK)$$ is nothing but a homothety of $$(XYJI)$$ with ratio $$\displaystyle \frac{1}{3}$$ with homothetic center is $$G$$. Is this ratio remind you about something? Yes, it is the ratio of $$\frac{GN}{GH}=\frac{1}{3}$$ where $$H$$ is the orthocenter of $$\triangle ABC$$ or we can say $$H$$ is a homothety of $$N$$ with the homothetic center is $$G$$. Therefore $$H$$ is the center of $$(TZLK)$$ and $$H\equiv O_3$$. Finally we have $$HO_1 \perp BE$$ and $$HO_2 \perp CD$$ as you desire.

• What are angle transformations? – Anand Sep 8 '20 at 5:58
• Oh it's just dealing with equal angle. $\angle A = \angle B$ and $\angle B = \angle C$ so $\angle A = \angle C$. It's nothing but my english. – Nguyễn Quân Sep 8 '20 at 6:00
• So what's the transformation here? Also, what do you mean by "there are idea of the problem that is obvious by angles transformation,"? Sorry for bothering you. – Anand Sep 8 '20 at 6:04
• Oh, got it! Thanks :) – Anand Sep 8 '20 at 6:11
• @NguyễnQuân is it okay, if I read your solution a bit later, I still want to try it on my own .. Thanks! – Sunaina Pati Sep 8 '20 at 8:21