INMO Mock Paper 2: 6 Questions and Solutions

Put $k=f(0)$. The equation at $x=0$ gives $f(f(y))=k+9y+6$, so $f$ is injective. At $y=0$, it gives $f(x+k)=f(x)+6$. Applying two shifts yields $f(x+f(y)+f(z))=f(x)+9(y+z)+12$. The same value is obtained from $f(x+f(y+z)+k)$. Injectivity therefore gives $f(y)+f(z)=f(y+z)+k$.

Hence $g(n)=f(n)-k$ is additive on the integers. By induction and the identity $g(-n)=-g(n)$, it has the form $g(n)=cn$ for the integer $c=g(1)$. Thus $f(n)=cn+k$. Substitution in the original equation gives $c^2y+ck=9y+6$ for all integers $y$. Therefore $c^2=9$ and $ck=6$. The two possibilities are $(c,k)=(3,2),(-3,-2)$. Direct substitution verifies both $f(n)=3n+2$ and $f(n)=-3n-2$.

Answer / conclusion: f(n)=3n+2 or f(n)=−3n−2.

Review the idea: Solving functional equations · Mathematical induction the first principle

First $(3,3,4)$ satisfies the equation. For any positive solution, the other root of $t^2-abt+a^2+b^2+2$ is $c^*=ab-c=(a^2+b^2+2)/c$, a positive integer. The replacement preserves the equation, and permutations do as well.

Conversely, order a solution as $a\le b\le c$. The case $a=1$ would give $b^2+c^2-bc+3=0$, impossible because $b^2+c^2\ge2bc$. The case $a=2$ gives $(b-c)^2+6=0$, also impossible. Thus $a\ge3$. For $F(t)=t^2-abt+a^2+b^2+2$, $$F(b)=a^2+2-(a-2)b^2.$$ If $a\ge4$, this is at most $a^2+2-(a-2)a^2=a^2(3-a)+2<0$. If $a=3,b\ge4$, it is $11-b^2<0$. Write $F(t)=(t-c)(t-c^*)$. Since $b\le c$ and $F(b)<0$, we cannot have $b=c$; hence $b-c<0$. The product $(b-c)(b-c^*)$ is negative, so its other factor is positive. Therefore $c^*<b<c$. The formula $c^*=(a^2+b^2+2)/c$ already shows $c^*>0$. Replacing $c$ therefore reduces the positive integer $a+b+c$.

The remaining case is $a=b=3$, where $c^2-9c+20=0$, giving c=4 or 5. The triple $(3,3,5)$ replaces its last coordinate by 4. After each replacement, reorder the entries and apply the same rule. The positive integer sum cannot decrease indefinitely. The only case where the reduction stops is $(3,3,4)$, since $(3,3,5)$ reduces once more to it. Each replacement can be undone by the same rule: $c\mapsto ab-c\mapsto ab-(ab-c)=c$. Thus reversing the complete descent proves that every solution is generated as stated.

Answer / conclusion: Exactly the positive triples generated from (3,3,4) by these operations.

Review the idea: Vietas formulas · Strong induction · Factorisation integer solutions

A cevian is a line joining a triangle vertex to a point on the opposite side. We use coordinates for the three cevians, with a short explanation of how their area ratio is calculated.

Coordinate-area tool. In ordinary coordinates, the triangle with vertices $(0,0),(p,q),(r,s)$ has area $|ps-qr|/2$. This follows by enclosing the triangle in a rectangle and subtracting the surrounding right-triangle areas; reversing the orientation only changes the sign inside the absolute value. Translating one vertex to the origin gives the same formula using the two coordinate differences from that vertex.

Take actual coordinates $A=(0,0),B=(b,0),C=(u,v)$, where $b>0$ and $v\ne0$. To keep the calculation short, name the actual point $(b\xi+u\eta,v\eta)$ by the pair $(\xi,\eta)$. Thus the pair labels of $A,B,C$ are $(0,0),(1,0),(0,1)$. Combining pairs in a fixed ratio combines their actual coordinates in the same ratio, so straight lines and the given side ratios are preserved by these labels. For two pair differences $(\Delta\xi_1,\Delta\eta_1)$, $(\Delta\xi_2,\Delta\eta_2)$, the coordinate-area formula gives twice the actual triangle area as $$\left|(b\Delta\xi_1+u\Delta\eta_1)v\Delta\eta_2-v\Delta\eta_1(b\Delta\xi_2+u\Delta\eta_2)\right|=|bv|\left|\Delta\xi_1\Delta\eta_2-\Delta\eta_1\Delta\xi_2\right|.$$ The $u$-terms cancel. Twice the area of $ABC$ is $|bv|$, so this common factor cancels when taking the area ratio.

We may therefore compute entirely with the pair labels. Put $L=k^2+k+1$. The side points are $D=(1/(k+1),k/(k+1))$, $E=(0,1/(k+1))$, $F=(k/(k+1),0)$. Their cevian equations, in these pair coordinates, are $y=kx$, $x+(k+1)y=1$, $(k+1)x+ky=k$. Solving in pairs gives $$X=(1/L,k/L),\quad Y=(k^2/L,1/L),\quad Z=(k/L,k^2/L).$$ Using the two differences $Y-X,Z-X$, the area ratio is $$\frac{[XYZ]}{[ABC]}=\frac{(k^2-1)(k^2-k)-(1-k)(k-1)}{L^2}=\frac{(k-1)^2L}{L^2}=\frac{(k-1)^2}{k^2+k+1}.$$ Here brackets denote triangle area.

Equating this explicit ratio to $1/4$ gives $4(k-1)^2=k^2+k+1$, or $k^2-3k+1=0$. Both roots $(3\pm\sqrt5)/2$ are positive and different from 1. Substitution verifies the required area, so these are exactly the two values.

Answer / conclusion: k=(3+√5)/2 or (3−√5)/2.

Review the idea: Triangle area ratios · Concurrency and collinearity

We prove by induction that any $2^{k+1}-1$ integers contain $2^k$ whose sum is divisible by $2^k$. For $k=0$, one entry suffices. Suppose the assertion is true for $k-1$. In a list of $2^{k+1}-1$ entries, exactly one of the numbers of even and odd entries is odd. Within each parity class, pair entries and leave only one unpaired entry overall. This produces $2^k-1$ disjoint pairs, each with even sum. Divide the pair sums by 2. By induction, select $2^{k-1}$ of these half-sums with total divisible by $2^{k-1}$. Their original pairs contain exactly $2^k$ entries, and their total is twice that sum, hence divisible by $2^k$.

For $k=5$, 63 entries therefore suffice. To prove that 62 do not suffice, take 31 zeros and 31 ones. A selection of 32 entries contains between 1 and 31 ones, so its sum is not divisible by 32. Hence the least $N$ is 63.

Answer / conclusion: 63.

Review the idea: Strong induction · Parity · Pigeonhole principle

At $a=b=c=1$, the inequality requires $3+3K\ge6$, so $K\ge1$. For $K=1$, the difference is $\sum_{\rm cyc}a^2(a-b)(a-c)$. By symmetry assume $a\ge b\ge c\ge0$. Grouping the first two terms gives $$(a-b)\bigl(a^2(a-c)-b^2(b-c)\bigr)+c^2(a-c)(b-c).$$ The bracket equals $(a-b)(a^2+ab+b^2-c(a+b))$, and its second factor is nonnegative because $c\le b$ gives $a^2+ab+b^2-c(a+b)\ge a^2$. Thus the whole difference is nonnegative.

If $a>b$, the first grouped term is strictly positive. Equality therefore requires $a=b$, after which the difference is $c^2(a-c)^2$. It vanishes when $c=0$ or $c=a$. Undoing the ordering, equality occurs when all three are equal, or when one is zero and the other two equal. These include the all-zero case. Hence the least coefficient is 1.

Answer / conclusion: K=1; equality when all equal or one zero and the other two equal.

Review the idea: Sum of squares · Symmetric polynomials

Order tool. For a prime $q$ not dividing $m$, let $d$ be the least positive exponent for which $m^d\equiv1\pmod q$; this exists by Fermat. If $m^N\equiv1$, write $N=kd+r$, $0\le r<d$. Then $m^r\equiv1$, so minimality gives $r=0$. Therefore $d\mid N$.

Suppose a prime $q\ne3$ divides $m^2+m+1$. It cannot divide $m$. Multiplication by $m-1$ gives $m^3\equiv1\pmod q$. Also $m\not\equiv1\pmod q$, since that would give $q\mid3$. The order $d$ divides 3 and is not 1, so $d=3$. Fermat gives $m^{q-1}\equiv1\pmod q$, since $q\nmid m$. By the order tool, $3\mid q-1$. Therefore $q\equiv1\pmod3$.

Now suppose all primes 1 modulo 3 are $p_1,\ldots,p_r$, allowing an empty list. Put $m=3p_1\cdots p_r$, taking an empty product as 1. The integer $m^2+m+1>1$ has a prime divisor $q$. The number $m^2+m+1$ is congruent to 1 modulo 3, so none of its prime divisors is 3; in particular $q\ne3$. It is also 1 modulo every $p_i$, so $q$ is absent from the purported complete list. The preceding order argument gives $q\equiv1\pmod3$, a contradiction. Hence there are infinitely many such primes.

Answer / conclusion: Infinitely many primes are 1 modulo 3.

Review the idea: Number theory theorems · Prime numbers