RMO Mock Paper 3: 6 Questions and Solutions

Start with $y=1$. The given equation becomes $$f(x+1)-2f(x)+f(x-1)=6.$$ To find a function with this property, note that $$(x+1)^2-2x^2+(x-1)^2=2.$$ Therefore $3x^2$ gives the required value 6. Adding a linear expression does not change it, because the corresponding difference of $Ax+B$ is zero. The values $f(0)=0$ and $f(1)=4$ suggest the candidate $q(x)=3x^2+x$: it has $q(0)=0$, $q(1)=4$, and the required difference.

We must prove that no other function works. Define $g(x)=f(x)-q(x)$. Subtracting the difference equations for $f$ and $q$ gives $$g(x+1)=2g(x)-g(x-1),\qquad g(0)=g(1)=0.$$ It follows that $g(2)=2g(1)-g(0)=0$. Whenever two consecutive values are zero, the next is zero by the same rule. Induction therefore gives $g(x)=0$ for every nonnegative integer $x$.

For negative integers, rearrange the rule as $$g(x-1)=2g(x)-g(x+1).$$ With $x=0$, this gives $g(-1)=2g(0)-g(1)=0$. With $x=-1$, it gives $g(-2)=0$, and repeating proves that every negative value is also zero. Hence $f(x)=3x^2+x$ for all integers $x$.

The equation with $y=1$ was only a necessary condition, so we finish by checking the full equation. For all integers $x,y$, $$3(x+y)^2+(x+y)+3(x-y)^2+(x-y)=6x^2+2x+6y^2=2(3x^2+x)+6y^2.$$ Thus the candidate satisfies every condition and is the unique answer.

Answer / conclusion: f(x)=3x²+x for every integer x.

Review the idea: Solving functional equations · Second order recurrences

Modulo 3, the equation gives $x^2+y^2\equiv0$. As each square is 0 or 1, both $x$ and $y$ are divisible by 3. Write $x=3u,y=3v$. Then $3(u^2+v^2)=2z^2$, so $3\mid z$ as well. Put $z=3w$; substitution gives $u^2+v^2=6w^2$, another integer solution.

If a nonzero solution existed, choose one minimising the positive integer $|x|+|y|+|z|$. The construction produces a nonzero solution with one third that sum, a contradiction. Therefore only $(0,0,0)$ is possible; it plainly satisfies the equation.

Answer / conclusion: Only (0,0,0).

Review the idea: Remainders · Proof methods

Write $a=BC$, $b=CA$, $c=AB$, and let $r_1,r_2$ be the inradii of triangles $ABD,ACD$, respectively. Triangle $ABD$ and triangle $ABC$ are both right triangles and share the acute angle at $B$, so they are similar by the angle-angle criterion. Their hypotenuses are $AB=c$ and $BC=a$, so the scale factor is $c/a$. A similarity scales all lengths, including an inradius, by the same factor. Thus $r_1=cr/a$. Likewise, $ACD$ and $ABC$ share their acute angle at $C$ and are right triangles, giving $r_2=br/a$.

Choose perpendicular coordinate axes through $D$, with $DB$ on the negative horizontal ray, $DC$ on the positive horizontal ray, and $DA$ on the positive vertical ray. The incentre of a right triangle lies inside the triangle and is at distance equal to its inradius from each leg. Therefore $E=(-r_1,r_1)$ and $F=(r_2,r_2)$.

The horizontal separation of $E,F$ is $r_1+r_2$; their vertical separation is $|r_2-r_1|$. Applying Pythagoras to these perpendicular separations gives $$EF^2=(r_1+r_2)^2+(r_2-r_1)^2=2(r_1^2+r_2^2).$$ Substitute the two inradius formulas and then use $b^2+c^2=a^2$, Pythagoras in $ABC$: $$EF^2=2r^2\frac{c^2+b^2}{a^2}=2r^2.$$ Since lengths are positive, taking square roots yields $EF=\sqrt2\,r$.

Answer / conclusion: EF=√2 r.

Review the idea: Similar triangles · Pythagoras and stewart · Parallel lines and angle bisectors

Arrange the numbers in the order $$1,30,2,29,3,28,\ldots,15,16.$$ For $1\le k\le15$, positions $2k-1,2k$ contain $k,31-k$, whose sum is 31. These are all pairs of distinct numbers in the range with sum 31. For $2\le k\le15$, positions $2k-2,2k-1$ contain $32-k,k$, whose sum is 32. These are all the distinct pairs with sum 32: $1+31$ uses a number outside the range, and $16+16$ does not use distinct elements. Thus two numbers are forbidden together exactly when their positions are consecutive in this list.

Pair positions $(1,2),(3,4),\ldots,(29,30)$. At most one position from each pair can be selected, so there can be at most 15 selected numbers. Choosing all odd positions attains 15, since no two are consecutive. Hence the greatest size is 15.

It remains to count all selections of that size. Write the selected positions in increasing order as $i_1,\ldots,i_{15}$. They satisfy $$1\le i_1,\qquad i_{k+1}\ge i_k+2,\qquad i_{15}\le30.$$ Before the $k$-th selected position, at least $k-1$ unselected positions are needed to separate it from the preceding selections. Remove these compulsory gaps by defining $j_k=i_k-(k-1)$. Then $$j_{k+1}-j_k=(i_{k+1}-i_k)-1\ge1,$$ and $j_1\ge1$, $j_{15}=i_{15}-14\le16$. Thus the $j_k$ form a choice of 15 distinct numbers from $1,\ldots,16$.

This correspondence is reversible: from any such increasing choice define $i_k=j_k+k-1$. Then $i_{k+1}-i_k\ge2$, $i_1\ge1$, and $i_{15}\le30$, so it gives an allowed selection. There are $\binom{16}{15}=16$ choices, one for each number omitted from $1,\ldots,16$. Therefore exactly 16 sets attain the maximum size 15.

Answer / conclusion: Maximum size 15; exactly 16 maximum sets.

Review the idea: Counting with bijections · Combinations and binomial coefficients

If $abc>0$, the zero-sum condition forces two variables to be negative and one positive. After relabelling write $(a,b,c)=(-u,-v,u+v)$, where $u,v>0$ and $u+v\le1$. Then $$abc=uv(u+v)\le\frac{(u+v)^3}{4}\le\frac14.$$ Equality in the first bound requires $u=v$, and in the second requires $u+v=1$. Hence the maximum $1/4$ occurs exactly at permutations of $(1,-1/2,-1/2)$. A zero or negative product cannot improve this positive maximum. Replacing all three variables by their negatives preserves the domain and sum, and negates the product. Therefore the minimum is $-1/4$, attained exactly at permutations of $(-1,1/2,1/2)$.

Answer / conclusion: Maximum 1/4 and minimum −1/4, with stated permutations.

Review the idea: Arithmetic geometric and harmonic means · Inequality rules and signs

Suppose a prime $p\equiv3\pmod4$ divides $x^2+y^2$. Coprimality implies $p\nmid y$, since otherwise $p\mid x$ too. Thus $t\equiv xy^{-1}\pmod p$ satisfies $t^2\equiv-1$. Raising to $(p-1)/2$, an odd integer, gives $t^{p-1}\equiv-1\pmod p$, contradicting Fermat’s little theorem. Hence every odd prime divisor of $N$ is 1 modulo 4. Also $4\nmid N$: coprime $x,y$ cannot both be even, and squares modulo 4 are 0 or 1.

The three smallest permitted distinct primes are therefore $2,5,13$. Any admissible $N$ is at least their product, 130. Finally $130=3^2+11^2$, with $\gcd(3,11)=1$, and its prime divisors are exactly $2,5,13$. Thus 130 is attained and is the minimum.

Answer / conclusion: 130 = 3²+11².

Review the idea: Number theory theorems · Unique prime factorisation