By Votja P. A.

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7. F. 41 (see Probl`eme propos´e 60). Note added in proof. After completing this paper, we learned that Problem CMJ 354 (College Math. J. 18(1987), 248) by Alvin Tirman asks for the determination of Pythagorean triangles with the property that the triangle formed by the altitude and median corresponding to the hypothenuse is also Pythagorean. It is immediate that the solution of this problem follows from paragraph 1. of our paper. 52 12 An arithmetic problem in geometry 1. The following old problem in its complete generality was studied in 1891 by Gy.

A PA ≥ 4 Here (by Ta a a=T+ PA ≥ 2 T+ p b+c a Ta Ta = T ), finally we get Ta b+c a , (11) where, as we noticed, T = T (ABC), Ta = T (P BC). G. 74). Let the triangle ABC be acute-angled, and let P ≡ O in (10). Then OA is an altitude in triangle OBC, so R2 − OA = √ We obtain the curious inequality a2 . 4 4R2 − a2 ≤ 3R. 98). 2 Finally, we give two applications of (2). Let ABC be acute-angled, and let AA , BB , CC be the altitudes, and O1 , O2 , O3 the midpoints of the segments BC, AC, AB - respectively.

E. (m + n)|2λmn(m − n); and since (m + n, 2mn(m − n)) = 1. e.   λ = s(m + n) case (ii) we get m|λ, so λ = sm  Therefore in an isosceles triangle r is integer only when i) b = s(m + n)(m2 + n2 ), a = 2n = 4mns(m + n); or 46 (17) ii) b = sm(m2 + n2 ), a = sm(m2 − n2 ). abc ab2 2nb2 b2 For R = = = = we have that R is integer only when 2q|b2 , where 4∆ 4∆ 4nq 2q b2 = n2 + q 2 . e. 4mn|λ. By summing, R is integer only if in i) λ = 2s(m2 − n2 ), while in ii), λ = 4smn. Then the corresponding sides a, b can be written explicitely.

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Diophantine Approximations and Value Distribution Theory by Votja P. A.


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