Team Selection Test

First, note that the focusing lines of a triangle are its Simson lines. We will make use of the following well-known lemmas about Simson lines and we state them without proof.

Lemma 1. Fix a triangle. Let $P$ be a point on the circumcircle and let $H$ be the orthocenter. The Simson line associated to $P$ passes through the midpoint of $[PH]$.

Lemma 2. Let $P_1,P_2$ be points on the circumcircle. Let ${\ell }_1,{\ell }_2$ be the Simson lines associated to $P_1,P_2$ respectively. Then, one has $$\angle ({\ell }_1,{\ell }_2)=\frac{1}{2}\cdot \overline{P_1{P}_2}$$

We now set a notation convention: Let $P$ be a point on the 9-point circle. Denote the reflection of $H$ across $P$ by $\bar{P}$. Note that $\bar{P}$ is on the circumcircle. Consider the Simson line associated to $\bar{P}$, which passes through $P$ (by Lemma 1). Let this line intersect the 9-point circle at $P^{\prime }$ other than $P$. For any point $Q$ on the 9-point circle, let $\bar{Q}$ and $Q^{\prime }$ be defined similarly.

In this notation, Lemma 2 appears as $$\angle (P{P}^{\prime },Q{Q}^{\prime })=\frac{1}{2}\cdot \overline{Q\bar{P}}\iff \overline{P^{\prime }{Q}^{\prime }}=2\cdot \overline{Q\bar{P}}$$

Exactly 3 of the Simson lines are tangent to the 9-point circle.

Indeed, line $R{R}^{\prime }$ is tangent to the 9-point circle iff $$R=R^{\prime }\iff \overrightarrow{PR}=2\cdot \overrightarrow{RP}\iff \overrightarrow{PR}=3\cdot \overrightarrow{RP}$$ where the last equation is satisfied by exactly 3 points $R$ on the circle, clearly. These 3 points are spaced equally around the circle.

Let the 9-point circle together with these 3 equally spaced points marked on it be called the marked 9-point circle. Therefore, given the marked 9-point circle of a triangle, all the Simson lines are determined. Take a triangle and consider its marked 9-point circle $\Omega$ and the equilateral triangle formed by the tangent lines at the marked points. Then, the marked 9-point circle of this equilateral triangle is the same as $\Omega$. Therefore, this equilateral triangle is equivalent to the original triangle. Furthermore, an equilateral triangle can be uniquely determined by its set of Simson lines, so the conclusion follows.

The last claim is justified as follows: Note that the sides of any triangle are among its Simson lines, thus given a full set of Simson lines, one must consider the equilateral triangles that can be formed by these lines in order to find the equilateral triangle that produced this set of Simson lines. Among these equilateral triangles, only the largest one produces this set of Simson lines.