Why Point of the Intouch Triangle

Today we will explore the properties of a certain well-known point with respect to the intouch triangle. The Why Point, originally seen in the famous 2011 G4, is a thoroughly explored configuration with a plethora of interesting properties. Among the many perspectives of the Why Point configuration is the Why Point of the Intouch triangle.

Here is the picture we shall be working with.

A summary of the key results pertaining to this configuration,

First, the circles and pass through the reflections and of across points and respectively.
- Now PoP at $D$ shows that $\overline{DI_a}$ is the radical axis of these two circles.
- Their second intersection $Y_d$ lies on $\omega$ , which is a straightforward angle chase.
- In particular, a key step is that by Midpoint Theorem the line through the reflection of $D$ across $B$ and point $F$ (which passes through the $D-$ antipode in $\omega$ ) is parallel to $\overline{BI}$ , which is the statement of STEMS 2020 B4.
Further, the circles and pass through the and Evan is Old Points respectively.
- Most notably, inversion sends this configuration to one involving the mixtillinear excircles which are mostly easier to handle.
- An easy angle chase shows that $PFEQ$ is cyclic from which Radical Center Theorem on circles $\omega$ , $(Y_dD'PQ)$ and $(PQFE)$ we have that lines $Y_dD'$ and $EF$ intersect on $BC$ .
- The converse of Pascal now implies that points $B$ , $C$ , $E$ , $F$ , $D'$ and $Y_d$ are conconic.
Shooting Lemma now yields the most famous Shortlist 2002 G7 which tells us that circle is tangent to at .
- The pairwise radical axes of circles $(BY_dC)$ , $(AY_eC)$ and $(AY_fB)$ concur at $X_{57}$ , the homothety center of the intouch and excentral triangles which is essentially RMM 2012/6.
To see why the point defined as the tangency point of (the circle through and which is tangent to ) and is the Why Point of ,
- Inversion about the incircle maps $(Y_dBC)$ to $(Y_dB_0C_0)$ where $B_0$ and $C_0$ are the midpoints of segments $DE$ and $DF$ , whose tangency point to the circumcircle is the Why Point as per Shortlist 2011 G4.
- Now, a homography mapping the centroid of the intouch triangle to it’s circumcenter while preserving the incircle tells us that lines $AY_d$ , $BY_e$ and $CY_f$ are also concurrent.
- Furthermore, if $D_1$ is the reflection of $D$ across line $\overline{AI}$ and similarly, lines $Y_dD_1,Y_eE_1$ and $Y_fF_1$ concur at the centroid of $\triangle DEF$ , which is Basic Problem 026, used by Google Deepmind.
Now, revisiting Fake USAMO 2020/3 we have that lines $AE_{Oa}$ , $BE_{Ob}$ and $CE_{Oc}$ also concur at this homothety center $X_{57}$ .
Throwing in $Ge$ to the picture, which we know is the symmedian point of $\triangle DEF$ we have that the reflections of $Ge$ across the sides of $\triangle DEF$ lie on $AE_{O_a}$ , $BE_{Ob}$ and $CE_{Oc}$ which solves Brazil MO 2013/6.
Also of interest is the tangent to $\omega$ at $Y_d$ . Pascal’s Theorem on concyclic hexagon $Y_dY_dD'X'X'D$ shows that the second tangent to $\omega$ at $X$ and the tangent to $\omega$ at $Y_d$ intersect on the line through $A$ parallel to line $BC$ which is the key result in MODS MO 2021/7.