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$A$ and $C$ are fixed points on fixed circles $O_1$ and $O_2$.
Point $B$ is moving on circle $O_1$.
Point $D$ is the intersection of the circle through $A,B,C$ with circle $O_2$.
Point $F,G$ are the intersection of line $BD$ with circles $O_1$ and $O_2$.
Point $P$ is on the line $BD$ such that $$\tag1\label1 \overrightarrow{PF}⋅\overrightarrow{PG}=\overrightarrow{PB}⋅\overrightarrow{PD}. $$ Show that the trajectory of $P$ is a hyperbola, which is tangent to the moving line $BD$ and the circles $O_1, O_2$.

hbghlyj
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  • What does this mean: "$\overrightarrow{PF}⋅\overrightarrow{PG}=\overrightarrow{PB}⋅\overrightarrow{PD}$"? If you had not put the arrows on, then I would interpret it as products of distances. But with the arrows, these are rays. What "dot" operation between rays are you referring to? Especially since each side is a ray "dotted" with itself. – Paul Sinclair Oct 30 '23 at 22:32
  • @PaulSinclair They are not rays, they are vectors. I use dot product of vectors rather than product of distances to ensure $P$ is unique. – hbghlyj Oct 30 '23 at 23:27
  • @PaulSinclair For example, on the 1D real line, let $P,F,G,B,D$ be the point $x,1,5,2,3$, then $\overrightarrow{PF}⋅\overrightarrow{PG}=\overrightarrow{PB}⋅\overrightarrow{PD}$ means$$(x-1)(x-5)=(x-2)(x-3)$$Since the quadratic terms $x^2$ cancel, this is a linear equation, so it has a unique solution $x=-1$. But if I write the equation in distances $PF⋅PG=PB⋅PD$, this means$$|x-1||x-5|=|x-2||x-3|$$then $P$ is not unique, because $(x-1)(x-5)=-(x-2)(x-3)$ has two solutions $x =\frac14 (11 - \sqrt{33})$. Altogether there are 3 points $P$ that satisfies $PF⋅PG=PB⋅PD$. – hbghlyj Oct 30 '23 at 23:32
  • But since all the vectors are on the same line, the angles between them are $0$, and thus the inner products are simply the products of the distances. It would be more clearly stated as just $PF⋅PG=PB⋅PD$, instead of switching to a different language (vectors) instead of the geometric language of the rest of the post. – Paul Sinclair Oct 31 '23 at 03:37
  • @PaulSinclair The point $P$ defined by inner products is unique, defined by distances is not unique. In my example, $(x−1)(x−5)=(x−2)(x−3)$ has a unique solution $x$, but $|x−1||x−5|=|x−2||x−3|$ has two more solutions. – hbghlyj Oct 31 '23 at 08:04
  • @PaulSinclair yes all the vectors are on the same line, the angles between them are $0$ or $\pi$, and $\cos0=1,\cos\pi=-1$, so the inner product $\overrightarrow{PF}⋅\overrightarrow{PG}$ may be positive or negative the products of the distances – hbghlyj Oct 31 '23 at 10:40

1 Answers1

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This is a partial solution: I showed that the trajectory of $P$ is tangent to the moving line $BD$. But I don't know how to show the trajectory of $P$ is a hyperbola and is tangent to circles $O_1,O_2$.

Make a copy $B',D'$ of $B,D$. Define $P$ be the intersection of $BB'$ and $DD'$. We will show the limit of $P$ is the same point as the point $P$ defined by equation $(1)$.

By this problem we have $$\frac{FF'}{BB'}=\frac{DD'}{GG'}$$ so $$\frac{PF}{PB'}=\frac{PD'}{PG}$$ As $B'\to B,D'\to D$, we get $$\frac{PF}{PB}=\frac{PD}{PG}$$ So this point $P$ is a solution of the equation $(1)$. But there is a unique point on line $BD$ that satisfies the equation $(1)$, since the equation $(1)$ is equivalent to $\frac{PF}{PB}=\frac{FD}{BG}$.

PS: As in my comment, if using unsigned distance, the equation $(1)$ have multiple solutions on line $BD$.

hbghlyj
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