MathDB

3

Part of 2016 Saint Petersburg Mathematical Olympiad

Problems(3)

tetrahedron whose midpoints of all edges lie in a sphere, concurrency of heights

Source: St Petersburg Olympiad 2016, Grade 11, P3

9/8/2018
In a tetrahedron, the midpoints of all the edges lie on the same sphere. Prove that it's altitudes intersect at one point.
geometry3D geometrytetrahedronspherealtitudesconcurrent
B_1,A_1 touchpoints with incircle, K\in AB: AK = KB_1, BK = KA_1, show <C>=60

Source: St Petersburg Olympiad 2016, Grade 10, P3

9/8/2018
The circle inscribed in the triangle ABCABC is tangent to side ACAC at point B1B_1, and to side BCBC at point A1A_1. On the side ABAB there is a point KK such that AK=KB1,BK=KA1AK = KB_1, BK = KA_1. Prove that ∠ACB≥60 \angle ACB\ge 60
geometryincircleangle bisector
P,Q \in AB, AC = AP,BC = BQ, R intersection of ... prove <ACB + < PRQ = 180^o,

Source: St Petersburg Olympiad 2016, Grade 9, P3

9/8/2018
On the side ABAB of the non-isosceles triangle ABCABC, let the points PP and QQ be so that AC=APAC = AP and BC=BQBC = BQ. The perpendicular bisector of the segment PQPQ intersects the angle bisector of the ∠C\angle C at the point RR (inside the triangle). Prove that ∠ACB+∠PRQ=180o\angle ACB + \angle PRQ = 180^o.
geometryperpendicular bisectorangle bisectorequal segments