# Midpoints, bisectors, orthocenter, incenter and circumcenter

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?

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

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?!)