MathDB

2

Part of 1999 Romania National Olympiad

Problems(3)

a^2 +b^2 +c^2 never prime 1999 Romania NMO VII p3

Source:

8/14/2024
Let a,b,ca, b, c be non zero integers,a≠c a\ne c such that ac=a2+b2c2+b2\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2} Prove that a2+b2+c2a^2 +b^2 +c^2 cannot be a prime number.
number theoryprime
min pos. root of (a+b)x^2-2(ab-1)x-(a+b) = 0

Source: 1999 Romania NMO IX p2

8/15/2024
For a,b>0a, b > 0, denote by t(a,b)t(a,b) the positive root of the equation (a+b)x2−2(ab−1)x−(a+b)=0.(a+b)x^2-2(ab-1)x-(a+b) = 0. Let M={(a.b)∣ a≠b   and   t(a,b)≤ab}M = \{ (a.b) | \, a \ne b \,\,\, and \,\,\,t(a,b) \le \sqrt{ab} \} Determine, for (a,b)∈M(a, b)\in M, the mmimum value of t(a,b)t(a,b).
algebrainequalities
MN x NP x PQ x QM >= AM x BN x CP xDQ , regular tetrahedron

Source: 1999 Romania NMO X p2

8/14/2024
On the sides (AB)(AB), (BC)(BC), (CD)(CD) and (DA)(DA) of the regular tetrahedron ABCDABCD, one considers the points MM, NN, PP, QQ, respectively Prove that MN⋅NP⋅PQ⋅QM≥AM⋅BN⋅CP⋅DQ.MN \cdot NP \cdot PQ \cdot QM \ge AM \cdot BN \cdot CP \cdot DQ.
geometry3D geometrytetrahedrongeometric inequalityGeometric Inequalities