93. The Appolonius' construct problems.

by Virgil Nicula, Aug 28, 2010, 7:57 PM

Lemma P.P.L. Given are two points $ A$ , $ B$ and a line $ d$ such that $ \{A,B\}\cap d = \emptyset$ , the line $ d$ doesn't

separate $ A$ , $ B$ and $ AB\cap d\ne \emptyset$ . Construct the circles which are tangent to $ d$ and pass through $ A$ , $ B$ .


Proof. Denote $ P\in AB\cap d$ and suppose w.l.o.g. $ A\in (PB)$ . Define the tangent point $ T\in d$ of the required circle $ w = C(O,R)$ .

Observe that $ O$ belongs to the bisector of the segment $ [AB]$ and $ OT\perp d$ . Consider the circle $ \rho$ with diameter $ [PB]$ and $ C\in\rho$

for which $ CA\perp AB$ . Observe that $ PT^2 = PA\cdot PB = PC^2$ , i.e. $ PT = PC$ . Thus, $ T$ belongs to the circle with center $ P$

and radius $ [PC]$ . Appear just two solutions. Now the construction of the required circles is easily.



Lemma P.P.C. Given are two points $ A$ , $ B$ and a circle $ w$ such that $ AB\cap w = \emptyset$ . Construct the circles which are tangent to $ w$ and $ \{A,B\}\subset w$ .

Proof. Construct a circle $ x$ which pass through $ A$ , $ B$ and is secant to $ w$ (in the points $ C$ , $ D$ ). Construct from $ E\in CD\cap AB$

the tangent lines (in the points $ T_1$ , $ T_2$ ) to the circle $ w$ . Then the required circles are the circumcircles of $ \triangle ABT_1$ , $ ABT_2$ .



Lemma P.L.C. Given are a point $ A$ , a line $ d$ and a circle $ w = C(O,R)$ such that $ AO > R$ , $ \delta_d(O) > R$ and

the line $ d$ doesn't separate $ O$ , $ A$ . Construct the circles which are tangent to $ w$ , $ d$ and which pass through $ A$ .


Proof. Denote the diameter $ [BC]$ of $ w$ for which $ BC\perp d$ and $ D\in BC\cap d$ . Suppose w.l.o.g. $ B\in (CD)$ . Define the tangent points $ X$ , $ Y$

of the required circle $ x$ with $ w$ , $ d$ respectively. Show easily that $ X\in CY$ and the quadrilateral $ DBXY$ is cyclically. Denote the second

intersection $ Z$ of $ AC$ with circle $ x$ . Observe that $ CA\cdot CZ = CX\cdot CY = CB\cdot CD$ , i.e. $ Z$ belongs to the line $ AC$ and to the circumcircle

of the triangle $ ABD$ . Now the construction of a circle $ x$ becomes the construction of a circle which pass through two given points $ A$ , $ Z$

and which is tangent to a given line $ d$ (see lemma P.P.L.).



Lemma P.C.C. Given are a point $ A$ , and two circles $ w_1 = C(O_1,R_1)$ and $ w_2 = C(O_2,R_2)$ such that $ AO_1 > R_1$ ,

$ AO_2 > R_2$ and $ O_1O_2 > R_1 + R_2$ . Construct the circles which pass through $ A$ and which are exterior tangent to $ w_1$ , $ w_2$ .


Proof. Construct the intersection of $ O_1O_2$ with a common exterior tangent $ T_1T_2$ of $ w_1$ , $ w_2$ , where $ T_1\in w_1$ , $ T_2\in w_2$ .

The required circle $ x$ is exterior tangent to $ w_1$ , $ w_2$ in $ X$ , $ Y$ respectively. Show easily that $ S\in XY$ and the quadrilateral $ XYT_1T_2$

is cyclically. Denote the second intersection $ Z$ of $ x$ with $ AS$ . Therefore, $ SZ\cdot SA = SX\cdot SY = ST_1\cdot ST_2$ , i.e. the point $ Z\in AS$

and to the circumcircle of $ \triangle AT_1T_2$ . Now the construction of a circle $ x$ becomes the construction of a circle which pass through

two given points $ A$ , $ Z$ and which is tangent to a given circle $ w_1$ (or $ w_2$ ) (see lemma P.P.C.).



Lemma L.C.C. Given are a line $ d$ and two circles $ w_1 = C(O_1,R_1)$ , $ w_2 = C(O_2,R_2)$ such that the line $ d$

doesn't separate $ O_1$ , $ O_2$ and $ w_1\cap d = w_2\cap d = \emptyset$ . Construct the circles which are exterior tangent to $ w_1$ , $ w_2$ and $ d$ .


Proof. Suppose w.l.o.g. that $ R_1\ge R_2$ . Consider the circle $ w_1'$ with center $ O_1$ and radius $ R_1 - R_2$ . Construct the line $ d'\parallel d$ for which

the distance between $ d$ , $ d'$ is equally to $ R_2$ and the line $ d$ "separates" the circles $ w_1$ , $ w_2$ from the line $ d'$ . Construct the circle $ x' = C(O,R)$

which pass through $ O_2$ and which is tangent to $ w_1'$ and $ d'$ (see lemma P.L.C.). Then the required circle is $ x = C(O,R - R_2)$ .



Remark. The construction of the circles which are tangent (exterior or interior) to three given circles, being possibly degenerated

to a line ($ R\rightarrow\infty$ ) or a point ($ R = 0$ ) is named the Appolonius' construct problem. Using the symbols P (point), L (line), C (circle)

there are ten Appolonius' construct problems : P.P.P. ; P.P.L. ; P.P.C. ; P.D.D. ; P.D.C. ; P.C.C. ; D.D.D. ; D.D.C. ; D.C.C. ; C.C.C.

In conclusion, the solving of this remarkable construct problem can illustrate on the following sketch of "reduction to previous problem" :

L.L.C. $ \implies$ P.L.L. $ \implies$ P.P.L. $ \implies$ P.P.P. ; L.C.C. $ \implies$ P.L.C. $ \implies$ P.P.L. ; C.C.C. $ \implies$ P.C.C. $ \implies$ P.P.C. $ \implies$ P.P.P.
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:22 PM

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