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In triangle $ABC$, let $r_A$ be the line that passes through the midpoint of $BC$ and is perpendicular to the internal bisector of $\angle{BAC}$. Define $r_B$ and $r_C$ similarly. Let $H$ and $I$ be the orthocenter and incenter of $ABC$, respectively. Suppose that the three lines $r_A$, $r_B$, $r_C$ define a triangle. Prove that the circumcenter of this triangle is the midpoint of $HI$

Solution:

Construct the medial triangle of $ABC$, $DEF$, with $D, E, F$ the midpoints of $BC, CA, AB$. Note the angle bisector of $\angle BAC$ is parallel to the angle bisector of $\angle EDF$. Thus, the triangle formed by $r_A, r_B, r_C$ is the excentral triangle of the medial triangle.

Let $S$, $N$ denote the incenter and circumcenter of the medial triangle. Then $S$ is the orthocenter of the triangle formed by $r_A, r_B, r_C$ with $N$ the Nine-Point Center of the same triangle, so the reflection of $N$ across $S$, $N'$ is the circumcenter of this triangle.

Also, $H$ is the reflection of $O$, the circumcenter of $ABC$, about $N$. Thus $HN'$ is parallel to $OS$, and $HN' = OS$.

Now consider a homothety about $G$, the centroid of $ABC$, of factor $-2$. $O$ is mapped to $H$. Since this maps the medial triangle $DEF$ to $ABC$, $S$, the incenter, maps to the incenter $I$ of $ABC$. Then $HI$ is parallel to $OS$, so it follows that $H, I, N'$ are collinear.

$HN' = OS$ from before, and $HI$ = $2OS$, so it follows that $N'$ is the midpoint of $HI$, as desired.

What would the design of this problem look like?

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  • $\begingroup$ "What would the design of this problem look like?" What does this mean? $\endgroup$ – darij grinberg Nov 4 '19 at 4:01
  • $\begingroup$ @darijgrinberg The geometric figure of this explanation $\endgroup$ – trombho Nov 4 '19 at 4:02
  • $\begingroup$ So the problem was solved in detail without a picture, and now we only need the picture?! $\endgroup$ – dan_fulea Nov 4 '19 at 5:58
  • $\begingroup$ @dan_fulea I think the problem is correct ... but answering your question, yes! $\endgroup$ – trombho Nov 4 '19 at 6:03
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Here is a possible picture for the already solved problem:

Stackexchange problem 3421067 picture dan_fulea


Well, how could you fill in the details without a faithful picture?! (And what is the "reflection of $N$ across $S$" by construction, the point $N$ is reflected with respect to the center $S$ or conversely?!)

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  • $\begingroup$ The solution is not mine! Edited wrong and had not seen $\endgroup$ – trombho Nov 5 '19 at 15:48

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