Casey theorem on five circles (generalized Ptolemy theorem)

alekk

Notations. A circle $k$ with center $U$ and radius $\rho$ will be abbreviated as "circle $k\left(U,\rho\right)$ ".

For any two circles $a$ and $b$ , we define a "distance" $d_{ab}$ between these two circles as follows: Let one of the two external common tangents of the circles $a$ and $b$ (it doesn't matter which one we take) touch these circles at two points $P$ and $Q$ ; then, we define $d_{ab}$ as $d_{ab}=PQ$ . Note that this distance $d_{ab}$ is thus only defined if the circles $a$ and $b$ have external common tangents.

Casey Theorem. Let five circles $C\left(O,r\right)$ , $C_{1}\left(O_{1},r_{1}\right)$ , $C_{2}\left(O_{2},r_{2}\right)$ , $C_{3}\left(O_{3},r_{3}\right)$ , $C_{4}\left(O_{4},r_{4}\right)$ be given. If the circles $C_{1}$ , $C_{2}$ , $C_{3}$ , $C_{4}$ are all interiorly tangent to the circle $C$ , and the points where they touch the circle $C$ are arranged in the same order as we have named them, then the following relation holds:

$d_{C_{1}C_{2}}\cdot d_{C_{3}C_{4}}+d_{C_{2}C_{3}}\cdot d_{C_{4}C_{1}}=d_{C_{1}C_{3}}\cdot d_{C_{2}C_{4}}$ .

Remark. The same result holds if the circles $C_{1}$ , $C_{2}$ , $C_{3}$ , $C_{4}$ are all exteriorly tangent to the circle $C$ .

**Posted:** Tue Aug 19, 2003 11:50 pm

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Valentin Vornicu

the proof uses Ptolemy's Theorem - since it is a generalization of it Smile

**Posted:** Wed Aug 20, 2003 1:25 am

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Lagrangia

Proof: Let's denote T1,T2,T3 and T4 the points of tangentcy of the circles C1, C2, C3 and C4 with C and we are going to compute d_CiCj the tangent distances between the circles Ci, Cj using the radius of the circles ri,rj, and the radius r of the circle C(O,r) and the distance TiTj between the points of tangentcy with C(O,r)

Applying the cosinus theorem in the triangles: TiOTj and OiOOj we get:

cos(TiOTj)=(2r^2-TiTj^2)/2r^2 and
OiOj^2=(r-ri)^2+(r-rj)^2-2(r-ri)(r-rj)cos(TiOTj)=(r-ri)^2+ (r-rj)^2-2(r-ri)(r-rj)*(2r^2-TiTj^2)/2r^2=(r-ri)^2+(r-rj)^2-2(r-ri)(r-rj)+(r-ri)(r-rj)*TiTj^2/r^2=(ri-rj)^2+(r-ri)(r-rj)*TiTj^2/r^2.

Let AB be the common exterior tangent of the circles: C(Oi,ri) and C(Oj,rj) so d_CiCj=AB.
From ABOjOi ( which is a right angled trapezoid) we get:

d_CiCj^2=AB^2=OiOj^2-(OiA-OjB)^2=(ri-rj)^2+(r-ri)(r-rj)TiTj^2/r^2-(ri-rj)^2=(r-ri)(r-rj)*TiTj^2/r^2 =>
d_CiCj=TiTj/r * sqrt((r-ri)(r-rj)).
The relation from the theorem becomes:
T1T2/r*sqrt((r-r1)(r-r2))*T3T4/r*sqrt((r-r3)(r-r4))+T2T3/r*sqrt((r-r2)(r-r3))*T4T1/r*sqrt((r-r4)(r-r1))=
T1T3/r*sqrt((r-r1)(r-r3))*T2T4/r*sqrt((r-r2)(r-r4)) which is T1T2*T3T4+T2T3*T4T1=T1T3*T2T4 which is Ptolemy thorem.. so Casey's theorem is proved!!!! Smile

cheers!

**Posted:** Wed Aug 20, 2003 4:23 am

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Is it true that if $d_{C_{1}C_{2}}\cdot d_{C_{3}C_{4}}+d_{C_{2}C_{3}}\cdot d_{C_{4}C_{1}}=d_{C_{1}C_{3}}\cdot d_{C_{2}C_{4}}$ for some circles $C_1(O_1,r_1),C_2(O_2,r_2),C_3(O_3,r_3),C_4(O_4,r_4)$ then there exists an circle $C(O,r)$ tangent to all these circles Rolling Eyes

**Posted:** Tue Mar 17, 2009 12:56 am