Casey's Theorem and its Applications Luis Gonz´alez Maracaibo. Venezuela July 2011 Abstract. We present a proof of the generalized Ptolemy's theorem, also known as Casey's theorem and its applications in the resolution of difficult geometry problems. 1 Casey's Theorem. Theorem 1. Two circles Γ1(r1) and Γ2(r2) are internally/externally tangent to a circle Γ(R) through A; B; respetively. The length δ12 of the common external tangent of Γ1; Γ2 is given by: AB q δ = (R ± r )(R ± r ) 12 R 1 2 Proof. Without loss of generality assume that r1 ≥ r2 and we suppose that Γ1 and Γ2 are internally tangent to Γ. The remaining case will be treated analogously. A common external tangent between Γ1 and Γ2 touches Γ1; Γ2 at A1;B1 and A2 is the orthogonal projection of O2 onto O1A1. (See Figure 1). By Pythagorean theorem for 4O1O2A2; we obtain 2 2 2 2 δ12 = (A1B1) = (O1O2) − (r1 − r2) Let 6 O1OO2 = λ. By cosine law for 4OO1O2; we get 2 2 2 (O1O2) = (R − r1) + (R − r2) − 2(R − r1)(R − r2) cos λ By cosine law for the isosceles triangle 4OAB; we get AB2 = 2R2(1 − cos λ) 1 Figure 1: Theorem 1 Eliminating cos λ and O1O2 from the three previous expressions yields AB2 ! δ 2 = (R − r )2 + (R − r )2 − (r − r )2 − 2(R − r )(R − r ) 1 − 12 1 2 1 2 1 2 2R2 Subsequent simplifications give AB q δ = (R − r )(R − r ) (1) 12 R 1 2 Analogously, if Γ1; Γ2 are externally tangent to Γ; then we will get AB q δ = (R + r )(R + r ) (2) 12 R 1 2 If Γ1 is externally tangent to Γ and Γ2 is internally tangent to Γ; then a similar reasoning gives that the length of the common internal tangent between Γ1 and Γ2 is given by AB q δ = (R + r )(R − r ) (3) 12 R 1 2 2 Theorem 2 (Casey). Given four circles Γi; i = 1; 2; 3; 4; let δij denote the length of a common tangent (either internal or external) between Γi and Γj. The four circles are tangent to a fith circle Γ (or line) if and only if for appropriate choice of signs, δ12 · δ34 ± δ13 · δ42 ± δ14 · δ23 = 0 The proof of the direct theorem is straightforward using Ptolemy's theorem for the quadrilateral ABCD whose vertices are the tangency points of Γ1(r1); Γ2(r2); Γ3(r3); Γ4(r4) with Γ(R). We susbtitute the lengths of its sides and digonals in terms of the lenghts of the tangents δij; by using the formulas (1); (2) and (3). For instance, assuming that all tangencies are external, then using (1), we get AB·CD+AD·BC q δ12 · δ34 + δ14 · δ23 = R2 (R − r1)(R − r2)(R − r3)(R − r4) AC·BD q q δ12 · δ34 + δ14 · δ23 = R2 (R − r1)(R − r3) · (R − r2)(R − r4) δ12 · δ34 + δ14 · δ23 = δ13 · δ42: Casey established that this latter relation is sufficient condition for the existence of a fith circle Γ(R) tan- gent to Γ1(r1); Γ2(r2); Γ3(r3); Γ4(r4): Interestingly, the proof of this converse is a much tougher exercise. For a proof you may see [1]. 2 Some Applications. I) 4ABC is isosceles with legs AB = AC = L: A circle ! is tangent to BC and the arc BC of the circumcircle of 4ABC: A tangent line from A to ! touches ! at P: Describe the locus of P as ! varies. Solution. We use Casey's theorem for the circles (A); (B); (C) (with zero radii) and !; all internally tangent to the circumcircle of 4ABC: Thus, if ! touches BC at Q; we have: L(BQ + CQ) L · CQ + L · BQ = AP · BC =) AP = = L BC The length AP is constant, i.e. Locus of P is the circle with center A and radius AB = AC = L: II) (O) is a circle with diameter AB and P; Q are two points on (O) lying on different sides of AB: T is the orthogonal projection of Q onto AB: Let (O1); (O2) be the circles with diameters T A; T B and P C; P D are the tangent segments from P to (O1); (O2); respectively. Show that PC + PD = P Q: [2]. 3 Figure 2: Application II Solution. Let δ12 denote the length of the common external tangent of (O1); (O2). We use Casey's theorem for the circles (O1); (O2); (P ); (Q); all internally tangent to (O): p δ TA · TB PC · QT + PD · QT = PQ · δ =) PC + PD = PQ · 12 = PQ · = P Q: 12 QT TQ III) In 4ABC; let !A;!B;!C be the circles tangent to BC; CA; AB through their midpoints and the arcs BC; CA; AB of its circumcircle (not containing A; B; C). If δBC ; δCA; δAB de- note the lengths of the common external tangents between (!B;!C ); (!C ;!A) and (!A;!B); respectively, then prove that a + b + c δ = δ = δ = BC CA AB 4 Solution. Let δA; δB; δC denote the lengths of the tangents from A; B; C to !A;!B;!C ; respectively. By Casey's theorem for the circles (A); (B); (C);!B; all tangent to the circumcircle of 4ABC; we get 1 δB · b = a · AE + c · CE =) δB = 2 (a + c) 1 Similarly, by Casey's theorem for (A); (B); (C);!C we'll get δC = 2 (a + b) 4 Now, by Casey's theorem for (B); (C);!B;!C ; we get δB · δC = δBC · a + BF · BE =) δ · δ − BF · BE (a + c)(a + b) − bc a + b + c δ = B C = = BC a 4a 4 1 By similar reasoning, we'll have δCA = δAB = 4 (a + b + c): IV) A circle K passes through the vertices B; C of 4ABC and another circle ! touches AB; AC; K at P; Q; T; respectively. If M is the midpoint of the arc BT C of K; show that BC; P Q; MT concur. [3] Solution. Let R; % be the radii of K and !; respectively. Using formula (1) of Theorem 1 for !; (B) and !; (C): Both (B); (C) with zero radii and tangent to K through B; C; we obtain: CQ2 · R2 CQ2 · R BP 2 · R2 BP 2 · R TB BP TC2 = = ;TB2 = = =) = (R − %)(R − 0) R − % (R − %)(R − 0) R − % TC CQ Let PQ cut BC at U: By Menelaus' theorem for 4ABC cut by UPQ we have UB BP AQ BP TB = · = = UC AP CQ CQ TC Thus, by angle bisector theorem, U is the foot of the T-external bisector TM of 4BT C: V) If D; E; F denote the midpoints of the sides BC; CA; AB of 4ABC: Show that the incircle (I) of 4ABC is tangent to (DEF ): (Feuerbach theorem). Solution. We consider the circles (D); (E); (F ) with zero radii and (I): The notation δXY stands for the length of the external tangent between the circles (X); (Y ); then c a b b − c a − c b − a δDE = ; δEF = ; δFD = ; δDI = ; δEI = ; δFI = 2 2 2 2 2 2 For the sake of applying the converse of Casey's theorem, we shall verify if, for some combination of signs + and −; we get ±c(b−a)±a(b−c)±b(a−c) = 0; which is trivial. Therefore, there exists a circle tangent to (D); (E); (F ) and (I); i.e. (I) is internally tangent to (DEF ): We use the same reasoning to show that (DEF ) is tangent to the three excircles of 4ABC: VI) 4ABC is scalene and D; E; F are the midpoints of BC; CA; AB: The incircle (I) and 9 point circle (DEF ) of 4ABC are internally tangent through the Feuerbach point Fe: Show that one of the segments FeD; FeE; FeF equals the sum of the other two. [4] 5 Solution. WLOG assume that b ≥ a ≥ c: Incircle (I; r) touches BC at M: Using formula (1) of Theorem R 1 for (I) and (D) (with zero radius) tangent to the 9-point circle (N; 2 ); we have: 2 R 2 s 2 DM · ( 2 ) R (b − c) FeD = R R =) FeD = · ( 2 − r)( 2 − 0) R − 2r 2 By similar reasoning, we have the expressions s s R (a − c) R (b − a) F E = · ;F F = · e R − 2r 2 e R − 2r 2 Therefore, the addition of the latter expressions gives s R b − c F E + F F = · = F D e e R − 2r 2 e VII) 4ABC is a triangle with AC > AB: A circle !A is internally tangent to its circumcircle ! and AB; AC: S is the midpoint of the arc BC of !; which does not contain A and ST is the tangent segment from S to !A: Prove that ST AC − AB = [5] SA AC + AB Solution. Let M; N be the tangency points of !A with AC; AB: By Casey's theorem for !A; (B); (C); (S); all tangent to the circumcircle !; we get ST · BC + CS · BN = CM · BS =) ST · BC = CS(CM − BN) If U is the reflection of B across AS; then CM − BN = UC = AC − AB: Hence ST · BC = CS(AC − AB)(?) By Ptolemy's theorem for ABSC; we get SA · BC = CS(AB + AC): Together with (?); we obtain ST AC − AB = SA AC + AB 6 VIII) Two congruent circles (S1); (S2) meet at two points. A line ` cuts (S2) at A; C and (S1) at B; D (A; B; C; D are collinear in this order). Two distinct circles !1;!2 touch the line ` and the circles (S1); (S2) externally and internally respectively.