Olympiad challenge room: original written problems

Let p=P(0). What is P(kp) modulo p?

Every value P(kp) is both a positive prime and divisible by p. Turn this into infinitely many roots of one polynomial.

Set p=P(0), a positive prime. Every nonconstant term of P(kp) is divisible by p, and the constant term is p. Hence p divides P(kp) for every integer k.

By the hypothesis, P(kp) is a positive prime, so it must equal p. The polynomial Q(x)=P(x)−p therefore has infinitely many distinct roots kp. A nonzero polynomial has at most its degree many distinct roots, so Q is the zero polynomial. Thus P is constantly p. Conversely, every constant positive prime polynomial has the required property.

Rewrite a/(b+c) as a²/[a(b+c)].

Use Engel form of Cauchy, then compare (a+b+c)² with 3(ab+bc+ca).

Engel form gives the sum at least (a+b+c)²/[2(ab+bc+ca)], since a(b+c)+b(c+a)+c(a+b)=2(ab+bc+ca)>0.

The identity a²+b²+c²−ab−bc−ca=((a−b)²+(b−c)²+(c−a)²)/2≥0 gives (a+b+c)²≥3(ab+bc+ca). Therefore the previous fraction is at least 3/2.

Equality in the second bound requires a=b=c. For these positive equal values each original fraction is 1/2, so equality indeed holds. No other equality case is possible.

Solve the 2-by-2 case. Then divide a larger board into four equal quadrants.

The original missing square lies in one quadrant. Place one central tromino so that the other three quadrants each have one square already occupied.

For n=1, removing one square from a 2-by-2 board leaves exactly one L-shaped tromino.

Assume every $2^{k}$-by-$2^{k}$ board with one missing square can be tiled. Divide a $2^{k+1}$-by-$2^{k+1}$ board into four quadrants. Exactly one contains the original missing square. Place a central L-tromino using the three centre-adjacent squares belonging to the other three quadrants.

Each quadrant now has exactly one square unavailable: the original missing square in one, or the central tromino square in each of the others. Apply the induction hypothesis separately to the four quadrants. These tilings and the central tile cover every remaining square exactly once, proving the result by induction.

Compute a₂ and test the expression at n=0.

Call the expression Iₙ. Substitute the recurrence to compare Iₙ₊₁ with Iₙ.

The recurrence gives a₂=13, so I₀=2·13−5²=1. For any n, Iₙ₊₁=aₙ₊₁aₙ₊₃−aₙ₊₂²=3aₙ₊₁aₙ₊₂−aₙ₊₁²−aₙ₊₂².

Since aₙ=3aₙ₊₁−aₙ₊₂, the same expression equals aₙaₙ₊₂−aₙ₊₁²=Iₙ. Thus induction gives Iₙ=1 for every n.

Any positive common divisor of aₙ and aₙ₊₁ divides aₙaₙ₊₂−aₙ₊₁²=1, so their greatest common divisor is 1. All terms are integers because the initial values and recurrence are integral.

Find f(0), then set n=1 to obtain a recurrence.

Compare f(n) with n², since (m+n)² contains the extra term 2mn.

Setting m=n=0 gives f(0)=0. Define g(n)=f(n)−n². Expanding (m+n)² shows g(m+n)=g(m)+g(n) on the nonnegative integers.

Let c=g(1). Induction using g(n+1)=g(n)+c gives g(n)=cn for every n≥0. Hence every possible solution has f(n)=n²+cn for one real constant c.

Conversely, for any real c, the formula f(n)=n²+cn gives f(m+n)=m²+n²+2mn+c(m+n)=f(m)+f(n)+2mn. All real c are therefore permitted. The proof relies on an integer domain; it does not assume continuity.

Move x and y to the left and add 1 to both sides.

The equation becomes (x−1)(y−1)=36. Show both factors are positive before listing all divisor pairs.

Rearranging gives (x−1)(y−1)=36. Since x,y are positive integers, both factors are nonnegative. Their product is positive, so each is positive.

Let d=x−1. Then d is a positive divisor of 36, and y−1=36/d. Thus all solutions are (d+1,36/d+1) for d∈{1,2,3,4,6,9,12,18,36}.

The nine ordered pairs are (2,37),(3,19),(4,13),(5,10),(7,7),(10,5),(13,4),(19,3),(37,2). Every listed pair works by reversing the factorisation, and every solution has been captured by a positive divisor.

Consider the n prefix sums Sₖ=a₁+⋯+aₖ.

If no prefix sum is 0 modulo n, put all n prefix sums into the n−1 nonzero remainder classes.

If a prefix sum Sₖ is divisible by n, the block a₁,…,aₖ already works. This also settles n=1.

Otherwise n≥2 and the n prefix sums occupy only n−1 nonzero residue classes modulo n. Pigeonhole gives Sᵢ≡Sⱼ for some i<j.

Subtracting gives aᵢ₊₁+⋯+aⱼ=Sⱼ−Sᵢ≡0 mod n. The block is consecutive and nonempty because j>i. Negative entries cause no difficulty: residues are defined for all integers.

Use BH perpendicular to AC and CH perpendicular to AB to find ∠HBC and ∠HCB.

Compute ∠BHC, then use reflection and the opposite-angle test for a cyclic quadrilateral.

Because the triangle is acute, H lies inside it. Perpendicularity gives ∠HBC=90°−∠C and ∠HCB=90°−∠B. The angle sum in BHC yields ∠BHC=180°−[(90°−C)+(90°−B)]=B+C=180°−A.

Reflection across BC fixes B and C and preserves angles, so ∠BH′C=∠BHC=180°−A. The points A and H′ lie on opposite sides of BC. Therefore the convex quadrilateral ABH′C has opposite angles ∠BAC and ∠BH′C summing to 180°.

By the converse cyclic-quadrilateral criterion, A,B,H′,C lie on one circle. This is the circumcircle of ABC, so it contains H′. The acute-triangle assumption keeps the ordinary angle and convexity argument valid as written.