Bmo: 2008 Solutions

: The proof relies on showing that two specific triangles involving these points are similar. By establishing similarity, the equality of subtended angles at the circle's circumference confirms the points lie on a common circle. Problem 4 (Integer Sequences) : For a sequence , find all is a perfect -th power for : Analysis starts by proving that

Unofficial but detailed solution overviews can often be found on community sites like the Art of Problem Solving (AoPS) or hosted on document platforms like Scribd . BMO 2008 Solutions Overview | PDF | Triangle - Scribd

Try linear solutions: f(x)=kx. Then f(f(x))=k^2 x = x ⇒ k=±1. Check k=1: LHS f(x f(y)+f(x)) = x y + x, RHS y x + x works. k=-1: f(x f(y)+f(x)) = -( -x y - x )? Let’s check: f(x f(y)+f(x)) = -[ x(-y) + (-x) ] = -[ -xy -x] = xy+x. RHS= y(-x)+x = -xy+x. Not equal unless x=0. So only f(x)=x works? But BMO 2008 solutions often include f(x)=-x? Let’s test: f(x)=-x: LHS= -[ x(-y) + (-x) ] = -[ -xy - x] = xy+x. RHS= y(-x) + x = -xy+x. So xy+x = -xy+x ⇒ 2xy=0 for all x,y ⇒ no. So only f(x)=x. But is that the only? Yes, after proving f additive, etc.

Let us explore select problems from the 2008 Round 1 paper. We will look at the methodology behind the solutions rather than just providing the final answer, as the process is what trains the mathematician. bmo 2008 solutions

Proving there are infinitely many pairs is divisible by Solution Resources

The final clean proof: Let tangents at C and D meet at X. Then X, C, A, D concyclic? Use power of a point and invert about A. The solution is well-documented in geometry olympiad handbooks.

For brevity in this article, the key insight is the leading to ( (3m-2008)(3n-2008) = 2008^2 ). : The proof relies on showing that two

The problem asks us to prove that for any integer $n$, $100 + n^2$ is never a perfect square.

Let ( x=0 ): ( f(0\cdot f(y) + f(0)) = y f(0) + 0 ) ⇒ ( f(f(0)) = y c ). But LHS constant, RHS varies with y unless c=0. So ( c=0 ). Thus ( f(0)=0 ).

In a random draw, the positions are symmetric. BMO 2008 Solutions Overview | PDF | Triangle

The first round was held in November 2007. It consisted of six problems, and official video solutions are hosted by the UK Mathematics Trust (UKMT) .

Among the annals of Olympiad history, the (Round 1 and Round 2) are often cited as classic examples of the "Olympiad style"—problems that seem impenetrable at first glance but yield elegant solutions with the right insight. This article provides a deep dive into the BMO 2008 solutions , analyzing the problems, the mathematical principles involved, and the strategies required to solve them under timed conditions.

Before diving into the solutions, it is vital to understand the format. The BMO is split into two rounds: