RMO Mock Paper 1: 6 Questions and Solutions

Since $1/x<1/12$, both $x$ and $y$ exceed 12. Multiplication and completion give $(x-12)(y-12)=144$. Put $d=x-12$, so $y=12+144/d$. The condition $x<y$ is exactly $d<12$. The positive divisors of 144 below 12 are $1,2,3,4,6,8,9$. Their corresponding sums $24+d+144/d$ are $169,98,75,64,54,50,49$. Exactly 169, 64 and 49 are squares. Thus the pairs are $(13,156),(16,48),(21,28)$. Each gives the stated reciprocal sum, so the list is complete.

Answer / conclusion: (13,156), (16,48), (21,28).

Review the idea: Factorisation integer solutions · Counting and summing divisors

The four prescribed roots and the degree bound give $P(x)-1=kx(x-1)(x-2)(x-3)$. At $x=4$, we obtain $24=24k$, so $k=1$. Set $t=x^2-3x$. Then $x(x-3)=t$ and $(x-1)(x-2)=t+2$. Therefore $$P(x)=1+t(t+2)=(t+1)^2=(x^2-3x+1)^2.$$ This is nonnegative and is zero exactly when $x^2-3x+1=0$, namely at $x=(3\pm\sqrt5)/2$. Both are real and satisfy the formula.

Answer / conclusion: Minimum 0 at (3 ± √5)/2.

Review the idea: Remainder and factor theorems · Sum of squares

In right triangle $ADB$, $AD=AB\cos\angle DAB$. In right triangle $AED$, $AE=AD\cos\angle DAB$. Consequently $AB\cdot AE=AD^2$. The same reasoning in $ADC$ and $AFD$ gives $AC\cdot AF=AD^2$; addition proves the first identity.

Both $\angle AED$ and $\angle AFD$ are right angles, so the four points lie on the circle of diameter $AD$. Its radius is $AD/2$. The chord $EF$ subtends the angle $\angle EAF=\angle BAC$. By the extended sine rule in triangle $AEF$, $EF=2(AD/2)\sin\angle BAC$, proving the second statement.

Answer / conclusion: Both stated identities hold.

Review the idea: Similar triangles · Circles and power of a point · Trigonometry in geometry

Suppose row $i$ contains $r_i$ marked cells. Any two marks in that row determine a pair of columns, so the row supplies $\binom{r_i}{2}$ column pairs. If the same column pair occurred in two rows, their four marked cells would form a forbidden rectangle. Conversely, every forbidden rectangle repeats one column pair. There are only $\binom72=21$ column pairs, and therefore $$\sum_{i=0}^{6}\binom{r_i}{2}\le21.$$

For every nonnegative integer $r$, $$\binom r2-(2r-3)=\frac{(r-2)(r-3)}2\ge0.$$ Indeed, if $r\le2$, both factors are nonpositive; if $r\ge3$, both are nonnegative. There is no integer strictly between 2 and 3. Writing $M=r_0+\cdots+r_6$ for the total number of marks, we obtain $$2M-21=\sum_{i=0}^{6}(2r_i-3)\le\sum_{i=0}^{6}\binom{r_i}{2}\le21.$$ Hence $M\le21$.

We now construct 21 marks. Label the rows and columns $0,1,\ldots,6$. In successive rows mark the following column sets: $$\{0,1,3\},\quad\{1,2,4\},\quad\{2,3,5\},\quad\{3,4,6\},\quad\{0,4,5\},\quad\{1,5,6\},\quad\{0,2,6\}.$$ The labelled grid below shows the construction.

Here is a direct check of all column pairs, written as two column labels: row 0 gives 01, 03, 13; row 1 gives 12, 14, 24; row 2 gives 23, 25, 35; row 3 gives 34, 36, 46; row 4 gives 04, 05, 45; row 5 gives 15, 16, 56; and row 6 gives 02, 06, 26. These 21 pairs are all different, so no rectangle occurs. Each of the seven rows has three marks, giving 21 marks. The greatest possible number is therefore $21$.

Answer / conclusion: 21 marked cells.

Review the idea: Combinations and binomial coefficients · Pigeonhole principle

Put $q=ab+bc+ca$. Squaring the constraint gives $a^2+b^2+c^2=9-2q$. Therefore $$(a^2+b^2+c^2)q=9q-2q^2=\frac{81}{8}-2\left(q-\frac94\right)^2\le\frac{81}{8}.$$ The bound is attained, for example, at $(a,b,c)=(2,1/2,1/2)$. Equality holds precisely for the nonnegative triples satisfying $a+b+c=3$ and $ab+bc+ca=9/4$, equivalently $a^2+b^2+c^2=9/2$. This description includes every equality case; it is a continuous family, not only permutations of the displayed example.

Answer / conclusion: 81/8; equality exactly when the given sum is 3 and ab+bc+ca=9/4.

Review the idea: Sum of squares · Symmetric polynomials

First, $(1,1)$ is a solution. If $(a,b)$ is a solution, replacing it by $(b,4b-a)$ preserves the equation, because $$b^2+(4b-a)^2+2-4b(4b-a)=a^2+b^2+2-4ab=0.$$ The recurrence begins $1,1,3,11,41,\ldots$. If $b\ge a>0$, then $4b-a\ge3b>b$. Thus, after the first equal pair, all terms are positive and strictly increase. The displayed identity proves that every consecutive pair, and its reversal, is a solution.

To prove that none are missing, take any positive integer solution and order it so that $a\le b$. If $a=1$, the equation becomes $(b-1)(b-3)=0$, so $b=1$ or $3$. Now suppose $a>1$. Regard the equation as a quadratic in $b$: $$F(t)=t^2-4at+a^2+2.$$ Since one root is $b$, the other root is $$b_\ast=4a-b=\frac{a^2+2}{b}.$$ The first expression shows that $b_\ast$ is an integer, and the second shows that it is positive. Substituting $t=a$ in $F(t)=(t-b)(t-b_\ast)$ gives $$(a-b)(a-b_\ast)=2-2a^2<0.$$ The factor $a-b$ is nonpositive. For the product to be negative it must be strictly negative, while $a-b_\ast$ must be positive. Hence $0<b_\ast<a<b$.

Because $b_\ast$ is also a root, $(b_\ast,a)$ is another positive integer solution. Its largest entry is $a$, strictly smaller than the previous largest entry $b$. Repeating this step cannot continue forever: a strictly decreasing sequence of positive integers must end. It therefore reaches a solution with smaller entry 1, namely $(1,1)$ or $(1,3)$.

Finally, the descent can be reversed uniquely. If the smaller pair is $(u,v)=(b_\ast,a)$, the previous pair was $(v,4v-u)$. This is exactly the rule defining consecutive recurrence terms. Both terminal pairs occur in the recurrence, so reversing all descent steps proves that every positive integer solution is one of the stated pairs or its reversal.

Answer / conclusion: Exactly the consecutive recurrence pairs and reversals.

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