Among the large number of findings I have arrived at with the help of GeoGebra is this feature:
If the two common interior tangents of two distant circles are perpendicular, each of which touches the same parabola at two points, then the circle passing through the four points of contact touches the same parabola in turn, and the distance between the center of the circle and the perpendicular point will be equal to twice the distance between the focus and the guide.
The circle passing through the points $P,Q,R,T$ touches the parabola $C$ at two points.
$AM=BM$
$MN=2OF$
I came up with this about four years ago but have no idea how to prove it
Can someone help me please
Best Answer
Let $K$ be the midpoint of $PR$. Note that the lines $AP$, $BR$, $MK$ are all perpendicular to $PR$. This shows that $AP \parallel RB \parallel MK$. As a consequence, $\dfrac{AM}{MB}=\dfrac{PK}{KR}=1$, i.e. $AM=MB$.
Let $B_1$ be the tangency point of the circle $G$ and the parabola $C$. Let $B_2$ be the projection of $B_1$ onto the directrix of $C$. Let $F_1$ be the reflection of $F$ in the directrix. It is well known that the tangent to $C$ at $B_1$ is the bisector of $B_2B_1F$. Since $BB_1$ is perpendicular to that tangent, it follows that $B_1B$ is the external bisector of $B_2B_1F$. In particular $B_1B \parallel B_2F$. Since $B_1B_2 \perp B_2O \perp BF$, we have $B_1B_2\parallel BF$. This shows that $B_1B_2FB$ is a parallelogram, hence $BF=B_1B_2$. Next, observe that $B_2FF_1$ is an isosceles triangle with angle at the base equal to $\angle B_2FO = \angle FB_2B_1$ and therefore $\triangle FF_1B_2 \sim \triangle FB_2B_1$. This shows that $\dfrac{B_2B_1}{B_2F}=\dfrac{B_2F}{FF_1}$, and using $B_2B_1=FB$, $B_2F=B_1B$, $FF_1=2OF$ we find $$OF = \frac{BB_1^2}{2FB}.$$
Similarly, if the circle $D$ is tangent to the parabola $C$ at $A_1$, then $OF=\dfrac{AA_1^2}{2FA}$.
Since $APNQ$, $BRNT$ are squares, $AB=\sqrt2(AP+BR)=\sqrt2(AA_1+BB_1)$. We have
$$OF=\frac{AA_1^2}{2FA}=\frac{BB_1^2}{2FB}=\frac{AA_1^2-BB_1^2}{2FA-2FB}=\frac{(AA_1+BB_1)(AA_1-BB_1)}{2AB}=\frac{(AA_1+BB_1)(AA_1-BB_1)}{2\sqrt2(AA_1+BB_1)}=\frac{AA_1-BB_1}{2\sqrt2}.$$
Now, observe that $$MN=AN-AM=AN-\frac{AN+BN}{2}=\frac{AN-BN}{2}=\frac{\sqrt2(AA_1-BB_1)}{2}=2OF.$$
To prove that the red circle is tangent to the parabola we can use the ideas presented above and show that the radius $r$ of the red circle satisfies $OF=\dfrac{r^2}{2FM}$. This is equivalent to $r^2=2OF \cdot FM$. Note that $$2OF\cdot FM= 2OF \cdot \frac{FA+FB}{2} = \frac{AA_1^2+BB_1^2}{2},$$ hence we have to show that $2r^2=AA_1^2+BB_1^2$.
Note that $PQRT$ is an isosceles trapezoid and therefore $\angle MPR = \angle TQM = \angle MTQ$. This shows that $M,N,P,T$ are concyclic, hence $90^\circ = \angle TNP = \angle TMP$. Since $MT=MP=r$, we have $$2r^2=PT^2=PN^2+TN^2=AP^2+BT^2=AA_1^2+BB_1^2,$$ as desired.