Download A History of Mathematical Notations (Dover Books on by Florian Cajori PDF

By Florian Cajori

This vintage examine notes the 1st visual appeal of a mathematical image and its foundation, the contest it encountered, its unfold between writers in numerous nations, its upward push to recognition, and its eventual decline or final survival. initially released in 1929 in a two-volume version, this huge paintings is gifted the following in a single volume.

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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.

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