MO Regionalrunde Klasse 9 Mock Paper 3: Questions, Hints and Solutions

The prime p cannot divide n, since then n⁴+1 would be 1 modulo p. We have $n^4\equiv-1\pmod p$, and because p is odd this is not 1. Squaring gives n⁸≡1. Let d be the least positive exponent with $n^d\equiv1\pmod p$. Division with remainder shows d divides every exponent with residue 1: if k=qd+r, then $n^k\equiv n^r\pmod p$, and minimality forces r=0. Thus d divides 8, but does not divide 4; among divisors of 8 this forces d=8. Fermat’s little theorem gives $n^{p-1}\equiv1\pmod p$, so the same argument yields 8|(p−1).

Conclusion: Every odd prime divisor is 1 modulo 8.

Review the idea: Number theory theorems · Euclidean division and remainders

Let S=x²+y²+z². Squaring x+y+z=0 gives xy+yz+zx=−S/2. Therefore $$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)=x^2y^2+y^2z^2+z^2x^2=S^2/4.$$ Finally $S^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)$. Substitution and rearrangement give the required identity, with no sign restrictions needed.

Conclusion: The identity holds for all real triples with zero sum.

Review the idea: Symmetric polynomials · Algebraic identities

Within a connected component, counting edge endpoints gives the sum of all degrees as twice the number of edges, which is even. A sum of integers is even only if an even number of its odd summands occur. Thus each component contains an even number of odd-degree vertices. If the two odd-degree vertices of the whole graph were in different components, each of those components would contain exactly one, a contradiction. They are therefore in the same component, which by definition means a path joins them.

Conclusion: The two odd-degree vertices lie in the same connected component.

Review the idea: Parity · Counting

Let $x=[PBC]/[ABC]$, y=[PCA]/[ABC] and z=[PAB]/[ABC]. These are positive and sum to 1. Triangles PBC and ABC have the same base BC, and A,P,D are collinear, so their altitudes have ratio PD/AD. Hence x=PD/AD and PA/PD=(AD−PD)/PD=1/x−1. Similarly the other ratios are 1/y−1 and 1/z−1. By Cauchy, $(1/x+1/y+1/z)(x+y+z)\ge(1+1+1)^2=9$. Subtracting 3 gives the bound 6. Equality requires x=y=z=1/3. Then AD=3PD, so AP:PD=2:1, and similarly on the other cevians. For the final midpoint step, triangles PBD and PCD share their altitude to line BC, so [PBD]:[PCD]=BD:DC. Triangles APB and DPB share the altitude from B to line AD, giving [APB]/[DPB]=AP/PD. Likewise [APC]/[DPC]=AP/PD. Hence [PAB]:[PAC]=BD:DC. Since the two areas are equal, BD=DC. Similarly E and F are side midpoints. Thus P is the intersection of the medians, namely the centroid. Thus equality holds exactly at the centroid.

Conclusion: Equality exactly when P is the centroid.

Review the idea: Triangle area ratios · Cauchy schwarz inequality