# European Girls' Mathematical Olympiad 2012-Day 1 Solutions 

Problem 1. Let $A B C$ be a triangle with circumcentre $O$. The points $D, E$ and $F$ lie in the interiors of the sides $B C, C A$ and $A B$ respectively, such that $D E$ is perpendicular to $C O$ and $D F$ is perpendicular to $B O$. (By interior we mean, for example, that the point $D$ lies on the line $B C$ and $D$ is between $B$ and $C$ on that line.)

Let $K$ be the circumcentre of triangle $A F E$. Prove that the lines $D K$ and $B C$ are perpendicular.
Origin. Netherlands (Merlijn Staps).
![](https://cdn.mathpix.com/cropped/2024_11_22_0830df84f0353d761a1ag-1.jpg?height=694&width=598&top_left_y=898&top_left_x=729)

Solution 1 (submitter). Let $\ell_{C}$ be the tangent at $C$ to the circumcircle of $\triangle A B C$. As $C O \perp \ell_{C}$, the lines $D E$ and $\ell_{C}$ are parallel. Now we find that

$$
\angle C D E=\angle\left(B C, \ell_{C}\right)=\angle B A C,
$$

hence the quadrilateral $B D E A$ is cyclic. Analogously, we find that the quadrilateral $C D F A$ is cyclic. As we now have $\angle C D E=\angle A=\angle F D B$, we conclude that the line $B C$ is the external angle bisector of $\angle E D F$. Furthermore, $\angle E D F=180^{\circ}-2 \angle A$. Since $K$ is the circumcentre of $\triangle A E F, \angle F K E=2 \angle F A E=2 \angle A$. So $\angle F K E+\angle E D F=180^{\circ}$, hence $K$ lies on the circumcircle of $\triangle D E F$. As $|K E|=|K F|$, we have that $K$ is the midpoint of the arc $E F$ of this circumcircle. It is well known that this point lies on the internal angle bisector of $\angle E D F$. We conclude that $D K$ is the internal angle bisector of $\angle E D F$. Together with the fact that $B C$ is the external angle bisector of $\angle E D F$, this yields that $D K \perp B C$, as desired.

Solution 2 (submitter). As in the previous solution, we show that the quadrilaterals $B D E A$ and $C D F A$ are both cyclic. Denote by $M$ and $L$ respectively the circumcentres of these quadrilaterals. We will show that the quadrilateral $K L O M$ is a parallelogram. The lines $K L$ and $M O$ are the perpendicular bisectors of the line segments $A F$ and $A B$, respectively. Hence both $K L$ and $M O$ are perpendicular to $A B$, which yields $K L \| M O$. In the same way we can show that the lines $K M$ and $L O$ are both perpendicular to $A C$ and hence parallel as well. We conclude that $K L O M$ is indeed a parallelogram. Now, let $K^{\prime}, L^{\prime}, O^{\prime}$ and $M^{\prime}$ be the respective projections of $K, L, O$ and $M$ to $B C$. We have to show that $K^{\prime}=D$. As $L$ lies on the perpendicular bisector of $C D$, we have that $L^{\prime}$ is the midpoint of $C D$. Similarly, $M^{\prime}$ is the midpoint of $B D$ and $O^{\prime}$ is the midpoint of $B C$. Now we are going to use directed lengths. Since $K L O M$ is a parallelogram, $M^{\prime} K^{\prime}=O^{\prime} L^{\prime}$. As

$$
O^{\prime} L^{\prime}=O^{\prime} C-L^{\prime} C=\frac{1}{2} \cdot(B C-D C)=\frac{1}{2} \cdot B D=M^{\prime} D
$$

we find that $M^{\prime} K^{\prime}=M^{\prime} D$, hence $K^{\prime}=D$, as desired.

Solution 3 (submitter). Denote by $\ell_{A}, \ell_{B}$ and $\ell_{C}$ the tangents at $A, B$ and $C$ to the circumcircle of $\triangle A B C$. Let $A^{\prime}$ be the point of intersection of $\ell_{B}$ and $\ell_{C}$ and define $B^{\prime}$ and $C^{\prime}$ analogously. As in the first solution, we find that $D E \| \ell_{C}$ and $D F \| \ell_{B}$. Now, let $Q$ be the point of intersection of $D E$ and $\ell_{A}$ and let $R$ be the point of intersection of $D F$ and $\ell_{A}$. We easily find $\triangle A Q E \sim \triangle A B^{\prime} C$. As $\left|B^{\prime} A\right|=\left|B^{\prime} C\right|$, we must have $|Q A|=|Q E|$, hence $\triangle A Q E$ is isosceles. Therefore the perpendicular bisector of $A E$ is the internal angle bisector of $\angle E Q A=\angle D Q R$. Analogously, the perpendicular bisector of $A F$ is the internal angle bisector of $\angle D R Q$. We conclude that $K$ is the incentre of $\triangle D Q R$, thus $D K$ is the angle bisector of $\angle Q D R$. Because the sides of the triangles $\triangle Q D R$ and $\triangle B^{\prime} A^{\prime} C^{\prime}$ are pairwise parallel, the angle bisector $D K$ of $\angle Q D R$ is parallel to the angle bisector of $\angle B^{\prime} A^{\prime} C^{\prime}$. Finally, as the angle bisector of $\angle B^{\prime} A^{\prime} C^{\prime}$ is easily seen to be perpendicular to $B C$ (as it is the perpendicular bisector of this segment), we find that $D K \perp B C$, as desired.

Remark (submitter). The fact that the quadrilateral $B D E A$ is cyclic (which is an essential part of the first two solutions) can be proven in various ways. Another possibility is as follows. Let $P$ be the midpoint of $B C$. Then, as $\angle C P O=90^{\circ}$, we have $\angle P O C=90^{\circ}-\angle O C P$. Let $X$ be the point of intersection of $D E$ and $C O$, then we have that $\angle C D E=\angle C D X=90^{\circ}-\angle X C D=90^{\circ}-\angle O C P$. Hence $\angle C D E=\angle P O C=\frac{1}{2} \angle B O C=\angle B A C$. From this we can conclude that $B D E A$ is cyclic.

Solution 4 (PSC). This is a simplified variant of Solution 1. $\angle C O B=2 \angle A$ (angle at centre of circle $A B C$ ) and $O B=O C$ so $\angle O B C=\angle B C O=90^{\circ}-\angle A$. Likewise $\angle E K F=2 \angle A$ and $\angle K F E=\angle F E K=90^{\circ}-\angle A$. Now because $D E \perp C O, \angle E D C=90^{\circ}-\angle D C O=90^{\circ}-\angle B C O=\angle A$ and similarly $\angle B D F=\angle A$, so $\angle F D E=180^{\circ}-2 \angle A$. So quadrilateral $K F D E$ is cyclic (opposite angles), so (same segment) $\angle K D E=$ $\angle K F E=90^{\circ}-\angle A$, so $\angle K D C=90^{\circ}$ and $D K$ is perpendicular to $B C$.

Problem 2. Let $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $\left[a_{1}, a_{2}, \ldots, a_{n}\right]$ and $\left[b_{1}, b_{2}, \ldots, b_{n}\right]$ the following holds:

$$
\max \left(\left|a_{1}-b_{1}\right|,\left|a_{2}-b_{2}\right|, \ldots,\left|a_{n}-b_{n}\right|\right)=1
$$

Origin. Poland (Tomasz Kobos).
Solution 1 (submitter). The largest possible $m$ is equal to $2^{n}$.
In order to see that the value $2^{n}$ can be indeed achieved, consider all binary vectors of length $n$ as rows of the table. We now proceed with proving that this is the maximum value.

Let $\left[a_{k}^{i}\right]$ be a feasible table, where $i=1, \ldots, m$ and $k=1, \ldots, n$. Let us define undirected graphs $G_{1}, G_{2}$, $\ldots, G_{n}$, each with vertex set $\{1,2, \ldots, m\}$, where $i j \in E\left(G_{k}\right)$ if and only if $\left|a_{k}^{i}-a_{k}^{j}\right|=1$ (by $E\left(G_{k}\right)$ we denote the edge set of the graph $G_{k}$ ). Observe the following two properties.
(1) Each graph $G_{k}$ is bipartite. Indeed, if it contained a cycle of odd length, then the sum of $\pm 1$ along this cycle would need to be equal to 0 , which contradicts the length of the cycle being odd.
(2) For every $i \neq j, i j \in E\left(G_{k}\right)$ for some $k$. This follows directly from the problem statement.

For every graph $G_{k}$ fix some bipartition $\left(A_{k}, B_{k}\right)$ of $\{1,2, \ldots, m\}$, i.e., a partition of $\{1,2, \ldots, m\}$ into two disjoint sets $A_{k}, B_{k}$ such that the edges of $G_{k}$ traverse only between $A_{k}$ and $B_{k}$. If $m>2^{n}$, then there are two distinct indices $i, j$ such that they belong to exactly the same parts $A_{k}, B_{k}$, that is, $i \in A_{k}$ if and only if $j \in A_{k}$ for all $k=1,2, \ldots, n$. However, this means that the edge $i j$ cannot be present in any of the graphs $G_{1}, G_{2}$, $\ldots, G_{n}$, which contradicts (2). Therefore, $m \leq 2^{n}$.

Solution 2 (PSC). In any table with the given property, the least and greatest values in a column cannot differ by more than 1 . Thus, if each value that is neither least nor greatest in its column is changed to be equal to either the least or the greatest value in its column (arbitrarily), this does not affect any $\left|a_{i}-b_{i}\right|=1$, nor does it increase any difference above 1 , so the table still has that given property. But after such a change, for any choice of what the least and greatest values in each column are, there are only two possible choices for each entry in the table (either the least or the greatest value in its column); that is, only $2^{n}$ possible distinct rows, and the given property implies that all rows must be distinct. As in the previous solution, we see that this number can be achieved.

Solution 3 (Coordinators). We prove by induction on $n$ that $m \leq 2^{n}$.
First suppose $n=1$. If real numbers $x$ and $y$ have $|x-y|=1$ then $\lfloor x\rfloor$ and $\lfloor y\rfloor$ have opposite parities and hence it is impossible to find three real numbers with all differences 1 . Thus $m \leq 2$.

Suppose instead $n>1$. Let $a$ be the smallest number appearing in the first column of the table; then every entry in the first column of the table lies in the interval $[a, a+1]$. Let $A$ be the collection of rows with first entry $a$ and $B$ be the collection of rows with first entry in $(a, a+1]$. No two rows in $A$ differ by 1 in their first entries, so if we list the rows in $A$ and delete their first entries we obtain a table satisfying the conditions of the problem with $n$ replaced by $n-1$; thus, by the induction hypothesis, there are at most $2^{n-1}$ rows in $A$. Similarly, there are at most $2^{n-1}$ rows in $B$. Hence $m \leq 2^{n-1}+2^{n-1}=2^{n}$. As before, this number can be achieved.

Solution 4 (Coordinators). Consider the rows of the table as points of $\mathbb{R}^{n}$. As the values in each column differ by at most 1 , these points must lie in some $n$-dimensional unit cube $C$. Consider the unit cubes centred on each of the $m$ points. The conditions of the problem imply that the interiors of these unit cubes are pairwise disjoint. But now $C$ has volume 1, and each of these cubes intersects $C$ in volume at least $2^{-n}$ : indeed, if the unit cube centred on a point of $C$ is divided into $2^{n}$ cubes of equal size then one of these cubes must lie entirely within $C$. Hence $m \leq 2^{n}$. As before, this number can be achieved.

Solution 5 (Coordinators). Again consider the rows of the table as points of $\mathbb{R}^{n}$. The conditions of the problem imply that these points must all lie in some $n$-dimensional unit cube $C$, but no two of the points lie in any smaller cube. Thus if $C$ is divided into $2^{n}$ equally-sized subcubes, each of these subcubes contains at most one row of the table, giving $m \leq 2^{n}$. As before, this number can be achieved.

Problem 3. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(y f(x+y)+f(x))=4 x+2 y f(x+y)
$$

for all $x, y \in \mathbb{R}$.

Origin. Netherlands (Birgit van Dalen).
Solution 1 (submitter). Setting $y=0$ yields

$$
f(f(x))=4 x
$$

from which we derive that $f$ is a bijective function. Also, we find that

$$
f(0)=f(4 \cdot 0)=f(f(f(0)))=4 f(0)
$$

hence $f(0)=0$. Now set $x=0$ and $y=1$ in the given equation and use (1) again:

$$
4=f(f(1))=2 f(1)
$$

so $f(1)=2$ and therefore also $f(2)=f(f(1))=4$. Finally substitute $y=1-x$ in the equation:

$$
f(2(1-x)+f(x))=4 x+4(1-x)=4=f(2) \quad \text { for all } x \in \mathbb{R}
$$

As $f$ is injective, from this it follows that $f(x)=2-2(1-x)=2 x$. It is easy to see that this function satisfies the original equation. Hence the only solution is the function defined by $f(x)=2 x$ for all $x \in \mathbb{R}$.

Solution 2 (Coordinators). Setting $y=0$ in the equation we see

$$
f(f(x))=4 x
$$

so $f$ is a bijection. Let $\kappa=f^{-1}(2)$ and set $x+y=\kappa$ in the original equation to see

$$
f(2 \kappa-2 x+f(x))=4 \kappa
$$

As the right hand side is independent of $x$ and $f$ is injective, $2 \kappa-2 x+f(x)$ is constant, i.e. $f(x)=2 x+\alpha$.
Substituting this into the original equation, we see that $2 x+\alpha$ is a solution to the original equation if and only if $4\left(y^{2}+x y+x\right)+(3+2 y) \alpha=4\left(y^{2}+x y+x\right)+2 y \alpha$ for all $x, y$, i.e. if and only if $\alpha=0$. Thus the unique solution to the equation is $f(x)=2 x$.

Problem 4. A set $A$ of integers is called sum-full if $A \subseteq A+A$, i.e. each element $a \in A$ is the sum of some pair of (not necessarily different) elements $b, c \in A$. A set $A$ of integers is said to be zero-sum-free if 0 is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of $A$.

Does there exist a sum-full zero-sum-free set of integers?

Origin. Romania (Dan Schwarz).
Remark. The original formulation of this problem had a weaker definition of zero-sum-free that did not require all nonzero integers to be sums of finite nonempty subsets of $A$.

Solution (submitter, adapted). The set $A=\left\{F_{2 n}: n=1,2, \ldots\right\} \cup\left\{-F_{2 n+1}: n=1,2, \ldots\right\}$, where $F_{k}$ is the $k^{\text {th }}$ Fibonacci number $\left(F_{1}=1, F_{2}=1, F_{k+2}=F_{k+1}+F_{k}\right.$ for $k \geq 1$ ) qualifies for an example. We then have $F_{2 n}=F_{2 n+2}+\left(-F_{2 n+1}\right)$ and $-F_{2 n+1}=\left(-F_{2 n+3}\right)+F_{2 n+2}$ for all $n \geq 1$, so $A$ is sum-full (and even with unique representations). On the other hand, we can never have

$$
0=\sum_{i=1}^{s} F_{2 n_{i}}-\sum_{j=1}^{t} F_{2 n_{j}+1}
$$

owing to the fact that Zeckendorf representations are known to be unique.
It remains to be shown that all nonzero values can be represented as sums of distinct numbers $1,-2,3$, $-5,8,-13,21, \ldots$. This may be done using a greedy algorithm: when representing $n$, the number largest in magnitude that is used is the element $m= \pm F_{k}$ of $A$ that is closest to 0 subject to having the same sign as $n$ and $|m| \geq|n|$. That this algorithm terminates without using any member of $A$ twice is a straightforward induction on $k$; the base case is $k=2(m=1)$ and the induction hypothesis is that for all $n$ for which the above algorithm starts with $\pm F_{\ell}$ with $\ell \leq k$, it terminates without having used any member of $A$ twice and without having used any $\pm F_{j}$ with $j>\ell$.

Remark (James Aaronson and Adam P Goucher). Let $n$ be a positive integer, and write $u=2^{n}$; we claim that the set

$$
\left\{1,2,4, \ldots, 2^{n-1},-u, u+1,-(2 u+1), 3 u+2,-(5 u+3), 8 u+5, \ldots\right\}
$$

is a sum-full zero-sum-free set. The proof is similar to that used for the standard examples.