olymmath

USAJMO 2016

United States of America Junior Mathematical Olympiad 2016.

Day 1

Problem 1

  • Day 1

The isosceles triangle △ABC\triangle ABC , with AB=ACAB=AC , is inscribed in the circle ω\omega . Let PP be a variable point on the arc BC⌢\stackrel{\frown}{BC} that does not contain AA , and let IBI_B and ICI_C denote the incenters of triangles △ABP\triangle ABP and △ACP\triangle ACP , respectively.

Prove that as PP varies, the circumcircle of triangle △PIBIC\triangle PI_BI_C passes through a fixed point.

Answer, solutions and source

Problem 2

  • Day 1

Prove that there exists a positive integer n<106n < 10^6 such that 5n5^n has six consecutive zeros in its decimal representation.

Proposed by Evan Chen

Answer, solutions and source

Problem 3 · Also USAMO 2016 Problem 1

  • Day 1

Let X1,X2,…,X100X_1, X_2, \ldots, X_{100} be a sequence of mutually distinct nonempty subsets of a set SS . Any two sets XiX_i and Xi+1X_{i+1} are disjoint and their union is not the whole set SS , that is, Xi∩Xi+1=∅X_i\cap X_{i+1}=\emptyset and Xi∪Xi+1≠SX_i\cup X_{i+1}\neq S , for all i∈{1,…,99}i\in\{1, \ldots, 99\} . Find the smallest possible number of elements in SS .

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Find, with proof, the least integer NN such that if any 20162016 elements are removed from the set 1,2,...,N{1, 2,...,N} , one can still find 20162016 distinct numbers among the remaining elements with sum NN .

Answer, solutions and source

Problem 5

  • Day 2

Let △ABC\triangle ABC be an acute triangle, with OO as its circumcenter. Point HH is the foot of the perpendicular from AA to line BC↔\overleftrightarrow{BC} , and points PP and QQ are the feet of the perpendiculars from HH to the lines AB↔\overleftrightarrow{AB} and AC↔\overleftrightarrow{AC} , respectively.

Given that AH2=2⋅AO2,AH^2=2\cdot AO^2, prove that the points O,P,O,P, and QQ are collinear.

Answer, solutions and source

Problem 6 · Also USAMO 2016 Problem 4

  • Day 2

Find all functions f:R→Rf:\mathbb{R}\rightarrow \mathbb{R} such that for all real numbers xx and yy , (f(x)+xy)⋅f(x−3y)+(f(y)+xy)⋅f(3x−y)=(f(x+y))2.(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.

Answer, solutions and source