MathDB

3

Part of 2002 Junior Balkan Team Selection Tests - Romania

Problems(5)

equilateral triangle is divided into 100 equilateral triangles by parallel

Source:

5/31/2020
A given equilateral triangle of side 1010 is divided into 100100 equilateral triangles of side 11 by drawing parallel lines to the sides of the original triangle. Find the number of equilateral triangles, having vertices in the intersection points of parallel lines whose sides lie on the parallel lines.
combinatoricscombinatorial geometryEquilateral
1x1 tiles of 4 colours in a 1xn rectangle

Source: 2002 Romania JBMO TST 1.3

5/31/2020
Consider a 1×n1 \times n rectangle and some tiles of size 1×11 \times 1 of four different colours. The rectangle is tiled in such a way that no two neighboring square tiles have the same colour. a) Find the number of distinct symmetrical tilings. b) Find the number of tilings such that any consecutive square tiles have distinct colours.
rectangleTilingtilescombinatoricsColoring
angle chasing inside a 20-80-80 triangle, perpendicular, CN =1/2 BC

Source: 2002 Romania JBMO TST2 p3

5/16/2020
Let ABCABC be an isosceles triangle such that AB=ACAB = AC and ∠A=20o\angle A = 20^o. Let MM be the foot of the altitude from CC and let NN be a point on the side ACAC such that CN=12BCCN =\frac12 BC. Determine the measure of the angle AMNAMN.
geometryanglesAngle Chasingequal segments
4x^3 = (a^2+b^2 + c^2)x + abc, computational geometry, max area of DECB

Source: 2002 Romania JBMO TST5 p3

5/16/2020
Let ABCABC be a triangle and a=BC,b=CAa = BC, b = CA and c=ABc = AB be the lengths of its sides. Points DD and EE lie in the same halfplane determined by BCBC as AA. Suppose that DB=c,CE=bDB = c, CE = b and that the area of DECBDECB is maximal. Let FF be the midpoint of DEDE and let FB=xFB = x. Prove that FC=xFC = x and 4x3=(a2+b2+c2)x+abc4x^3 = (a^2+b^2 + c^2)x + abc.
geometrymaxareamidpointsidelengths
tangents of one circle intersect on another circle

Source: 2002 Romania JBMO TST4 p3

5/16/2020
Let C1(O1)C_1(O_1) and C2(O2) C_2(O_2) be two circles such that C1C_1 passes through O2O_2. Point MM lies on C1C_1 such that M∉O1O2M \notin O_1O_2. The tangents from MM at O2O_2 meet again C1C_1 at AA and BB. Prove that the tangents from AA and BB at C2C_2 - others than MAMA and MBMB - meet at a point located on C1C_1.
geometrycirclesTangentsconcurrentconcurrency