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?
Best Answer
Here is a possible picture for the already solved problem:
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?!)