MO Regionalrunde Klasse 9 Mock Paper 2: Questions, Hints and Solutions

We have $\gcd(a+b,a+2b)=\gcd(a+b,b)=1$. If two coprime positive integers have square product, each is a square: every prime occurs in only one factor and must have even exponent. Thus a+b=x² and a+2b=y² for positive coprime integers x,y. Subtraction gives $b=y^2-x^2$ and $a=2x^2-y^2$. Positivity is precisely $x\lt y\lt \sqrt2\,x$. Conversely take coprime positive integers x,y in this interval and use these formulas. Then a,b are positive, and their gcd is $\gcd(x^2,y^2-x^2)=1$. The two original factors are x²,y², so their product is a square. This proves completeness.

Conclusion: a=2x²−y², b=y²−x² for coprime positive x<y<√2 x.

Review the idea: Unique prime factorisation · Greatest common divisor

Let Q(x)=P(x)−x³. Since $(x+1)^3-x^3=3x^2+3x+1$, the equation becomes Q(x+1)−Q(x)=0. A nonconstant polynomial Q of degree d≥1 with leading coefficient c has difference Q(x+1)−Q(x) of degree d−1 and leading coefficient cd, by the binomial expansion. This cannot be the zero polynomial. Therefore Q is constant. From P(0)=2 we get Q=2, so $P(x)=x^3+2$. Substitution verifies the identity and initial value.

Conclusion: Only P(x)=x³+2.

Review the idea: Polynomial functions · Binomial expansions and generating functions

Put $S_0$=0 and $S_k=a_1+\cdots+a_k$. Choose m in {0,…,n−1} for which $S_m$ is smallest; since $S_n$=$S_0$, it is also no larger than $S_n$. Start with $a_{m+1}$, continuing cyclically. A partial sum before the end of the original list is $S_j$−$S_m$≥0. A partial sum that wraps around is $(S_n-S_m)+S_j=S_j-S_m\ge0$. The full sum is 0. Thus every initial partial sum in the chosen shift is nonnegative.

Conclusion: Start after a minimum of the original partial sums.

Review the idea: Sequences and sums · Proof methods

Let the centres be $O_1$,$O_2$. The radii $O_1A$ and $O_2B$ are perpendicular to AB and both point towards line AB. The centres and T are collinear, with T between the centres. Put $\alpha=\angle AO_1T$; then $\angle BO_2T=180^\circ-\alpha$, since $O_1A$ and $O_2B$ are parallel and $O_1T$,$O_2T$ point oppositely. Isosceles triangles $O_1AT$ and $O_2BT$ give $\angle ATO_1=(180^\circ-\alpha)/2$ and $\angle BTO_2=\alpha/2$. A and B lie on the same side of the line of centres, so these two angles and angle ATB partition the straight angle $O_1TO_2$. Hence angle ATB is 180°−90°=90°.

Conclusion: Angle ATB is a right angle.

Review the idea: Circles and power of a point · Angles