MathDB

4

Part of 2009 Ukraine National Mathematical Olympiad

Problems(8)

Show that ∠AKM = ∠KPC - [UKRMO 2009 Grade 8]

Source:

1/21/2011
In the triangle ABCABC given that ∠ABC=120∘.\angle ABC = 120^\circ . The bisector of ∠B\angle B meet ACAC at MM and external bisector of ∠BCA\angle BCA meet ABAB at P.P. Segments MPMP and BCBC intersects at KK. Prove that ∠AKM=∠KPC.\angle AKM = \angle KPC .
geometryangle bisector
On the equation k+m^k+n^{m^k}}=2009^n - [UKRMO 2009 Grade 8]

Source:

1/21/2011
а) Prove that for any positive integer nn there exist a pair of positive integers (m,k)(m, k) such that k+mk+nmk=2009n.{k + m^k + n^{m^k}} = 2009^n. b) Prove that there are infinitely many positive integers nn for which there is only one such pair.
number theory proposednumber theory
Prove that xt<1/3 - [UKRMO 2009 Grade 9]

Source:

1/23/2011
Let x≤y≤z≤tx \leq y \leq z \leq t be real numbers such that xy+xz+xt+yz+yt+zt=1.xy + xz + xt + yz + yt + zt = 1.
a) Prove that xt<13,xt < \frac 13,
b) Find the least constant CC for which inequality xt<Cxt < C holds for all possible values xx and t.t.
inequalitiesinequalities proposed
The midpoint of CD lies on BQ - [UKRMO 2009 Grade 9]

Source:

1/23/2011
In the trapezoid ABCDABCD we know that CD⊥BC,CD \perp BC, and CD⊥AD.CD \perp AD . Circle ww with diameter ABAB intersects ADAD in points AA and P,P, tangent from PP to ww intersects CDCD at M.M. The second tangent from MM to ww touches ww at Q.Q. Prove that midpoint of CDCD lies on BQ.BQ.
geometrytrapezoidcircumcircleperpendicular bisectorgeometry proposed
Functional Equation - [UKRMO 2009 Grade 10]

Source:

1/23/2011
Find all functions f:R→Rf : \mathbb R \to \mathbb R such that f(x+xy+f(y))=(f(x)+12)(f(y)+12)∀x,y∈R.f\left(x+xy+f(y)\right)= \left( f(x)+\frac 12 \right) \left( f(y)+\frac 12 \right) \qquad \forall x,y \in \mathbb R.
functionalgebrafunctional equationalgebra unsolved
The least possible number of vertices -[UKRMO 2009 Grade 11]

Source:

1/23/2011
Let GG be a connected graph, with degree of all vertices not less then m≥3m \geq 3, such that there is no path through all vertices of GG being in every vertex exactly once. Find the least possible number of vertices of G.G.
pigeonhole principlegraph theorycombinatorics proposedcombinatorics
Find the ratio of areas of triangles - [UKRMO 2009 Grade 10]

Source:

1/23/2011
Let ABCDABCD be a parallelogram with ∠BAC=45∘,\angle BAC = 45^\circ, and AC>BD.AC > BD . Let w1w_1 and w2w_2 be two circles with diameters ACAC and DC,DC, respectively. The circle w1w_1 intersects ABAB at EE and the circle w2w_2 intersects ACAC at OO and CC, and ADAD at F.F. Find the ratio of areas of triangles AOEAOE and COFCOF if AO=a,AO = a, and FO=b.FO = b .
ratiogeometryparallelogramtrigonometrygeometry unsolved
Find all polynomials - [UKRMO 2009 Grade 11]

Source:

1/23/2011
Find all polynomials P(x)P(x) with real coefficients such that for all pairwise distinct positive integers x,y,z,tx, y, z, t with x2+y2+z2=2t2x^2 + y^2 + z^2 = 2t^2 and gcd⁡(x,y,z,t)=1,\gcd(x, y, z, t ) = 1, the following equality holds 2P2(t)+2P(xy+yz+zx)=P2(x+y+z).2P^2(t ) + 2P(xy + yz + zx) = P^2(x + y + z) .
Note. P2(k)=(P(k))2.P^2(k)=\left( P(k) \right)^2.
algebrapolynomialalgebra unsolved