# Strange point lies on common tangent of 9-point circle and incircle

Let $$ABC$$ be a triangle, with medial triangle $$DEF$$ and intouch triangle $$PQR$$. Let $$J$$ be the midpoint of $$\overline{AD}$$, and let $$BJ$$ meet $$AP$$ at $$K$$. Let $$X$$ be the point on ray $$\overrightarrow{CB}$$ such that $$CX=CA$$. Let the line through $$K$$ parallel to $$BC$$ meet $$AX$$ at $$U$$. Let $$RU$$ meet $$BC$$ at $$T$$. Prove that $$T$$ lies on the common tangent of the nine-point circle and the incircle of $$\triangle ABC$$.

I have a proof with barycentric coordinates. We can compute $$K=(2(s-b):s-c:s-b)$$, $$X=(0:b:a-b)$$, $$U=(a-b+c:b:a-b)$$ and $$T=(0:a-c:a-b)$$. The common tangent of the nine-point circle and incircle has equation $$(a-b)(a-c)x+(b-c)(b-a)y+(b-c)(a-c)z=0,$$ and the result follows.

However, as you might guess, I'm interested in a synthetic solution to this problem.

• Oh boy! what a problem, so many conditions xD. I'm usually a big fan of classical geometry problems so I might give it a shot. Sounds fun! – Cristian Baeza Mar 17 '18 at 20:39
• This is the Feuerbach Point. – Ed Pegg Jul 16 '18 at 17:33
• @EdPegg Unfortunately not. The Feuerbach point is where the incircle and nine-point circle touch. $T$ is just some point on their common tangent. It is not a triangle centre. Its definition is not symmetrical -- when choosing $AD$ we had three medians to choose from, and there are two choices of $X$ (the other having $BX=BA$). – Rosie F Apr 8 '20 at 16:50
• Also asked on (AoPS)[artofproblemsolving.com/community/c6h2256593p17425626] and artofproblemsolving.com/community/c1266869 – brainjam Nov 9 '20 at 1:58

The question has been asked and answered at Art of Problem Solving. User amar_04 gives a synthetic proof. Reproduced here, in case the URL breaks.

Just use Menelaus repeatedly

$$\textbf{LEMMA 1:-}$$ The Tangent to the Incircle at the Feuerbach Point and the Nagel Line of $$\Delta ABC$$ are Isotomic Lines. Proof:- See Theorem 10 here

$$\textbf{LEMMA 2:-}$$ Let $$\tau_1,\tau_2$$ be two Isotomic Lines in $$\angle A$$ cutting $$BC$$ at $$X,Y$$. Then $$BX=CY$$

Proof:- Let $$\tau_1$$ cut $$\overline{AB},\overline{AC}$$ at $$\{P,M\}$$ and let $$\tau_2$$ cut $$\overline{AB},\overline{AC}$$ at $$Q,N$$. Apply Menelaus Theorem on $$\Delta ABC$$ cut by the transversals $$\tau_1,\tau_2$$ \begin{align*}\frac{CX}{XB}\cdot\frac{BP}{PA}\cdot\frac{AM}{MC}=-1=\frac{BY}{YC}\cdot\frac{CN}{NA}\cdot\frac{AQ}{QB}&\implies\frac{CX}{XB}=\frac{BY}{YC}\\ &\implies\frac{CB}{XB}=\frac{CB}{YC}\\ &\implies BX=CY\end{align*}

$$\textbf{LEMMA 3:-}$$ If the Tangent $$(\ell)$$ to the Incircle at the Feuerbach point of $$\Delta ABC$$ cuts $$\overline{BC}$$ at $$X$$. Then $$BX=\frac{a(b-a)}{b+c-2a}$$

Proof:- Let $$G,N_a$$ be the Centroid and Nagel Point of $$\Delta ABC$$ and let $$\overline{AG}\cap\overline{BC}=\{M\}$$ and $$\overline{AN_a}\cap\overline{BC}=\{P\}$$. Let the Nagel Line of $$\Delta ABC$$ cut $$\overline{BC}$$ at $$Y$$. Now applying Menelaus Theorem on $$\Delta AMP$$ cut by the Nagel Line gives

\begin{align*}\frac{PY}{YM}\cdot\frac{MG}{GA}\cdot\frac{AN_a}{N_aP}=-1 &\implies\frac{PY}{YM}=\frac{2(s-a)}{a}\\ &\implies\frac{PM}{MY}=\frac{2s-3a}{a}\\ &\implies\frac{\frac{a}{2}-(s-b)}{MY}=\frac{2s-3a}{a}\\ &\implies MY=\frac{ab-ac}{2(b+c-2a)}\\ &\implies CY=MY+CM\\ &=\frac{ab-ac}{2(b+c-2a)}+\frac{a}{2}\\ &=\frac{a(b-a)}{b+c-2a}\end{align*} Now by $$\textbf{LEMMA 1}$$ we get that $$\ell$$ and the Nagel Line are Isotomic Lines and if $$\ell\cap\overline{BC}=X$$. Then from $$\textbf{LEMMA 2}$$ we get $$CY=BX=\frac{a(b-a)}{b+c-2a}$$.

Now come back to the Problem at hand. Applying Menelaus Theorem on $$\Delta ADP$$ cut by the transversal $$\overline{BI}$$ we get $$\frac{DB}{BP}\cdot\frac{PK}{AK}\cdot\frac{AJ}{JD}=-1\implies\frac{AK}{PK}=\frac{a}{2(s-b)}=\frac{AU}{XU}$$ Again by Menelaus Theorem we get $$\frac{XT}{TB}\cdot\frac{BR}{RA}\cdot\frac{AU}{XU}=-1\implies > \frac{BX}{TB}=\frac{2s-3a}{a}\implies TB=\frac{a(b-a)}{b+c-2a}$$ So by $$\textbf{LEMMA 3}$$ get that $$T\in$$ on the Tangent to the Incircle at the Feuerbach Point. $$\blacksquare$$