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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 $\triangle ABC$.

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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.

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  • $\begingroup$ 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! $\endgroup$ – Cristian Baeza Mar 17 '18 at 20:39
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    $\begingroup$ This is the Feuerbach Point. $\endgroup$ – Ed Pegg Jul 16 '18 at 17:33
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    $\begingroup$ @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 $BX=BA$). $\endgroup$ – Rosie F Apr 8 '20 at 16:50
  • $\begingroup$ Also asked on (AoPS)[artofproblemsolving.com/community/c6h2256593p17425626] and artofproblemsolving.com/community/c1266869 $\endgroup$ – brainjam Nov 9 '20 at 1:58
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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. $\blacksquare$

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