MathDB

3

Part of 2012 Tuymaada Olympiad

Problems(4)

Point in a triangle, with angle constraints

Source: Tuymaada 2012, Problem 3, Day 1, Seniors

7/21/2012
Point PP is taken in the interior of the triangle ABCABC, so that ∠PAB=∠PCB=14(∠A+∠C).\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C). Let LL be the foot of the angle bisector of ∠B\angle B. The line PLPL meets the circumcircle of △APC\triangle APC at point QQ. Prove that QBQB is the angle bisector of ∠AQC\angle AQC.
Proposed by S. Berlov
geometrycircumcirclesymmetrytrapezoidincentergeometric transformationreflection
Inequality with abc=1

Source: Tuymaada 2012, Problem 7, Day 2, Seniors

7/21/2012
Prove that for any real numbers a,b,ca,b,c satisfying abc=1abc = 1 the following inequality holds 12a2+b2+3+12b2+c2+3+12c2+a2+3≤12.\dfrac{1} {2a^2+b^2+3}+\dfrac {1} {2b^2+c^2+3}+\dfrac{1} {2c^2+a^2+3}\leq \dfrac {1} {2}.
Proposed by V. Aksenov
inequalitiesinequalities proposed
Array with distinct sums of rows and columns

Source: Tuymaada 2012, Problem 3, Day 1, Juniors

7/21/2012
Prove that N2N^2 arbitrary distinct positive integers (N>10N>10) can be arranged in a N×NN\times N table, so that all 2N2N sums in rows and columns are distinct.
Proposed by S. Volchenkov
inductionlinear algebramatrixcombinatorics proposedcombinatorics
Radius of circle contained in quadrilateral

Source: Tuymaada 2012, Problem 7, Day 2, Juniors

7/21/2012
A circle is contained in a quadrilateral with successive sides of lengths 3,6,53,6,5 and 88. Prove that the length of its radius is less than 33.
Proposed by K. Kokhas
inequalitiesgeometrytrigonometrygeometry proposed