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本帖最后由 hbghlyj 于 2023-1-29 20:30 编辑 arimoto.lolipop.jp/Kerawala.pdf
1. We have two circles with radius $R$ and $r$, respectively : $S \equiv x^{2}+y^{2}-R^{2}=0$ and $s \equiv(x-d)^{2}+y^{2}-r^{2}=0$ in the x-y plane. Although Kerawala treated various cases in a general set up, we restrict here our study to the standard case where the large circle $S$ completely contains the small circle $s$ inside, that is we will assume $R-r>|d|$. If the chord joining the points $A_{i}\left(R \cos \theta_{i}, R \sin \theta_{i}\right)$ and $A_{i+1}\left(R \cos \theta_{i+1}, R \sin \theta_{i+1}\right)$ on the circle $S$ touches the circle $s$, we must have
(1.1) $R \cos \frac{1}{2}\left(\theta_{i+1}-\theta_{i}\right)-d \cos \frac{1}{2}\left(\theta_{i+1}+\theta_{i}\right)=r$,
which is easily found by drawing the picture. On writing $t_{i}=\tan \frac{\theta_{i}}{4}$, we have the following lemma.
Lemma 1.1
(1.2) $(R-d+r)\left(t_{i}+t_{i+1}\right)^{2}+(R+d+r)\left(1-t_{i} t_{i+1}\right)^{2}=2 R\left(1+t_{i} t_{i+1}\right)^{2}$
$i=0,1,2 \cdots$ |
Proof) Substituting
$$
\cos \left(\frac{\theta_{i+1}-\theta_{i}}{2}\right)=\frac{1-\tan ^{2} \frac{\theta_{i+1}-\theta_{i}}{4}}{1+\tan ^{2} \frac{\theta_{i+1}-\theta_{i}}{4}}=\frac{\left(1+t_{i} t_{i+1}\right)^{2}-\left(t_{i+1}-t_{i}\right)^{2}}{\left(1+t_{i} t_{i+1}\right)^{2}+\left(t_{i+1}-t_{i}\right)^{2}}
$$
$$
\cos \left(\frac{\theta_{i+1}+\theta_{i}}{2}\right)=\frac{1-\tan ^{2} \frac{\theta_{i+1}+\theta_{i}}{4}}{1+\tan ^{2} \frac{\theta_{i+1}+\theta_{i}}{4}}=\frac{\left(1-t_{i} t_{i+1}\right)^{2}-\left(t_{i+1}+t_{i}\right)^{2}}{\left(1-t_{i} t_{i+1}\right)^{2}+\left(t_{i+1}+t_{i}\right)^{2}}
$$
into (1.1), we have
$$
R\left\{\left(1+t_{i} t_{i+1}\right)^{2}-\left(t_{i+1}-t_{i}\right)^{2}\right\}-d\left\{\left(1-t_{i} t_{i+1}\right)^{2}-\left(t_{i+1}+t_{i}\right)^{2}\right\}
$$
$$
=r\left\{\left(1-t_{i} t_{i+1}\right)^{2}+\left(t_{i+1}+t_{i}\right)^{2}\right\}
$$
from which we have the desired result. ■
Now let $2 u^{2}=R-d+r, 2 v^{2}=R+d+r, w^{2}=R$. Then we rewrite (1.2) as
(1.3) $u^{2}\left(t_{i}+t_{i+1}\right)^{2}+v^{2}\left(1-t_{i} t_{i+1}\right)^{2}=w^{2}\left(1+t_{i} t_{i+1}\right)^{2}$
And similarly for $t_{i}$ and $t_{i-1}$ we have
$$
u^{2}\left(t_{i}+t_{i-1}\right)^{2}+v^{2}\left(1-t_{i} t_{i-1}\right)^{2}=w^{2}\left(1+t_{i} t_{i-1}\right)^{2}
$$
No loss of generality if we set $\theta_{0}=0$ or $t_{0}=0$, then we have
(1.4) $t_{1}^{2} u^{2}=w^{2}-v^{2}$
Hence considering the quadratic equation of the unknown $x:$.
(1.5) $u^{2}\left(t_{i}+x\right)^{2}+v^{2}\left(1-t_{i} x\right)^{2}=w^{2}\left(1+t_{i} x\right)^{2}$,
we have two roots $x=t_{i-1}$ and $x=t_{i+1}$. Since (1.5) becomes
$$
\begin{aligned}
&\left(u^{2}+\left(v^{2}-w^{2}\right) t_{i}^{2}\right) x^{2}+2 t_{i}\left(u^{2}-v^{2}-w^{2}\right) x+\left(u^{2} t_{i}^{2}+v^{2}-w^{2}\right), \\
=& u^{2}\left(1-t_{1}^{2} t_{i}^{2}\right) x^{2}+2 t_{i}\left(u^{2}-v^{2}-w^{2}\right) x+u^{2}\left(t_{i}^{2}-t_{1}^{2}\right)=0,
\end{aligned}
$$
we have by using relation between coefficients and zeros of the above equation,
(1.7) $t_{i-1}+t_{i+1}=\frac{-u^{2}+v^{2}+w^{2}}{u^{2}} \frac{2 t_{i}}{1-t_{i}^{2} t_{1}^{2}}$
and
(1.8) $t_{i-1} t_{i+1}=\frac{t_{i}^{2}-t_{1}^{2}}{1-t_{i}^{2} t_{1}^{2}}$
so that
(1.9) $\frac{t_{i-1}+t_{i+1}}{1-t_{i-1} t_{i+1}}=\frac{-u^{2}+v^{2}+w^{2}}{u^{2}-v^{2}+w^{2}} \frac{2 t_{i}}{1-t_{i}^{2}}$.
We will notice here the relation between $t_{i-1}$ and $t_{i+1}$ is similar to the relation (1.3) between $t_{i}$ and $t_{i+1}$. In fact, we can get the exactly same form as (1.3) except for the coefficients $a^{2}, b^{2}, c^{2}$ instead of $u^{2}, v^{2}, w^{2}$.
Lemma 1.2
(1.10) $a^{2}\left(t_{i-1}+t_{i+1}\right)^{2}+b^{2}\left(1-t_{i-1} t_{i+1}\right)^{2}=c^{2}\left(1+t_{i-1} t_{i+1}\right)^{2}$,
where $a^{-1}=R+d, \quad b^{-1}=R-d, \quad c^{-1}=r, \quad i=0,1,2 \cdots$ |
By (1.4), $t_{1}^{2}=\frac{w^{2}-v^{2}}{u^{2}}, 1+t_{1}^{2}=\frac{u^{2}+w^{2}-v^{2}}{u^{2}}=\frac{2(R-d)}{R-d+r}$ and
$1-t_{1}^{2}=\frac{u^{2}-w^{2}+v^{2}}{u^{2}}=\frac{2 r}{R-d+r}$. We have from (1.7)
$t_{i-1}+t_{i+1}=\frac{-u^{2}+v^{2}+w^{2}}{u^{2}} \frac{2 t_{i}}{1-t_{i}^{2} t_{1}^{2}}=\frac{2(R+d)}{R-d+r} \frac{2 t_{i}}{1-t_{i}^{2} t_{1}^{2}}$. From (1.8)
$1-t_{i-1} t_{i+1}=1-\frac{t_{i}^{2}-t_{1}^{2}}{1-t_{i}^{2} t_{1}^{2}}=\frac{\left(1-t_{i}^{2}\right)\left(1+t_{1}^{2}\right)}{1-t_{i}^{2} t_{1}^{2}}$ and $1+t_{i-1} t_{i+1}=\frac{\left(1+t_{i}^{2}\right)\left(1-t_{1}^{2}\right)}{1-t_{i}^{2} t_{1}^{2}}$.
Hence
$$
a^{2}\left(t_{i-1}+t_{i+1}\right)^{2}+b^{2}\left(1-t_{i-1} t_{i+1}\right)^{2}-c^{2}\left(1+t_{i-1} t_{i+1}\right)^{2}
$$
$$
=\left(\frac{1}{R+d}\right)^{2} \frac{4(R+d)^{2} \times 4 t_{i}^{2}}{(R-d+r)^{2}\left(1-t_{i}^{2} t_{1}^{2}\right)^{2}}+\left(\frac{1}{R-d}\right)^{2} \frac{\left(1-t_{i}^{2}\right)^{2}}{\left(1-t_{i}^{2} t_{1}^{2}\right)^{2}} \frac{4(R-d)^{2}}{(R-d+r)^{2}}
$$
$$
-\frac{1}{r^{2}} \frac{\left(1+t_{i}^{2}\right)^{2}}{\left(1-t_{i}^{2} t_{1}^{2}\right)^{2}} \frac{4 r^{2}}{(R-d+r)^{2}}=0\quad■
$$
From (1.10) we have counterparts of (1.7)-(1.9):
(1.11) $t_{i-2}+t_{i+2}=\frac{-a^{2}+b^{2}+c^{2}}{a^{2}} \frac{2 t_{i}}{1-t_{i}^{2} t_{2}^{2}}$
(1.12) $t_{i-2} t_{i+2}=\frac{t_{i}^{2}-t_{2}^{2}}{1-t_{i}^{2} t_{2}^{2}}$
(1.13) $\frac{t_{i-2}+t_{i+2}}{1-t_{i-2} t_{i+2}}=\frac{-a^{2}+b^{2}+c^{2}}{a^{2}-b^{2}+c^{2}} \frac{2 t_{i}}{1-t_{i}^{2}}$.
[1] Kerawala, S.M. Poncelet Porism in Two Circles, Bull.Calcutta Math.Soc.39,(1947) 85-105 |
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