This post contains RMO 2018 solutions, problems, and discussions.

- Let \(ABC\) be a triangle with integer sides in which \(AB < AC\). Let the tangent to the circumcircle of triangle \(ABC\) at \(A\) intersect the line \(BC\) at \(D\). Suppose \(AD\) is also an integer. Prove that gcd\((AB,AC) >1\).
- Let \(n\) be a natural number. Find all real numbers \(x\) satisfying the equation \(\displaystyle\sum_{k=1}^{n} \frac{kx^k}{1+x^{2k}}\)\(=\frac{n(n+1)}{4}\)
- For a rational number \(r\), its period is the length of the smallest repeating block in its decimal expansion. For example, the number \(r=0.123123123..\) has period \(3\). If \(S\) denotes the set of all rational numbers \(r\) of the form \(r=0.\overline{abcdefgh}\) having period \(8\), find the sum of all elements of \(S\).
- Let \(E\) denotes the set of \(25\) points \((m,n)\) in the\(xy\)-plane, where \(m\),\(n\) are natural numbers, \(1 \leq m \leq 5\), \(1 \leq m \leq 5\). Suppose the points of \(E\) are arbitrarily coloured using two colours, red and blue. Show that there always exist four points in the set \(E\) of the form \((a,b)\), \((a+k,b)\), \((a+k,b+k)\), \((a,b+k)\) for some positive integer \(k\) such that at least three of these four points have the same colour. ( That is there always exist four points in the set \(E\) which form the vertices of a square with sides parallel to axes and having at least three points of the same colour.)
- Find all natural numbers \(n\) such that \(1+[\sqrt{2n}]\) divides \(2n\). (For any real number \(x\), \([x]\) denotes the largest integer not exceeding x.)
- Let \(ABC\) be an acute-angled triangle with \(AB<AC\). Let \(I\) be the incentre of triangle \(ABC\), and let \(D\),\(E\),\(F\) be the points at which its incircle touches the side \(BC\),\(CA\),\(AB\) respectively. Let \(BI\), \(CI\) meet the line \(EF\) at \(Y\),\(X\), respectively. Further assume that both \(X\) and \(Y\) are outside the triangle \(ABC\). Prove that
- \(B\),\(C\),\(Y\),\(X\) are concyclic; and
- \(I\) is also the incentre of triangle \(DYX\).

## RMO 2018 Solutions

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**Problem 2**

**Problem 4**

Problem 4 Part 2

**Problem 5**

**Problem 6**

Hi sir is this question paper of today that is 7th October 2018

Yes

Problem 2 follows by a simple AM-GM inequality. Upper bound the given quantity Q by replacing x by absolute value of (x), then divide the Numerator and Denominator by |x| to obtain

\[Q\leq \sum_{k=1}^{n}\frac{k}{|x|^k+|x|^{-k}}. \]

Using the AM-GM inequality, we have

\[ Q\leq \frac{1}{2}\sum_{k=1}^{n}k=\frac{n(n+1)}{4},\]

hence the equality must hold throughout and we have the only solution \(x=1\).

In problem number 4 why the diagonal is RBRBR it can be RRBRR or any other combination . We just don’t want 3 consecutive monochromatic

watch Part 2