Gazeta Matematică - A Bridge Over Three Centuries by Vasile Berinde, Eugen Păltănea, Romanian Mathematical

By Vasile Berinde, Eugen Păltănea, Romanian Mathematical Society

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If a, b, c are positive rational numbers, with a c b and = VT + VC, then N/a, are on a ratio with rational numbers. Solution. \,/a = c. We obtain: VT, = — N/T) N7C, a+b c 2/a — b + c = vi:ta — b + c 2a 21ci, — We cannot have a + b — c = 0 <=> b c = a <=> together with fa = -VT+ would imply b = 0. 10. The orthogonal projection of the intersection point of the diagonals of an inscribing quadrilateral on its sides divide these sides in ratios whose product is equal to the unity. Solution. Using the annotations in the figure we have: Problems for the 7th form AANO ADQO 39 AN_ 73 y DO BP BO ABPO N AAMO =Az„ 1A AO CQ CO Tv= = ACQO ABNO NB BO ADMO ,ACP() DM DO = CP CO By multiplying these relations we have: AN PB CQ DM = 1.

Determine a number of the form abba, knowing that in any base of a numerical system it is a perfect cube. Solution. Let's have x the base of a numerical system into which the given number N is written. It results: N = ax3+ bx2+ bx + a = a(x +1)(x2— x +1) + bx(x +1) = (x + 1) [ax2+ (b — a)x + . As N must be a cube in any base, it results a,x2+ (b (V) x E N, x > 2. We obtain a = 1 and b a = 2. The sought number is N = 1331. 12. Irrespective of the base of a numerical system, there is no number of n> 2 digits equal to the product of its digits.

We have m(BFC) = 2y t x+y+t 2 2 — m(BGE) = 90° BF I AC; III. ABCE orthodiagonal quadrilateral = [AC] diameter and IV. [AC] diameter 2x = z, are similarly obtained. 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.

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