MathDB

1

Part of 2015 Romania Team Selection Tests

Problems(5)

Isogonal conjugates

Source: Romania TST 2015 Day 1 Problem 1

4/9/2015
Let ABCABC be a triangle, let OO be its circumcenter, let A′A' be the orthogonal projection of AA on the line BCBC, and let XX be a point on the open ray AA′AA' emanating from AA. The internal bisectrix of the angle BACBAC meets the circumcircle of ABCABC again at DD. Let MM be the midpoint of the segment DXDX. The line through OO and parallel to the line ADAD meets the line DXDX at NN. Prove that the angles BAMBAM and CANCAN are equal.
Isogonal conjugateorthocenterCircumcentergeometry
Sum divisibility !

Source: Romania TST 2015 Day 2 Problem 1

6/4/2015
Let aa be an integer and nn a positive integer . Show that the sum :
∑k=1na(k,n)\sum_{k=1}^{n} a^{(k,n)} is divisible by nn , where (x,y)(x,y) is the greatest common divisor of the numbers xx and yy .
SumDivisibilitynumber theoryRomanian TSTGCD
Tangent circles !!!

Source: Romania TST 2015 Day 3 Problem 1

6/4/2015
Two circles γ\gamma and γ′\gamma' cross one another at points AA and BB . The tangent to γ′\gamma' at AA meets γ\gamma again at CC , the tangent to γ\gamma at AA meets γ′\gamma' again at C′C' , and the line CC′CC' separates the points AA and BB . Let Γ\Gamma be the circle externally tangent to γ\gamma , externally tangent to γ′\gamma' , tangent to the line CC′CC', and lying on the same side of CC′CC' as BB . Show that the circles γ\gamma and γ′\gamma' intercept equal segments on one of the tangents to Γ\Gamma through AA .
circlesInversionTangentsRomanian TSTgeometry
Triangles of equal perimeters !!

Source: Romania TST 2015 Day 4 Problem 1

6/4/2015
Let ABCABC and ABDABD be coplanar triangles with equal perimeters. The lines of support of the internal bisectrices of the angles CADCAD and CBDCBD meet at PP. Show that the angles APCAPC and BPDBPD are congruent.
equal anglesperimeterRomanian TSTgeometry
Congruent angles and Miquel's point !!

Source: Romania TST 2015 Day 5 Problem 1

6/4/2015
Let ABCABC be a triangle. Let P1P_1 and P2P_2 be points on the side ABAB such that P2P_2 lies on the segment BP1BP_1 and AP1=BP2AP_1 = BP_2; similarly, let Q1Q_1 and Q2Q_2 be points on the side BCBC such that Q2Q_2 lies on the segment BQ1BQ_1 and BQ1=CQ2BQ_1 = CQ_2. The segments P1Q2P_1Q_2 and P2Q1P_2Q_1 meet at RR, and the circles P1P2RP_1P_2R and Q1Q2RQ_1Q_2R meet again at SS, situated inside triangle P1Q1RP_1Q_1R. Finally, let MM be the midpoint of the side ACAC. Prove that the angles P1RSP_1RS and Q1RMQ_1RM are equal.
Miquel pointequal anglesgeometryRomanian TSTBritishMathematicalOlympiadSpiral Similarity