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I'm interested in the following problem:

$S$ and $S'$ are fixed points, and $L$ is a fixed line.
For every point $P$ in the plane, let the line $S'P$ intersect the line $L$ at $Z$, and let $P'$ be the intersection of the line through $S'$ parallel to $SP$ and the line $SZ$.
Let $SP$ intersect $L$ at $A$, and let $Q$ be the inversion of $P$ with respect to the circle with center $A$ and passing through $S$.
Translate point $P'$ by the vector $\vec{S'S}$ to point $P''$.
Show that the ratio $QS:SP''$ is constant.

To prove $QS:SP''$ is constant, I only need to prove the intersection of the line $SS'$ and $P'Q$ is a fixed point (so that the ratio $QS:SP''$ will be equal to the ratio of the fixed point dividing $SS'$.)
So I need to prove the following:

$S$ and $S'$ are fixed points, and $L$ is a fixed line.
For every point $P$ in the plane, let the line $S'P$ intersect the line $L$ at $Z$, and let $P'$ be the intersection of the line through $S'$ parallel to $SP$ and the line $SZ$.
Let $SP$ intersect $L$ at $A$, and let $Q$ be the inversion of $P$ with respect to the circle with center $A$ and passing through $S$.
Show that the intersection of the line $SS'$ and $P'Q$ is a fixed point.

I observe the fixed point is on the line $L$. So I only need to prove the intersection of the line $SS'$ and $P'Q$ is on $L$, then it will be a fixed point.
But I don't know how to prove that the intersection of the line $SS'$ and $P'Q$ is on $L$.
By the definition of inversion, I have $SA^2=PA\cdot QA$.

hbghlyj
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1 Answers1

1

Ok, I have worked it out.

I observe the fixed point is on the line $L$. So I only need to prove the intersection of the line $SS'$ and $P'Q$ is on $L$, then it will be a fixed point.
To prove that the intersection of the line $SS'$ and $P'Q$ is on $L$, I need to prove that $P'Q$ and $L$ intersect the line $SS'$ at the same point, so I need to prove the ratio that these two points divide $SS'$ are equal, by Menelaus' theorem, I need to prove: $$\frac{S'Z}{ZP}\frac{PA}{AS}\frac{SQ}{S'P'}=1$$ using $\frac{S'Z}{ZP}=\frac{S'P'}{SP}$ the above becomes, after canceling $S'P'$, $$\frac{SQ}{SP}\frac{PA}{AS}=1$$ This follows from the definition of inversion. QED.

hbghlyj
  • 2,115