MO Landesrunde Klasse 9 Mock Paper 1: Questions, Hints and Solutions

If p divided b, it would also divide a² and hence a, contradicting coprimality. Thus 2b is invertible modulo p. Define $t\equiv(a-b)/(2b)\pmod p$, where division means multiplication by that inverse. A calculation gives $t^2+t+1\equiv(a^2+3b^2)/(4b^2)\equiv0$. Hence $t^3-1=(t-1)(t^2+t+1)\equiv0$. We cannot have t=1 modulo p, since that would give 3≡0, contrary to p>3. The least positive exponent producing 1 is therefore 3: it divides 3 by division with remainder and is not 1. To recall Fermat’s congruence here, multiplication by the nonzero residue t permutes the residues 1,…,p−1. Multiplying them and cancelling their product modulo the prime p gives $t^{p-1}\equiv1$. Thus $t^{p-1}\equiv1$, so the same division argument forces 3|(p−1).

Conclusion: Every such prime is 1 modulo 3.

Review the idea: Number theory theorems · Greatest common divisor

Put $Q=xy+yz-zx$. Expansion gives $(x-y+z)^2=1-2Q$, so $Q=(1-(x-y+z)^2)/2$. The square is at least zero. To bound it above, use the identity $$3(x^2+y^2+z^2)-(x-y+z)^2=(x+y)^2+(y+z)^2+(x-z)^2\ge0.$$ Thus $0\le(x-y+z)^2\le3$, giving $-1\le Q\le1/2$. The maximum 1/2 occurs exactly when $x-y+z=0$, subject to the given unit-sphere condition; for example $(1/\sqrt2,1/\sqrt2,0)$ works. The minimum −1 requires all three squares in the identity to vanish, which means x=z=−y. The condition $x^2+y^2+z^2=1$ then gives $(x,y,z)=\pm(1,-1,1)/\sqrt3$. Direct substitution verifies both extremes.

Conclusion: Minimum −1 at ±(1,−1,1)/√3; maximum 1/2 exactly when x−y+z=0 on the unit sphere.

Review the idea: Cauchy schwarz inequality · Sum of squares

We use directed angles between lines modulo 180 degrees: reversing a ray does not change such an angle, and the cyclic-quadrilateral criterion says equal directed angles subtending a chord imply concyclicity. From circle AEF, $\angle EPF\equiv\angle EAF$; from circle BFD, $\angle FPD\equiv\angle FBD$. Adding gives $\angle EPD\equiv\angle EAF+\angle FBD$. Because AE lies on AC, AF and BF on AB, and BD on BC, the right side is the sum of directed line angles from AC to AB and from AB to BC, hence the angle from AC to BC. Since CE lies on AC and CD lies on BC, this equals $\angle ECD$ modulo 180 degrees. Therefore C and P subtend the same chord ED, so C,D,E,P lie on one circle. All needed chords have distinct endpoints by the assumptions.

Conclusion: C,D,E,P lie on a common circle.

Review the idea: Cyclic and tangential quadrilaterals · Angles

Choose a longest path $v_1,v_2,\ldots,v_k$. Every neighbour of its endpoint $v_k$ must already be on the path, since an outside neighbour could extend it. There are at least three distinct such neighbours among the earlier vertices. Two, say $v_i,v_j$ with i<j, have indices of the same parity. The segment $v_i,v_{i+1},\ldots,v_j$, followed by the edges to $v_k$ and back to $v_i$, forms a cycle with $j-i+2$ edges. No vertex repeats because j<k, and the length is even because j−i is even.

Conclusion: A longest path yields an even cycle.

Review the idea: Parity · Pigeonhole principle

Suppose the entire list were $p_1,\ldots,p_k$, with each prime 3 modulo 4. This list is nonempty because 3 is one such prime. Let $N=4p_1\cdots p_k-1$, an odd integer greater than 1 and congruent to 3 modulo 4. No listed prime divides N, since N is −1 modulo each. Every prime divisor of N is odd, hence 1 or 3 modulo 4. If all were 1 modulo 4, their product, including multiplicities, would also be 1 modulo 4. Therefore N has a prime divisor 3 modulo 4 not on the list, a contradiction.

Conclusion: There are infinitely many primes congruent to 3 modulo 4.

Review the idea: Prime numbers · Proof methods

A polynomial of degree 0 or 1 has zero second difference, so cannot work. If P has degree d≥2 and leading coefficient c, expand $(x\pm1)^d=x^d\pm dx^{d-1}+d(d-1)x^{d-2}/2+\cdots$. The binomial theorem then shows that its second difference has leading term $cd(d-1)x^{d-2}$: odd first-order terms cancel, leaving the second-order term. Since the result is the constant 2, d=2 and 2c=2, so c=1. Thus $P(x)=x^2+ax+b$. Completing the square gives $P(x)=(x+a/2)^2+b-a^2/4$, nonnegative for all x precisely when $b\ge a^2/4$. Conversely every polynomial in that family has second difference 2 and is nonnegative.

Conclusion: P(x)=x²+ax+b with real a,b and b≥a²/4.

Review the idea: Polynomial functions · Sum of squares