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Geometry question from EGMO book by Evan Chan $(BAMO$ $2013/3).$

Let $H$ be the orthocenter of an acute triangle $ABC$. Consider the circumcenters of triangles $ABH, BCH,$ and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.

[Outline]

For now let us call $(ABH)$ as $O_C$, $(ACH)$ as $O_B$, $(CBH)$ as $O_A$. Well, since $O_C$, $O_B$ are perpendicular to A (perpendicular bisector of $AH$), this implies that $O_CO_B ||CB$. Similarly the other sides. This basically shows that $\angle CAB= \angle O_CO_AO_B$. Similar results from other side give that the 2 triangles are similar. We may now use the extended law of sines on $\triangle BHC$ to get that the circumradii of the triangles are infact equal. To prove that the triangles are congruent, the given conditions suffice and thus we are done.

Well thats all cool but the hints told us that we could set up a homothety and show that $O_BO_C = 2\cdot (\frac{1}{2} BC)= BC$. Hmmmm. My scaled diagram gives me the vibes that infact with a homothety centred at $N_9$ and scaling factor -1, $C$ becomes $O_C$ and the others follow similarly. Also, since $N_9$ bisects $O$ and $H$.This motivates me even more to set up the mentioned homothety as it can be easily proved that H is infact the orthocenter of $\triangle O_CO_AO_B$. But since i have never used homothety and this happens to be my first question ever using it. Could someone please walk me through this using a well written solution? Thanks a lot.

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enter image description here

Let $X,Y,Z$ be the circumcenters. Then, XB = XH, ZB = ZH and

$$∠BZX = \frac12∠BZH = ∠BAH = a,\>\>\>\>\>∠BXZ = \frac12∠BXH = ∠BCH = a$$

Then, BXHZ is a rhombus; so is CXHY. As a result, BZ = XH = CY, BZ || XH || CY and BCYZ is a parallelogram, which leads to BC = ZY. Similarly, AB = YX and AC = ZX. Thus, the triangles XYZ and ABC are congruent.

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  • $\begingroup$ Yes @Quanto i get that the problem can infact be solved which is what i had tried to depict with the outline. However could you please provide a solution that specifically uses homothety. $\endgroup$ – Aayam Mathur Jun 25 '20 at 19:13
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As you said, the triangle $O_A O_B O_C$ has its sides respectively parallel to those of $ABC$. This implies that it is the image of $ABC$ under some dilation or translation $h$.

Let $O$ be the circumcenter of $ABC$. Then it is easy to see that it is the orthocenter of $O_A O_B O_C$. Therefore $h(H) = O$.

At the same time, $H$ is the circumcenter of $O_A O_B O_C$. Therefore $h(O) = H$.

Given that $h$ exchanges $O$ and $H$, it must be the dilation of coefficient $-1$ centered at the midpoint $N$ of $O$ and $H$.

Added. It may be that the intended proof was to show separately that each triangle is homothetic to the triangle of midpoints of $ABC$. $ABC$ is homothetic to it through the center of gravity $G$. $O_A O_B O_C$ is homothetic to it through $O$. This last fact follows from the fact that the two triangles are in perspective from $O$ and have parallel sides. Incidentally, this gives another proof that the four circumradii are equal.

Taken together, all of what's been said shows that $O, H, N, G$ are aligned and allows you to determine what their relative positions are: the dilation of ratio $-1/2$ through $G$ followed by the dilation through $O$ with ratio $2$ is equal to the dilation through $N$ with ratio $-1$.

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