RMO Mock Paper 5: 6 Questions and Solutions

Because $a\le b\le c$, we have $a^2c\le abc=2(a+b+c)\le6c$. Dividing by positive $c$ gives $a^2\le6$, so $a=1$ or 2. If $a=1$, then $(b-2)(c-2)=6$. Neither $b=1$ nor $b=2$ solves the original equation, so both factors are positive. The ordered factor pairs are $(1,6),(2,3)$, giving $(1,3,8),(1,4,5)$. If $a=2$, then $(b-1)(c-1)=3$, and $2\le b\le c$ gives only $(b-1,c-1)=(1,3)$, hence $(2,2,4)$. Direct substitution verifies all three triples.

Answer / conclusion: (1,3,8), (1,4,5), (2,2,4).

Review the idea: Factorisation integer solutions · Triangle inequalities

At zero, $P(0)=P(0)^2$, so the constant term is 0 or 1. Suppose a nonconstant polynomial $Q$ with $Q(0)=1$ satisfies the identity. Let $kx^m$ be its lowest nonconstant term, with $k\ne0,m\ge1$. In $Q(x)^2$, the coefficient of $x^m$ is $2k$. In $Q(x^2)$, that coefficient is zero: when $m$ is odd there is no such power, and when it is even the coefficient would come from degree $m/2<m$, which is absent. This contradiction shows that such $Q$ must be constant 1.

Our polynomial is nonzero and nonconstant because $P(2)=16$. Therefore $P(0)=0$, and we may write $P(x)=x^mQ(x)$, $m\ge1$, $Q(0)\ne0$. Cancelling the polynomial factor $x^{2m}$ gives $Q(x^2)=Q(x)^2$; its constant term is 1. The preceding argument forces $Q=1$. Thus $P=x^m$, and $2^m=16$ gives $m=4$. The polynomial $x^4$ plainly satisfies both conditions.

Answer / conclusion: P(x)=x⁴.

Review the idea: Polynomial functions · Polynomial roots and multiplicity

We use ordinary coordinates and slopes. The slope of a nonvertical line is its vertical change divided by its horizontal change. For two perpendicular lines with finite nonzero slopes, the slopes are negative reciprocals: turning a line through $90^\circ$ interchanges the horizontal and vertical changes and reverses one sign. Parallel nonvertical lines have equal slopes.

Choose $A=(0,0)$, $B=(b,0)$, and $C=(u,v)$, where $b>0$ and $v>0$. The angles at $A$ and $B$ are acute, so the perpendicular projection of $C$ onto $AB$ lies strictly between $A,B$. Hence $0<u<b$. Write $D=(d,e)$. Because the triangle is acute, its altitude foot $D$ lies strictly inside segment $BC$; thus $u<d<b$ and $e>0$.

The slope of $BC$ is $v/(u-b)$, while the slope of $AD$ is $e/d$. Their perpendicularity gives $$\frac{e}{d}\cdot\frac{v}{u-b}=-1,\qquad ev=d(b-u).$$ The line $AC$ has equation $y=(v/u)x$. Since $DF$ is perpendicular to the horizontal line $AB$, it is the vertical line $x=d$. Therefore $$F=\left(d,\frac{vd}{u}\right).$$

Since $DE\perp AC$, its slope is $-u/v$; as it passes through $D$, its equation is $$y-e=-\frac{u}{v}(x-d).$$ At $E$, the coordinate $y$ is zero because $E$ lies on line $AB$. Solving for its horizontal coordinate gives $$x_E=d+\frac{ev}{u}=d+\frac{d(b-u)}{u}=\frac{bd}{u}.$$ Hence $E=(bd/u,0)$. These computations allow points on the entire side lines, so they include the possible extensions.

Finally, the horizontal difference $d-bd/u=d(u-b)/u$ is nonzero, because $d>0$ and $u<b$. The slope of $EF$ is therefore $$\frac{vd/u-0}{d-bd/u}=\frac{v}{u-b},$$ exactly the slope of $BC$. The two distinct points $E,F$ thus determine a line parallel to $BC$, as required.

Answer / conclusion: EF is parallel to BC.

Review the idea: Similar triangles · Trigonometry in geometry

Choose a distinguished red position. Reading clockwise from it gives four positive gap lengths $g_1,g_2,g_3,g_4$ summing to 8. Choose which two are even in $\binom42=6$ ways. Their smallest positive values are 2 and 2; the odd gaps have smallest values 1 and 1. These minima total 6. The remaining 2 must be added to exactly one of the four gaps, preserving parity, giving four choices. Thus there are $6\cdot4=24$ ordered gap lists.

There are 12 choices of the labelled starting position, so we have counted $12\cdot24$ pairs consisting of a colouring and a distinguished red position. Each valid colouring has exactly four choices of its distinguished red position, even if it has rotational symmetry. Therefore the number of colourings is $12\cdot24/4=72$.

Answer / conclusion: 72 colourings.

Review the idea: Circular permutations · Combinations with repetition

Put $a=x^2$, $b=y^2$, $c=z^2$. These numbers are positive and $abc=(xyz)^2=1$. Set $s=a+b+c$, $q=ab+bc+ca$, and $T=1+s$. The three denominators $1+a+b,1+b+c,1+c+a$ are then $T-c,T-a,T-b$, respectively.

Their common denominator is positive. Multiplying its factors one at a time gives $$\begin{aligned}D&=(T-a)(T-b)(T-c)\\&=(T^2-(a+b)T+ab)(T-c)\\&=T^3-(a+b+c)T^2+(ab+bc+ca)T-abc\\&=T^3-sT^2+qT-1.\end{aligned}$$ To add the three fractions, the numerator of each becomes the product of the other two denominators. Thus their common numerator is $$\begin{aligned}N&=(T-a)(T-b)+(T-b)(T-c)+(T-c)(T-a)\\&=3T^2-2(a+b+c)T+(ab+bc+ca)\\&=3T^2-2sT+q.\end{aligned}$$ Substituting $T=1+s$ simplifies these to $$D=s^2+2s+q+sq,\qquad N=s^2+4s+3+q.$$

The sum of the original fractions is $N/D$. Since $D>0$, proving this is at most 1 is equivalent to proving $$D-N=sq-2s-3=s(q-2)-3\ge0.$$ By AM–GM applied to $a,b,c$, $$s\ge3\sqrt[3]{abc}=3.$$ Applying AM–GM to $ab,bc,ca$, whose product is $(abc)^2=1$, gives $$q\ge3\sqrt[3]{(ab)(bc)(ca)}=3.$$ Hence $$s(q-2)-3\ge s-3\ge0,$$ proving the inequality.

For equality, both comparisons must be equalities. Since $s>0$, the first requires $q=3$; the second requires $s=3$. Equality in AM–GM for $a,b,c$ gives $a=b=c$, and their product 1 gives $a=b=c=1$. Positivity of $x,y,z$ then gives $x=y=z=1$. Conversely, each fraction is $1/3$ at these values, so equality holds.

Answer / conclusion: Equality exactly when x=y=z=1.

Review the idea: Arithmetic geometric and harmonic means · Symmetric polynomials

We can group the possible values by their square residues instead of testing 169 ordered pairs. For $x=0,1,2,3,4,5,6$, squaring and reducing modulo 13 gives, respectively, $$0,1,4,9,3,12,10.$$ For instance, $4^2=16\equiv3$, $5^2=25\equiv12$, and $6^2=36\equiv10\pmod{13}$. The remaining values are $13-6,13-5,\ldots,13-1$. Because $(13-k)^2\equiv k^2\pmod{13}$, they repeat the six nonzero residues, each once.

The table lists every possible square residue $r=x^2\pmod{13}$, all values of $x$ producing it, and the required residue $1-r$ for $y^2$. The table also indicates when the required residue is not a square, so no $y$ is possible.

For each row, any listed $x$ can be paired with any listed $y$; this is why we multiply their counts. The only admissible ordered square-residue pairs are $(0,1),(1,0),(4,10),(10,4)$. The total number of ordered pairs is therefore $$1\cdot2+2\cdot1+2\cdot2+2\cdot2=12.$$ The table covers all 13 values of $x$, and the square-residue list also covers all 13 values of $y$, so no other pairs are possible.

Answer / conclusion: 12 ordered pairs.

Review the idea: Complete and reduced residue systems · Number theory theorems