RMO Mock Paper 2: 6 Questions and Solutions

Put $t=n+2$. The gcd becomes $\gcd(t^2-2t+3,t^2+3)=\gcd(t^2+3,2t)$. If an odd prime divides both numbers, it divides $t$ and hence 3; therefore only the odd prime 3 can occur. When $3\mid t$, $t^2+3$ is divisible by 3 but not 9. When $3\nmid t$, it is not divisible by 3. If $t$ is even, $t^2+3$ is odd, so no factor 2 occurs. If $t$ is odd, $2t$ has exactly one factor 2, and $t^2+3$ is even, so the gcd has exactly one factor 2. Thus the answer is $(2\text{ if }n\text{ is odd, else }1)\,(3\text{ if }n\equiv1\pmod3,\text{ else }1)$. These factors are coprime, so the description is complete.

Answer / conclusion: Possible gcds 1,2,3,6, with the stated independent factors.

Review the idea: Greatest common divisor · Unique prime factorisation

The equation is symmetric in $x,y$, and its right side contains their sum. This suggests setting $s=x+y$ and $d=x-y$. The identity $$3(x+y)^2+(x-y)^2=4(x^2+xy+y^2)$$ changes the equation into $$3s^2+d^2=28s,\qquad d^2=28s-3s^2.$$ Because $x,y$ are positive integers, $s\ge2$. Since $d^2\ge0$, we have $s(28-3s)\ge0$, so $s\le28/3$. Thus the integer $s$ satisfies $2\le s\le9$.

Positivity gives another bound: $$|d|=|x-y|=x+y-2\min(x,y)\le s-2.$$ Consequently $$28s-3s^2=d^2\le(s-2)^2,$$ which simplifies to $s^2-8s+1\ge0$, or $(s-4)^2\ge15$. Among the integers from 2 to 9, the values 2 through 7 have $|s-4|\le3$, so their squares are at most 9. Only $s=8,9$ remain.

If $s=8$, then $d^2=32$, which lies strictly between $5^2$ and $6^2$, so no integer $d$ works. If $s=9$, then $d^2=9$, giving $d=3$ or $-3$. Recover the original numbers using $$x=\frac{s+d}{2},\qquad y=\frac{s-d}{2}.$$ This gives $(x,y)=(6,3)$ or $(3,6)$. Finally, $36+18+9=63=7(6+3)$, so both pairs satisfy the original equation.

Answer / conclusion: (3,6) and (6,3).

Review the idea: Symmetric polynomials · Factorisation integer solutions

The angle-bisector theorem gives $BD:DC=13:15$. Since $BC=14$, $DC=15/2$. The points $D$ and $B$ lie on the same ray from $C$, so the power of $C$ is $CD\cdot CB=(15/2)14=105>0$. The line $CA$ already meets the circle at $A$, with $CA=15$. Its other intersection therefore has directed distance $CE=105/15=7$ along the same ray. Since $0<7<15$, it is indeed on the interior of $CA$, distinct from $A$. Hence $AE=15-7=8$, and $AE:EC=8:7$.

Answer / conclusion: AE:EC = 8:7.

Review the idea: Parallel lines and angle bisectors · Circles and power of a point

Let the entries be $a_1,\ldots,a_9$, with indices reduced modulo 9. Each entry is included in three sums, so the nine sums total $3(1+\cdots+9)=135$. If every sum were at most 15, all would be 15. Subtracting consecutive sums would give $a_i=a_{i+3}$, contrary to distinctness. The largest sum is therefore at least 16.

This bound is attained by the clockwise order $1,5,8,2,6,7,3,4,9$. Its nine clockwise sums are $14,15,16,15,16,14,16,14,15$, all at most 16. Hence the required minimum is 16.

Answer / conclusion: 16; construction [1, 5, 8, 2, 6, 7, 3, 4, 9]

Review the idea: Counting · Proof methods

We separate each fraction into a simple part and a nonnegative correction. Write $D=a+2b>0$, so $a=D/3+2(a-b)/3$. Squaring this identity and dividing by $D$ gives $$\frac{a^2}{a+2b}=\frac{a+2b}{9}+\frac{4(a-b)}9+\frac{4(a-b)^2}{9(a+2b)}.$$ Make the same calculation with $(a,b)$ replaced by $(b,c)$ and then $(c,a)$. Adding the three equations, the linear differences cancel because $(a-b)+(b-c)+(c-a)=0$, while the first numerators add to $3s$. Hence the left side of the required inequality equals $$\frac{s}{3}+\frac49\left(\frac{(a-b)^2}{a+2b}+\frac{(b-c)^2}{b+2c}+\frac{(c-a)^2}{c+2a}\right).$$

Now $2s-(a+2b)=a+2c>0$, so $0<a+2b<2s$. Taking reciprocals reverses this comparison; multiplying by the nonnegative square $(a-b)^2$ yields $$\frac{(a-b)^2}{a+2b}\ge\frac{(a-b)^2}{2s},$$ with equality exactly when $a=b$. Similarly, $2s-(b+2c)=2a+b>0$ and $2s-(c+2a)=2b+c>0$, giving the corresponding bounds for the other two fractions. Substituting them proves $$\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\ge\frac{s}{3}+\frac{2\bigl((a-b)^2+(b-c)^2+(c-a)^2\bigr)}{9s}.$$

If any difference is nonzero, its comparison is strict. Thus equality requires $a=b=c$. Direct substitution shows that these values do give equality.

Answer / conclusion: Equality exactly at a=b=c.

Review the idea: Sum of squares · Cauchy schwarz inequality

Suppose $P=FG$ with nonconstant integer polynomials. Since $\deg P=4$, one factor, say $F$, has degree at most 2. At each of $0,3,6,9$, $P$ is 1, so $F$ has value either 1 or $-1$. All four arguments are congruent modulo 3, so the four values of $F$ are also congruent modulo 3. But 1 and $-1$ are different residues modulo 3. Thus all four values equal one fixed sign $\varepsilon$. The polynomial $F-\varepsilon$, of degree at most 2, has four distinct roots; it is identically zero. Then $F$ is constant, a contradiction. Therefore no such factorisation exists.

Answer / conclusion: Irreducible as a product of nonconstant integer polynomials.

Review the idea: Irreducible polynomials · Remainder and factor theorems