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2021 EGMO Problems

Day 1

Problem 1

The number 2021 is fantabulous. For any positive integer $m$, if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?

Solution

Problem 2

Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation \[f(xf(x)+y) = f(y) + x^2\]holds for all rational numbers $x$ and $y$.

Here, $\mathbb{Q}$ denotes the set of rational numbers.

Solution

Problem 3

Let $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $ABC$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively such that $\angle EMA = \angle BCA$ and $\angle ANF = \angle ABC$. Prove that the points $E, F, N, M$ lie on a circle.

Solution

Day 2

Problem 4

Let $ABC$ be a triangle with incenter $I$ and let $D$ be an arbitrary point on the side $BC$. Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$. Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$. Prove that the reflection of $A$ across the line $EF$ lies on the line $BC$.

Solution

Problem 5

A plane has a special point $O$ called the origin. Let $P$ be a set of 2021 points in the plane such that

(i) no three points in $P$ lie on a line and

(ii) no two points in $P$ lie on a line through the origin.

A triangle with vertices in $P$ is fat if $O$ is strictly inside the triangle. Find the maximum number of fat triangles.

Solution

Problem 6

Does there exist a nonnegative integer $a$ for which the equation \[\left\lfloor\frac{m}{1}\right\rfloor + \left\lfloor\frac{m}{2}\right\rfloor + \left\lfloor\frac{m}{3}\right\rfloor + \cdots + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a\]has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers?

The expression $\lfloor x\rfloor$ denotes the integer part (or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor = 1, \lfloor\pi\rfloor =\lfloor 22/7 \rfloor = 3, \lfloor 42\rfloor = 42,$ and $\lfloor 0 \rfloor = 0$.

Solution

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