Casey s Theorem and its Applications

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1 Casey s Theorem and its Applications Luis González Maracaibo. Venezuela July 011 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 Γ 1 (r 1 ) and Γ (r ) are internally/externally tangent to a circle Γ(R) through A, B, respetively. The length δ 1 of the common external tangent of Γ 1, Γ is given by: δ 1 = AB R (R ± r 1 )(R ± r ) Proof. Without loss of generality assume that r 1 r and we suppose that Γ 1 and Γ are internally tangent to Γ. The remaining case will be treated analogously. A common external tangent between Γ 1 and Γ touches Γ 1, Γ at A 1, B 1 and A is the orthogonal projection of O onto O 1 A 1. (See Figure 1). By Pythagorean theorem for O 1 O A, we obtain δ 1 = (A 1 B 1 ) = (O 1 O ) (r 1 r ) Let O 1 OO = λ. By cosine law for OO 1 O, we get (O 1 O ) = (R r 1 ) + (R r ) (R r 1 )(R r ) cos λ By cosine law for the isosceles triangle OAB, we get AB = R (1 cos λ) 1

2 Figure 1: Theorem 1 Eliminating cos λ and O 1 O from the three previous expressions yields δ 1 = (R r 1 ) + (R r ) (r 1 r ) (R r 1 )(R r ) ( 1 AB R ) Subsequent simplifications give δ 1 = AB R (R r 1 )(R r ) (1) Analogously, if Γ 1, Γ are externally tangent to Γ, then we will get δ 1 = AB R (R + r 1 )(R + r ) () If Γ 1 is externally tangent to Γ and Γ is internally tangent to Γ, then a similar reasoning gives that the length of the common internal tangent between Γ 1 and Γ is given by δ 1 = AB R (R + r 1 )(R r ) (3)

3 Theorem (Casey). Given four circles Γ i, i = 1,, 3, 4, let δ ij denote the length of a common tangent (either internal or external) between Γ i and Γ j. The four circles are tangent to a fith circle Γ (or line) if and only if for appropriate choice of signs, δ 1 δ 34 ± δ 13 δ 4 ± δ 14 δ 3 = 0 The proof of the direct theorem is straightforward using Ptolemy s theorem for the quadrilateral ABCD whose vertices are the tangency points of Γ 1 (r 1 ), Γ (r ), Γ 3 (r 3 ), Γ 4 (r 4 ) with Γ(R). We susbtitute the lengths of its sides and digonals in terms of the lenghts of the tangents δ ij, by using the formulas (1), () and (3). For instance, assuming that all tangencies are external, then using (1), we get δ 1 δ 34 + δ 14 δ 3 = ( AB CD+AD BC R ) (R r 1 )(R r )(R r 3 )(R r 4 ) δ 1 δ 34 + δ 14 δ 3 = ( ) AC BD (R r R 1 )(R r 3 ) (R r )(R r 4 ) δ 1 δ 34 + δ 14 δ 3 = δ 13 δ 4. Casey established that this latter relation is sufficient condition for the existence of a fith circle Γ(R) tangent to Γ 1 (r 1 ), Γ (r ), Γ 3 (r 3 ), Γ 4 (r 4 ). Interestingly, the proof of this converse is a much tougher exercise. For a proof you may see [1]. Some Applications. I) ABC is isosceles with legs AB = AC = L. A circle ω is tangent to BC and the arc BC of the circumcircle of ABC. A tangent line from A to ω touches ω at P. Describe the locus of P as ω varies. Solution. We use Casey s theorem for the circles (A), (B), (C) (with zero radii) and ω, all internally tangent to the circumcircle of ABC. Thus, if ω touches BC at Q, we have: L CQ + L BQ = AP BC = AP = L(BQ + CQ) BC The length AP is constant, i.e. Locus of P is the circle with center A and radius AB = AC = L. = 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 (O 1 ), (O ) be the circles with diameters T A, T B and P C, P D are the tangent segments from P to (O 1 ), (O ), respectively. Show that P C + P D = P Q. []. 3

4 Figure : Application II Solution. Let δ 1 denote the length of the common external tangent of (O 1 ), (O ). We use Casey s theorem for the circles (O 1 ), (O ), (P ), (Q), all internally tangent to (O). P C QT + P D QT = P Q δ 1 = P C + P D = P Q δ1 QT = P Q T A T B T Q = P Q. III) In ABC, let ω A, ω B, ω 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 δ BC, δ CA, δ AB denote the lengths of the common external tangents between (ω B, ω C ), (ω C, ω A ) and (ω A, ω B ), respectively, then prove that δ BC = δ CA = δ AB = a + b + c 4 Solution. Let δ A, δ B, δ C denote the lengths of the tangents from A, B, C to ω A, ω B, ω C, respectively. By Casey s theorem for the circles (A), (B), (C), ω B, all tangent to the circumcircle of ABC, we get δ B b = a AE + c CE = δ B = 1 (a + c) Similarly, by Casey s theorem for (A), (B), (C), ω C we ll get δ C = 1 (a + b) 4

5 Now, by Casey s theorem for (B), (C), ω B, ω C, we get δ B δ C = δ BC a + BF BE = δ BC = δ B δ C BF BE a = (a + c)(a + b) bc 4a = a + b + c 4 By similar reasoning, we ll have δ CA = δ AB = 1 (a + b + c). 4 IV) A circle K passes through the vertices B, C of ABC and another circle ω 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, ϱ be the radii of K and ω, respectively. Using formula (1) of Theorem 1 for ω, (B) and ω, (C). Both (B), (C) with zero radii and tangent to K through B, C, we obtain: T C = CQ R (R ϱ)(r 0) = CQ R R ϱ, T B = BP R (R ϱ)(r 0) = BP R R ϱ = T B T C = BP CQ Let P Q cut BC at U. By Menelaus theorem for ABC cut by UP Q we have UB UC = BP AP AQ CQ = BP CQ = T B T C Thus, by angle bisector theorem, U is the foot of the T-external bisector T M of BT C. V) If D, E, F denote the midpoints of the sides BC, CA, AB of ABC. Show that the incircle (I) of ABC is tangent to (DEF ). (Feuerbach theorem). Solution. We consider the circles (D), (E), (F ) with zero radii and (I). The notation δ XY stands for the length of the external tangent between the circles (X), (Y ), then δ DE = c, δ EF = a, δ F D = b, δ b c DI =, δ a c EI =, δ b a F I = 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 ABC. VI) ABC is scalene and D, E, F are the midpoints of BC, CA, AB. The incircle (I) and 9 point circle (DEF ) of ABC are internally tangent through the Feuerbach point F e. Show that one of the segments F e D, F e E, F e F equals the sum of the other two. [4] 5

6 Solution. WLOG assume that b a c. Incircle (I, r) touches BC at M. Using formula (1) of Theorem 1 for (I) and (D) (with zero radius) tangent to the 9-point circle (N, R ), we have: F e D = DM ( R ) R ( R r)( R 0) = F (b c) ed = R r By similar reasoning, we have the expressions R (a c) R (b a) F e E =, F e F = R r R r Therefore, the addition of the latter expressions gives R F e E + F e F = R r b c = F e D VII) ABC is a triangle with AC > AB. A circle ω A is internally tangent to its circumcircle ω and AB, AC. S is the midpoint of the arc BC of ω, which does not contain A and ST is the tangent segment from S to ω A. Prove that ST SA = AC AB AC + AB [5] Solution. Let M, N be the tangency points of ω A with AC, AB. By Casey s theorem for ω A, (B), (C), (S), all tangent to the circumcircle ω, 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 SA = AC AB AC + AB 6

7 VIII) Two congruent circles (S 1 ), (S ) meet at two points. A line l cuts (S ) at A, C and (S 1 ) at B, D (A, B, C, D are collinear in this order). Two distinct circles ω 1, ω touch the line l and the circles (S 1 ), (S ) externally and internally respectively. If ω 1, ω are externally tangent, show that AB = CD. [6] Solution. Let P ω 1 ω and M, N be the tangency points of ω 1 and ω with an external tangent. Inversion with center P and power P B P D takes (S 1 ) and the line l into themselves. The circles ω 1 and ω go to two parallel lines k 1 and k tangent to (S 1 ) and the circle (S ) goes to another circle (S ) tangent to k 1, k. Hence, (S ) is congruent to its inverse (S ). Further, (S ), (S ) are symmetrical about P = P C P A = P B P D. By Casey s theorem for ω 1, ω, (D), (B), (S 1 ) and ω 1, ω, (A), (C), (S ) we get: DB = P B P D MN, AC = P A P C MN Since P C P A = P B P D = AC = BD = AB = CD. IX) ABC is equilateral with side length L. Let (O, r) and (O, R) be the incircle and circumcircle of ABC. P is a point on (O, r) and P 1, P, P 3 are the projections of P onto BC, CA, AB. Circles T 1, T and T 3 touch BC, CA, AB through P 1, P, P and (O, R) (internally), their centers lie on different sides of BC, CA, AB with respect to A, B, C. Prove that the sum of the lengths of the common external tangents of T 1, T and T 3 is a constant value. Solution. Let δ 1 denote the tangent segment from A to T 1. By Casey s theorem for (A), (B), (C), T 1, all tangent to (O, R), we have L BP 1 + L CP 1 = δ 1 L = δ 1 = L. Similarly, we have δ = δ 3 = L. By Euler s theorem for the pedal triangle P 1 P P 3 of P, we get: [P 1 P P 3 ] = p(p, (O)) [ABC] = R r [ABC] = 3 4R 4R 16 [ABC] Therefore, we obtain AP AP 3 + BP 3 BP 1 + CP 1 CP = sin 60 ([ABC] [P 1P P 3 ]) = L. ( ) By Casey s theorem for (B), (C), T, T 3, all tangent to (O, R), we get δ δ 3 = L = BC δ 3 + CP BP 3 = L δ 3 + (L AP 1 )(L AP ) By cyclic exchange, we have the expressions: L = L δ 31 + (L BP 3 )(L BP 1 ), L = L δ 1 + (L CP 1 )(L CP ) 7

8 Figure 3: Application VII Adding the three latter equations yields 3L = L(δ 3 + δ 31 + δ 1 ) + 3L 3L + AP 3 AP + BP 3 BP 1 + CP 1 CP Hence, combining with ( ) gives δ 3 + δ 31 + δ 1 = 3L L = L 3 Proposed Problems. 1) Purser s theorem: ABC is a triangle with circumcircle (O) and ω is a circle in its plane. AX, BY, CZ are the tangent segments from A, B, C to ω. Show that ω is tangent to (O), if and only if ±AX BC ± BY CA ± CZ AB = 0 8

9 ) Circle ω touches the sides AB, AC of ABC at P, Q and its circumcircle (O). Show that the midpoint of P Q is either the incenter of ABC or the A-excenter of ABC, according to whether (O), ω are internally tangent or externally tangent. 3) ABC is A-right with circumcircle (O). Circle Ω B is tangent to the segments OB, OA and the arc AB of (O). Circle Ω C is tangent to the segments OC, OA and the arc AC of (O). Ω B, Ω C touch OA at P, Q, respectively. Show that: AB AC = AP AQ 4) Gumma, We are given a cirle (O, r) in the interior of a square ABCD with side length L. Let (O i, r i ) i = 1,, 3, 4 be the circles tangent to two sides of the square and (O, r) (externally). Find L as a fuction of r 1, r, r 3, r 4. 5) Two parallel lines τ 1, τ touch a circle Γ(R). Circle k 1 (r 1 ) touches Γ, τ 1 and a third circle k (r ) touches Γ, τ, k 1. We assume that all tangencies are external. Prove that R = r 1 r. 6)Victor Thébault ABC has incircle (I, r) and circumcircle (O). D is a point on AB. Circle Γ 1 (r 1 ) touches the segments DA, DC and the arc CA of (O). Circle Γ (r ) touches the segments DB, DC and the arc CB of (O). If ADC = ϕ, show that: r 1 cos ϕ + r sin ϕ = r References [1] I. Shariguin, Problemas de Geometrié (Planimetrié), Ed. Mir, Moscu, [] Vittasko, Sum of two tangents, equal to the distance of two points, AoPS, [3] My name is math, Tangent circles concurrent lines, AoPs, [4] Mathquark, Point [Feuerbach point of a triangle; FY + FZ = FX], AoPS, [5] Virgil Nicula, ABC and circle tangent to AB, AC and circumcircle, AoPS, [6] Shoki, Iran(3rd round)009, AoPS,

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