1

Part of 1997 Vietnam Team Selection Test

Problems(2)

Arbitrary integers not smaller than k

Source: Vietnam TST 1997, Problem 4

f : \mathbb{N} \to \mathbb{Z}

is defined by f(0) \equal{} 2, f(1) \equal{} 503 and f(n \plus{} 2) \equal{} 503f(n \plus{} 1) \minus{} 1996f(n) for all

n \in\mathbb{N}

s_1

s_2

\ldots

s_k

be arbitrary integers not smaller than

k

p(s_i)

be an arbitrary prime divisor of

f\left(2^{s_i}\right)

, ( i \equal{} 1, 2, \ldots, k). Prove that, for any positive integer

t

t\le k

), we have 2^t \Big | \sum_{i \equal{} 1}^kp(s_i) if and only if

2^t | k

functionalgebrapolynomialinductionnumber theoryrelatively primenumber theory unsolved

Prove that there is a unique point satisfying conditions

Source: Vietnam TST 1997, Problem 1

ABCD

be a given tetrahedron, with BC \equal{} a, CA \equal{} b, AB \equal{} c, DA \equal{} a_1, DB \equal{} b_1, DC \equal{} c_1. Prove that there is a unique point

P

satisfying PA^2 \plus{} a_1^2 \plus{} b^2 \plus{} c^2 \equal{} PB^2 \plus{} b_1^2 \plus{} c^2 \plus{} a^2 \equal{} PC^2 \plus{} c_1^2 \plus{} a^2 \plus{} b^2 \equal{} PD^2 \plus{} a_1^2 \plus{} b_1^2 \plus{} c_1^2 and for this point

P

we have PA^2 \plus{} PB^2 \plus{} PC^2 \plus{} PD^2 \ge 4R^2, where

R

is the circumradius of the tetrahedron

ABCD

. Find the necessary and sufficient condition so that this inequality is an equality.

geometry3D geometrytetrahedroncircumcircleinequalitiesanalytic geometrygeometry unsolved