MathDB

1

Part of 2011 Saudi Arabia BMO TST

Problems(4)

\prod (x_k^2+ (k + 2)x_k + k^2 + k + 1) =(3/4)^n (n!)^2

Source: 2011 Saudi Arabia BMO TST 1.1 - Balkan MO

12/29/2021
Let nn be a positive integer. Find all real numbers x1,x2,...,xnx_1,x_2 ,..., x_n such that ∏k=1n(xk2+(k+2)xk+k2+k+1)=(34)n(n!)2\prod_{k=1}^{n}(x_k^2+ (k + 2)x_k + k^2 + k + 1) =\left(\frac{3}{4}\right)^n (n!)^2
Productalgebrafactorial
P(x)+P(y)+P(z)+P(x+y+z)=P(x+y)+P(y+z)+P(z+x)

Source: 2011 Saudi Arabia BMO TST 2.1 - Balkan MO

12/29/2021
Find all polynomials PP with real coefficients such that for all x,y,z∈Rx, y ,z \in R, P(x)+P(y)+P(z)+P(x+y+z)=P(x+y)+P(y+z)+P(z+x)P(x)+P(y)+P(z)+P(x+y+z)=P(x+y)+P(y+z)+P(z+x)
algebrapolynomial
OP^4 + ( MN/2 )^4 = MP^2 x NP^2 in square ABCD

Source: 2011 Saudi Arabia BMO TST 3.1 - Balkan MO

12/30/2021
Let ABCDABCD be a square of center OO. The parallel to ADAD through OO intersects ABAB and CDCD at MM and NN and a parallel to ABAB intersects diagonal ACAC at PP. Prove that OP4+(MN2)4=MP2⋅NP2OP^4 + \left(\frac{MN}{2} \right)^4 = MP^2 \cdot NP^2
geometrysquare
equiangular hexagon whose sidelengths are n + 1, n + 2 ,..., n + 6

Source: 2011 Saudi Arabia BMO TST 4.1 - Balkan MO

12/30/2021
Prove that for any positive integer nn there is an equiangular hexagon whose sidelengths are n+1,n+2,...,n+6n + 1, n + 2 ,..., n + 6 in some order.
hexagongeometryconsecutive