MathDB

2

Part of 2003 Cuba MO

Problems(2)

<PEQ = 2<APB , circles, tangents, cuba - romania

Source: 2017 Romania JBMO TST5 p3 - Cuban Olympiad 2003

5/16/2020
Let AA be a point outside the circle ω\omega . The tangents from AA touch the circle at BB and CC. Let PP be an arbitrary point on extension of ACAC towards CC, QQ the projection of CC onto PBPB and EE the second intersection point of the circumcircle of ABPABP with the circle ω\omega . Prove that ∠PEQ=2∠APB\angle PEQ = 2\angle APB
geometrycirclescircumcircleangles
p/q = 1-1/2+1/3+...+1/1335

Source: 2003 Cuba MO 2.2

9/15/2024
Prove that if pq=1−12+13−14+...−11334+11335\frac{p}{q}=1-\frac{1}{2} + \frac{1}{3}- \frac{1}{4} + ... -\frac{1}{1334} + \frac{1}{1335} where p,q∈Z+p, q \in Z_+ then pp is divisible by 20032003.
number theorydivides