By Hardy G. H.
Hardy's natural arithmetic has been a vintage textbook on the grounds that its book in1908. This reissue will deliver it to the eye of an entire new iteration of mathematicians.
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Extra info for A Course of Pure Mathematics
It is also clear that sin( 21 QJC) QI sin QB ′ C sin QJI = = = . QJ sin QIJ sin QA′ C sin( 21 QIC) Taking into account that sin B ′ QC : sin QB ′ C = B ′ C : QC and sin A′ QC : sin QA′ C = A′ C : QC we get B ′ C A′ C OI QI : = : = 1. 83. a) The conditions of the problem imply that no three lines meet at one point. Let lines AB, AC and BC intersect the fourth line at points D, E, and F , respectively (Fig. 17). Figure 17 (Sol. 83) Denote by P the intersection point of circumscribed circles of triangles ABC and CEF distinct from point C.
Let, for definiteness sake, ∠KCA = ϕ ≤ 45◦ . Then AC(cos ϕ − sin ϕ) √ BK = AC sin(45◦ − ϕ) = 2 and AC(cos ϕ + sin ϕ) √ . DK = AC sin(45◦ + ϕ) = 2 Clearly, AC cos ϕ = CK and AC sin ϕ = AK. 7. Since ∠B1 AA1 = ∠A1 BB1 , it follows that points A, B, A1 and B1 lie on one circle. Parallel lines AB and A1 B1 intercept on it equal chords AB1 and BA1 . Hence, AC = BC. 8. On side BC of triangle ABC construct outwards an equilateral triangle A1 BC. Let P be the intersection point of line AA1 with the circumscribed circle of triangle A1 BC.
Since ∠F BA = β = ∠AED, quadrilateral ABF E is an inscribed one and, therefore, ∠F AE = ∠F BE = α − β. , AF is the bisector of angle ∠A. 46. Since ED = CB, EN = CM and ∠DEC = ∠BCA = 30◦ (Fig. 15), it follows that △EDN = △CBM . Let ∠M BC = ∠N DE = α, ∠BM C = ∠EN D = β. Figure 15 (Sol. 46) It is clear that ∠DN C = 180◦ − β. Considering triangle BN C we get ∠BN C = 90◦ − α. Since α + β = 180◦ − 30◦ = 150◦ , it follows that ∠DN B = ∠DN C + ∠CN B = (180◦ − β) + (90◦ − α) = 270◦ − (α + β) = 120◦ . Therefore, points B, O, N and D, where O is the center of the hexagon, lie on one circle.