By Vasile Berinde, Eugen Păltănea, Romanian Mathematical Society

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It follows that 10 < a < 31, 10 < b < 31 and 1 < d < 8. Moreover, 100d = x — y = a2 — b2 = (a — b) (a + b). Then a and b have the same parity. Let us suppose first that a and b are both even, a = 2n and b = 2m. Then n and m must satisfy: 5 < n < 15, 5 < m < 15 and (n — m) (n + m) = 25d. Since 0 < n — m < 10 and 10 < n + m < 30, it results that we have two possibilities: i) 51(n + m) and 51(n — in) , then 51n and 51m . Hence n E {10, 15} and m E {5,10} . It is easy to see that (20, 10) , (30, 20) and (30, 10) are the only solutions (in this case) to our problem.

17. _ V2(x2 +y2). When the equality holds? Solution. On the cartesian coordinates xOy system let's have the points A(1, 0), B(0, 1) and C(1, 1). We have to demonstrate that for M(x, y) belonging to the surface OACB we have MA+ MB > f • MO. For M on the surface OACB intersected with the disc of center 0 and radius 1 D(0; 1) we have: MA + MB > AB = 12. > v4-0 with equality for M E [AB] n D(0;1) = {A, B} . For M on the surface OACB, M [OB] U [OA] we have: MB • OA + MA • OB > AB • OM (Ptolemy's inequality) <=> <=> MB +MA> 12-• MO.

Consider LABC an arbitrary triangle, P a point on the circumcircle of AABC, A1, B1 and C1 the projections of P on the sides BC, CA and AB, respectively, AA' B'C' the median triangle of LABC, H the orthocenter of AABC and Ha, Hb, Hethe orthocenters of the triangles AABiCi, ABC]. Ai , ACA1B1i respectively. We must prove that the lines AiBiCi and HaHbli, are parallel. It is known that the Simpson's line with respect to the previous point P passes through the middle of the side that links point P with orthocenter of AABC.