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AD · DC P D2 Solution 2. Let O1 , O2 , O3 , O4 be the centres of Γ1 , Γ2 , Γ3 , Γ4 , respectively, and let A , B , C , D be the midpoints of P A, P B, P C, P D, respectively. Since Γ1 , Γ3 are externally tangent at P , it follows that O1 , O3 , P are collinear. Similarly we see that O2 , O4 , P are collinear. O1 D φ1 θ4 θ1 A O4 φ4 C P φ2 θ3 θ2 O2 φ3 O3 B Put θ1 = ∠O4 O1 O2 , θ2 = ∠O1 O2 O3 , θ3 = ∠O2 O3 O4 , θ4 = ∠O3 O4 O1 and φ1 = ∠P O1 O4 , φ2 = ∠P O2 O3 , φ3 = ∠P O3 O2 , φ4 = ∠P O4 O1 . By the law of sines, we have O1 O2 : O1 O3 = sin φ3 : sin θ2 , O3 O4 : O1 O3 = sin φ1 : sin θ4 , O3 O4 : O2 O4 = sin φ2 : sin θ3 , O1 O2 : O2 O4 = sin φ4 : sin θ1 .
P Then we ‘escape’ from the point P . The followings are examples for n = 9, 10, 11. The angles around the black points are not right. n=9 n = 10 n = 11 The ‘escaping lines’ are not straight in these figures. However, in fact, we can make them straight when we draw sufficiently large figures. 26 C4. Let x1 , . . , xn and y1 , . . , yn be real numbers. Let A = (aij )1≤i,j≤n be the matrix with entries 1, if xi + yj ≥ 0; aij = 0, if xi + yj < 0. Suppose that B is an n × n matrix with entries 0, 1 such that the sum of the elements in each row and each column of B is equal to the corresponding sum for the matrix A.
Let Γ1 , Γ2 , Γ3 , Γ4 be distinct circles such that Γ1 , Γ3 are externally tangent at P , and Γ2 , Γ4 are externally tangent at the same point P . Suppose that Γ1 and Γ2 ; Γ2 and Γ3 ; Γ3 and Γ4 ; Γ4 and Γ1 meet at A, B, C, D, respectively, and that all these points are different from P . Prove that P B2 AB · BC = . AD · DC P D2 Solution 1. Γ1 A D Γ4 θ1 θ2 θ8 θ7 Γ2 P θ θ4 3 B θ6 θ5 Γ3 C Figure 1 Let Q be the intersection of the line AB and the common tangent of Γ1 and Γ3 . Then ∠AP B = ∠AP Q + ∠BP Q = ∠P DA + ∠P CB.