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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. enter image description here

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

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    $\begingroup$ you can certainly coordinate bash it by assuming your parabola is $y = ax^2+b$ - what's stopping you there $\endgroup$
    – dezdichado
    Commented Dec 29, 2023 at 20:20

2 Answers 2

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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.

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  • $\begingroup$ Please, have a look at my answer. $\endgroup$
    – Jean Marie
    Commented Jan 1 at 7:51
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Here is an analytical approach.

Consider the family of circles with center $(0,a)$ and radius $\sqrt{a}$, i.e., with common equation :

$$(C_a): \ \ \ x^2+(y-a)^2=a, \ \ \ a \ge 0 \tag{1}$$

It is not difficult to prove that the envelope of circles $(C_a)$ is the parabola $(P)$ with equation :

$$(P): \ \ \ y=x^2-\frac14$$

Consider now the notations of the figure below reproducing the desired configuration :

enter image description here

Let $O$, resp. $O'$ the centers of $(C_a)$, resp $(C_{a'})$.

Let $H$ be the internal center of similitude of this system of circles.

The property of your configuration is equivalent to the fact that triangles $OAH$ and $O'A'H$ are right isosceles triangles.

First consequence : being given $a$, the value of $a'$ can be expressed as a function of $a$ :

$$a'=(\sqrt{a}+\sqrt{2})^2=a+2+2\sqrt{2a}\tag{2}$$

(see Proof 1 below).

Second consequence :

$$\vec{OA}=\pmatrix{OA/\sqrt{2}\\OA/\sqrt{2}}=\pmatrix{\sqrt{\frac{a}{2}}\\ \sqrt{\frac{a}{2}}} \ \text{and} \ $$ $$ \vec{O'A'}=\pmatrix{O'A'/\sqrt{2}\\ - O'A'/\sqrt{2}}=\pmatrix{\sqrt{\frac{a'}{2}}\\ -\sqrt{\frac{a'}{2}}}$$

which implies that the coordinates of $A$ and $A'$ are :

$$A=(\sqrt{\tfrac{a}{2}},a+\sqrt{\tfrac{a}{2}}) \ \text{and} \ A'=(\sqrt{\tfrac{a'}{2}},a'-\sqrt{\tfrac{a'}{2}}).$$

Let $O''$ be the point with coordinates

$$O''=(0,a+1+\sqrt{2a}).$$

$O''$ is at the same distance from $A$ and from $A'$ ; this common distance is equal to $\sqrt{a+1+\sqrt{2a}}$ (see Proof 2 below).

Therefore $O''$ is the center of the (unique) circle centered on the $y$ axis and passing through $A$ and $A'$ ; moreover this circle belongs to the family described by equation (1).


Proof 1 of relationship (2) :

$$OO'=OH+HO' \iff a'-a=OA \sqrt{2} + OA' \sqrt{2}$$

$$(\sqrt{a'})^2-(\sqrt{a})^2=\sqrt{2}(\sqrt{a}+\sqrt{a'})$$

$$\sqrt{a'}-\sqrt{a}=\sqrt{2}$$

Proof 2 of the fact that

$$O''A=O''A'=\sqrt{a+1+\sqrt{2a}} \ \iff \ (O''A)^2=(O''A')^2=a+1+\sqrt{2a}$$

Let us prove that $(O''A)^2=a+1+\sqrt{2a}$ (the other identity with $(O''A')^2$ <follows the same logic), i.e., establish that

$$(\sqrt{\tfrac{a}{2}})^2+(a+1+\sqrt{2a}-a-\sqrt{\tfrac{a}{2}})^2=a+1+\sqrt{2a}$$

itself equivalent to :

$$\tfrac{a}{2}+(1+\tfrac{\sqrt{2}}{2}\sqrt{a})^2=a+1+\sqrt{2a}$$

which is readily verified.

Remark : The origin of coordinates is the focus of parabola (P).

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  • $\begingroup$ Thank you very much for the analytical approach to the solution. I thought the analytical approach would be very difficult, but it is easier than I expected. I will come back later to see updates to the post, God willing. $\endgroup$ Commented Dec 31, 2023 at 9:46

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