Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set

$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-1 \\
x_{n, k+1} & =\frac{1}{2}\left(x_{n k}+x_{1 k}\right)
\end{aligned}
$$

Show that for $n$ odd, $x_{j k}$ is not an integer for some $j, k$. Does the same conclusion hold for $n$ even?

Solution:
We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$. If exactly $p$
consequtive numbers $x_{j k}$ equal $M_{k}$, then exactly $p-1$ consequtive numbers $x_{j, k+1}$ equal $M_{k+1}$ which is equal to $M_{k}$. So after a finite number of steps we arrive at the situation $M_{k+1}<M_{k}$. In a corresponding manner we see that $m_{k+1}>m_{k}$ for some $k$ 's. If all the numbers in all the sequences are integers, then all $m_{k}$ 's and $M_{k}$ 's are integers. So after a finite number of steps $m_{k}=M_{k}$, and all numbers $x_{j k}$ are equal. Then $x_{1, k-1}+x_{2, k-1}=x_{2, k-1}+x_{3, k-1}=\cdots=x_{n-1, k-1}+x_{n, k-1}=x_{n, k-1}+x_{1, k-1}$. If $n$ is odd, then $x_{1, k-1}=x_{3, k-1}=\cdots=x_{n, k-1}$ and $x_{1, k-1}=x_{n-1, k-1}=\cdots=x_{2, k-1}$. But then we could show in a similar way that all numbers $x_{j, k-2}$ are equal and finally that all numbers $x_{j, 1}$ are equal, contrary to the assumption. If $n$ is even, then all $x_{i, k}$ 's can be integers. Take, for instance, $x_{1,1}=x_{3,1}=\cdots=x_{n-1,1}=0, x_{2,1}=x_{4,1}=\cdots=x_{n, 1}=2$. Then every $x_{j, k}=1, k \geq 2$.


Let $a, b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that

$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$

Solution:
By the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as

$$
A=\frac{r}{2}(a+b+c)
$$

and, by the sine theorem, as

$$
A=\frac{1}{2} a b \sin \gamma=\frac{1}{4} \frac{a b c}{R}
$$

Combining these, we obtain

$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{a+b+c}{a b c}=\frac{2 A}{r} \cdot \frac{1}{4 R A}=\frac{1}{2 r R} \geq \frac{1}{R^{2}}
$$


Solution:
Assume $a \leq b \leq c$. Then $b=a+x$ and $c=a+x+y, x \geq 0, y \geq 0$. Now $a b c-(a+b-c)(a-b+c)(-a+b+c)=a(a+x)(a+x+y)-(a-y)(a+2 x+y)(a+y)=a x^{2}+$ $a x y+a y^{2}+2 x y^{2}+y^{3} \geq 0$. So $a b c(a+b+c) \geq(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=16 A^{2}$, where the last inequality is implied by Heron's formula. When we substitute $A=\frac{a b c}{4 R}$ (see Solution 1) we obtain, after simplification,

$$
a+b+c \geq \frac{a b c}{R^{2}}
$$

which is equivalent to the claim.


Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals

$$
\frac{25}{8} k-211
$$

Solution:
Let

$$
a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9
$$

Set

$$
f(a)=\prod_{k=0}^{n} a_{k}
$$

Since

$$
f(a)=\frac{25}{8} a-211 \geq 0
$$

$a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}(8,25)=1$, so $8 \mid a$. On the other hand,

$$
f(a) \leq 9^{n-1} a_{n} \leq 10^{n} a_{n} \leq a
$$

So

$$
\frac{25}{8} a-211 \leq a
$$

or $a \leq \frac{8}{17} \cdot 211=\frac{1688}{17}<100$. The only multiples of 8 between 66 and 100 are $72,80,88$, and 96. Now $25 \cdot 9-211=17=7 \cdot 2,25 \cdot 10-211=39 \neq 8 \cdot 0,25 \cdot 11-211=64=8 \cdot 8$, and $25 \cdot 12-211=89 \neq 9 \cdot 6$. So 72 and 88 are the numbers asked for.


Let $a, b$, and $c$ be positive real numbers. Prove that

$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$

Solution:
Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality

$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^{4}+b^{4}-2 a^{2} b^{2}+b^{4}+c^{4}-2 b^{2} c^{2}+c^{4}+a^{4}-2 a^{2} c^{2} \\
+a b\left(a^{2}+b^{2}-2 c^{2}\right)+b c\left(b^{2}+c^{2}-2 a^{2}\right)+c a\left(c^{2}+a^{2}-2 b^{2}\right) \\
=\left(a^{2}-b^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2} \\
+a b(a-b)^{2}+b c(b-c)^{2}+c a(c-a)^{2} \\
+a b\left(2 a b-2 c^{2}\right)+b c\left(2 b c-2 a^{2}\right)+c a\left(2 c a-2 b^{2}\right)
\end{gathered}
$$

The six first terms on the right hand side are non-negative and the last three can be written as

$$
\begin{gathered}
2 a^{2} b^{2}-2 a b c^{2}+2 b^{2} c^{2}-2 a^{2} b c+2 c^{2} a^{2}-2 a b^{2} c \\
=a^{2}\left(b^{2}+c^{2}-2 b c\right)+b^{2}\left(a^{2}+c^{2}-2 a c\right)+c^{2}\left(a^{2}+b^{2}-2 a b\right) \\
=a^{2}(b-c)^{2}+b^{2}(c-a)^{2}+c^{2}(a-b)^{2} \geq 0
\end{gathered}
$$

So the original inequality is true.


Solution:
The inequality is equivalent to

$$
\begin{gathered}
2\left(a^{2}(a+b)(a+c)+b^{2}(b+c)(b+a)+c^{2}(c+a)(c+b)\right) \\
\geq(a+b+c)(a+b)(b+c)(c+a)
\end{gathered}
$$

The left hand side can be factored as $2(a+b+c)\left(a^{3}+b^{3}+c^{3}+a b c\right)$. Because $a+b+c$ is positive, the inequality is equivalent to

$$
2\left(a^{3}+b^{3}+c^{3}+a b c\right) \geq(a+b)(b+c)(c+a)
$$

After expanding the right hand side and subtracting $2 a b c$, we get the inequality

$$
2\left(a^{3}+b^{3}+c^{3}\right) \geq\left(a^{2} b+b^{2} c+c^{2} a\right)+\left(a^{2} c+b^{2} a+c^{2} b\right)
$$

still equivalent to the original one. But we now have two instances of the well-known inequality $x^{3}+y^{3}+z^{3} \geq x^{2} y+y^{2} z+z^{2} x$ or $x^{2}(x-y)+y^{2}(y-z)+z^{2}(z-x) \geq 0$. [Proof: We may assume $x \geq y, x \geq z$. If $y \geq z$, write $z-x=z-y+y-z$ to obtain the equivalent and true inequality $\left(y^{2}-z^{2}\right)(y-z)+\left(x^{2}-z^{2}\right)(x-y) \geq 0$, if $z \geq y$, similarly write $x-y=x-z+z-y$, and get $\left(x^{2}-z^{2}\right)(x-z)+\left(x^{2}-y^{2}\right)(z-y) \geq 0$.]


Solution:
The original inequality is symmetric in $a, b, c$. So we may assume $a \geq b \geq c$, which implies

$$
\frac{1}{b+c} \geq \frac{1}{c+a} \geq \frac{1}{a+b}
$$

The power mean inequality gives

$$
\frac{a^{2}+b^{2}+c^{2}}{3} \geq\left(\frac{a+b+c}{3}\right)^{2}
$$

We combine this and the Chebyshev inequality to obtain

$$
\begin{gathered}
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \\
\geq \frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \\
\geq \frac{2}{9}(a+b+c)^{2}\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)
\end{gathered}
$$

To complete the proof, we have to show that

$$
2(a+b+c)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geq 9
$$

But this is equivalent to the harmonic-arithmetic mean inequality

$$
\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq \frac{x+y+z}{3}
$$

with $x=a+b, y=b+c, z=c+a$.

05.3 There are 2005 young people sitting around a (large!) round table. Of these at most 668 are boys. We say that a girl $G$ is in a strong position, if, counting from $G$ to either direction at any length, the number of girls is always strictly larger than the number of boys. (G herself is included in the count.) Prove that in any arrangement, there always is a girl in a strong position.


Solution: Assume the number of girls to be $g$ and the number of boys $b$. Call a position clockwise fairly strong, if, counting clockwise, the number of girls always exceeds the number of boys. No girl immediately followed by a boy has a fairly strong position. But no pair consisting of a girl and a boy following her has any effect on the fairly strongness of the other positions. So we may remove all such pairs. so we are left with at least $g-b$ girls, all in a clockwise fairly strong position. A similar count of counterclockwise fairly strong positions can be given, yielding at least $g-b$ girls in such a position. Now a sufficient condition for the existence of a girl in a strong position is that the sets consisting of the girls in clockwise and counterclockwise fairly strong position is nonempty. This is certainly true if $2(g-b)>g$, or $g>2 b$. With the numbers in the problem, this is true.


The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathcal{C}_{1}$ passing through $C$ touch $\mathcal{C}_{1}$ at $F$ and $G$. Prove that $D$, $E, F$, and $G$ are concyclic.

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-57.jpg?height=571&width=529&top_left_y=1254&top_left_x=615)

Figure 15.

Solution:
(See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are equal, and $C H \| B E$. So $C$ is the midpoint of the arc $D E$. This again implies the equality of the angles $C E B$ and $B A E$, as well as $C E=C D$. So the triangles $A E C, C E B$, having also a common angle $E C B$, are similar. So

$$
\frac{C B}{C E}=\frac{C E}{A C}
$$

and $C B \cdot A C=C E^{2}=C D^{2}$. But by the power of a point theorem, $C B \cdot C A=C G^{2}=C F^{2}$. We have in fact proved $C D=C E=C F=C G$, so the four points are indeed concyclic.

06.1 Let $B$ and $C$ be points on two fixed rays emanating from a point $A$ such that $A B+A C$ is constant. Prove that there exists a point $D \neq A$ such that the circumcircles of the triangels $A B C$ pass through $D$ for every choice of $B$ and $C$.

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-58.jpg?height=459&width=560&top_left_y=513&top_left_x=611)

Figure 16 .


Solution: (See Figure 16.) Let $E$ and $F$ be the points on rays $A B$ and $A C$, respectively, such that $A E=A F=A B+A C$. Let the perpendicular bisectors of the segments $A E$ and $A F$ intersect at $D$. It is easy to see, for instance from the right triangles with $A D$ as the common hypothenuse and the projections of $A D$ on $A B$ and $A C$ as legs, that $D$ lies on the angle bisector of angle $B A C$. Moreover, $\angle A D F=180^{\circ}-2 \cdot \angle C A D=180^{\circ}-\angle B A C$. The triangle $A D F$ is isosceles, so $\angle B A D=\angle D A C=\angle C F D$ and $A D=D F$ in the triangles $A B D$ and $D C F$. Moreover, we know that $C F=A F-A C=A B$. The triangles $A D B$ and $F D C$ are congruent (sas). So $\angle B D A=\angle C D F$. But this implies $\angle B D C=\angle A D F=$ $180^{\circ}-\angle B A C$. This is sufficient for $A B D C$ to be an inscribed quadrilateral, and the claim has been proved.


The real numbers $x, y$ and $z$ are not all equal and satisfy

$$
x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}=k
$$

Determine all possible values of $k$.

Solution:
Let $(x, y, z)$ be a solution of the system of equations Since

$$
x=k-\frac{1}{y}=\frac{k y-1}{y} \quad \text { and } \quad z=\frac{1}{k-y}
$$

the equation

$$
\frac{1}{k-y}+\frac{y}{k y-1}=k
$$

to be simplified into

$$
\left(1-k^{2}\right)\left(y^{2}-k y+1\right)=0
$$

is true. So either $|k|=1$ or

$$
k=y+\frac{1}{y}
$$

The latter alternative, substituted to the original equations, yields immediately $x=y$ and $z=y$. So $k= \pm 1$ is the only possibility. If $k=1$, for instance $x=2, y=-1$ and $z=\frac{1}{2}$ is a solution; if $k=-1$, a solution is obtained by reversing the signs for a solution with $k=1$. So $k=1$ and $k=-1$ are the only possible values for $k$.


A sequence of positive integers $\left\{a_{n}\right\}$ is given by

$$
a_{0}=m \quad \text { and } \quad a_{n+1}=a_{n}^{5}+487
$$

for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible.

Solution: Consider the expression $x^{5}+487$ modulo 4. Clearly $x \equiv 0 \Rightarrow x^{5}+487 \equiv 3$, $x \equiv 1 \Rightarrow x^{5}+487 \equiv 0 ; x \equiv 2 \Rightarrow x^{5}+487 \equiv 3$, and $x \equiv 3 \Rightarrow x^{5}+487 \equiv 2$. Square numbers are always $\equiv 0$ or $\equiv 1 \bmod 4$. If there is an even square in the sequence, then all subsequent numbers of the sequence are either $\equiv 2$ or $\equiv 3 \bmod 4$, and hence not squares. If there is an odd square in the sequence, then the following number in the sequence can be an even square, but then none of the other numbers are squares. So the maximal number of squares in the sequence is two. In this case the first number of the sequence has to be the first square, since no number of the sequence following another one satisfies $x \equiv 1 \bmod 4$. We have to find numbers $k^{2}$ such that $k^{10}+487=n^{2}$. We factorize $n^{2}-k^{10}$. Because 487 is a prime, $n-k^{5}=1$ and $n+k^{5}=487$ or $n=244$ and $k=3$. The only solution of the problem thus is $m=3^{2}=9$.


The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used.

Solution:
Denote by $R_{i}$ the number of colours used to colour the squares of the $i$ 'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$ 'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appears. By the arithmetic-geometric inequality, $r_{k}+c_{k} \geq 2 \sqrt{r_{k} c_{k}}$. Since colour $k$ appears at most $c_{k}$ times on each of the $r_{k}$ columns on which it can be found, $c_{k} r_{k}$ must be at least the total number of occurences of colour $k$, which equals 100 . So $r_{k}+c_{k} \geq 20$. In the sum $\sum_{i=1}^{100} R_{i}$, each colour $k$ contributes $r_{k}$ times and in the sum $\sum_{j=1}^{100} C_{j}$ each colour $k$ contributes $c_{k}$ times. Hence

$$
\sum_{i=1}^{100} R_{i}+\sum_{j=1}^{100} C_{j}=\sum_{k=1}^{100} r_{k}+\sum_{k=1}^{100} c_{k}=\sum_{k=1}^{100}\left(r_{k}+c_{k}\right) \geq 2000
$$

But if the sum of 200 positive integers is at least 2000, at least one of the summands is at least 10. The claim has been proved.


Find one solution in positive integers to the equation

$$
x^{2}-2 x-2007 y^{2}=0
$$

Solution:
The equation can be written in the form

$$
x(x 2)=223 \cdot(3 y)^{2}
$$

Here the prime number 223 must divide $x$ or $x 2$. In fact, for $x=225$ we get $x(x 2)=$ $15^{2} \cdot 223$, which is equivalent to $223 \cdot(3 y)^{2}$ for $y=5$. Thus, $(x, y)=(225,5)$ is one solution.


A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle.

Solution: Take any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then the whole segment between them, $P$ included, is in the same rectangle. If the two points, say $Q$ and $R$, are on perpendicular lines, the perpendicular segments $R P$ and $P Q$ are also in the same rectangle. So in any case, $P$ is in one of the rectangles.


The number $10^{2007}$ is written on a blackboard, Anne and Berit play a game where the player in turn makes one of two operations:

(i) replace a number $x$ on the blackboard by two integer numbers a and $b$ greater than 1 such that $x=a b$;

(ii) erase one or both of two equal numbers on the blackboard.

The player who is not able to make her turn loses the game. Who has a winning strategy?

Solution:
We describe a winning strategy for Anne. Her first move is

$$
10^{2007} \rightarrow 2^{2007}, 5^{2007}
$$

We want to show that Anne can act in such a way that the numbers on the blackboard after each of her moves are of the form

$$
2^{\alpha_{1}}, \ldots, 2^{\alpha_{k}}, 5^{\alpha_{1}}, \ldots, 5^{\alpha_{k}}
$$

This is the case after Anne's first move. If Berit for example replaces $2^{\alpha_{j}}$ by $2^{\beta_{1}}$ and $2^{\beta_{2}}$, then Anne would replace $5^{\alpha_{j}}$ by $5^{\alpha_{1}}$ and $5^{\alpha_{2}}$. If Berit for example erases $5^{\alpha_{j}}$ or two $5^{\alpha_{j}}$ 's, (which means that there is an $\alpha_{i}=\alpha_{j}$ ) then Anne would erase $2^{\alpha_{j}}$ or $22^{\alpha_{j}}$ 's. Thus for each move Berit makes, Anne can answer with a 'symmetric' move. Since the game is finite, Berit must be the first player failing to make a move. Thus Anne has a winning strategy.


A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and AC. Show that $A P / P C=2 \cdot A B / B C$.

Solution:
First we show that if we fix the points $A, B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_{1}$ and $T_{1}$, and the tangents to the other circle, meeting that circle at points $S_{2}$ and $T_{2}$. Then, according to the power of a point theorem

$$
A S_{1}^{2}=A T_{1}^{2}=A B \cdot A C=A S_{2}^{2}=A T_{2}^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-61.jpg?height=527&width=592&top_left_y=502&top_left_x=1200)

This implies that all the tangent points $S_{1}, T_{1}, S_{2}$ and $T_{2}$ lie on the same circle with center $A$. Let $Q$ be the intersection of $S_{1} T_{1}$ and $S_{2} T_{2}$. Then by applying again the theorem of a power of a point but now with respect to the circle with center A, we have that $Q S_{1} \cdot Q T_{1}=Q S_{2} \cdot Q T_{2}$. But this in turn means that the point $Q$ has the same power with respect to the two circles we started with, and hence lies on the radical axis of those two circles, that is, the line $B C$ (the radical axis is the locus of points of equal power with respect to two given circles). So $Q$ is the intersection of $A C$ and both $S_{1} T_{1}$ and $S_{2} T_{2}$, which proves that the intersection point defined in the problem is the same for both circles.

Since the location of $P$ is independent of the circle through $B$ and $C$ we can, without loss of generality, choose the circle with $B C$ as diameter. Let $O$ be the center of this circle, $R$ its radius, $d=A O$, and $r=P O$. Then the triangles $A S O$ and $S P O$ are similar, so $O S / A O=P O / O S$, that is, $R / d=r / R$, or $R^{2}=d r$. Then finally we have

$$
\frac{A P}{P C}=\frac{d-r}{R+r}=\frac{d^{2}-d r}{d R+d r}=\frac{d^{2}-{ }^{R} 2}{d r+r^{2}}=\frac{d-R}{R}=2 \cdot \frac{d-R}{2 R}=2 \cdot \frac{A B}{B C}
$$


Determine all real numbers $A, B$ and $C$ such that there exists a real function $f$ that satisfies

$$
f(x+f(y))=A x+B y+C
$$

for all real $x$ and $y$.

Solution:
Let $A, B$ and $C$ be real numbers and $f$ a function such that $f(x+f(y))=$ $A x+B y+C$ for all $x$ and $y$. Let $z$ be a real number and set $x=z-f(0)$ and $y=0$. Then

$$
f(z)=f(z-f(0)+f(0))=A(z-f(0))+B \cdot 0+C=A z-A f(0)+C
$$

so there are numbers $a$ and $b$ such that $f(z)=a z+b$ for all $z$. Now $f(x+f(g))=$ $a x+a^{2} y+(a+1) b$, and $(A, B, C)=\left(a, a^{2},(a+1) b\right)$, where $a$ and $b$ are arbitrary real numbers, that is, $(A, B, C)=\left(a, a^{2}, c\right)$, where $a \neq-1$ and $c$ are arbitrary, or $(A, B, C)=(-1,1,0)$


Assume that $n \geq 3$ people with different names sit around a round table. We call any unordered pair of them, say $M$ and $N$, dominating, if

(i) $M$ and $N$ do not sit on adjacent seats, and

(ii) on one (or both) of the arcs connecting $M$ and $N$ along the table edge, all people have names that come alphabetically after the names of $M$ and $N$.

Determine the minimal number of dominating pairs.

Solution: We will show by induction that the number of dominating pairs (hence also the minimal number of dominating pairs) is $n-3$ for $n \geq 3$. If $n=3$, all pairs of people sit on adjacent seats, so there are no dominating pairs. Assume that the number of dominating pairs is $n-3$ for some $n>3$. If there are $n+1$ people around the table, let the person whose name is alphabetically last leave the table. The two people sitting next to that person, who formed a dominating pair, no longer do. On the other hand, any other dominating pair remains a dominating pair in the new configuration of $n$ people, and any dominating pair in the new configuration was also a dominating pair in the old. The number of dominating pairs in the new configuration is $n-3$, so the number in the old was $(n+1)-3$.


Let $A B C$ be a triangle and let $D$ and $E$ be points on $B C$ and $C A$, respectively, such that $A D$ and $B E$ are angle bisectors of $A B C$. Let $F$ and $G$ be points on the circumcircle of $A B C$ such that $A F$ and $D E$ are parallel and $F G$ and $B C$ are parallel. Show that

$$
\frac{A G}{B G}=\frac{A C+B C}{A B+C B}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-62.jpg?height=460&width=462&top_left_y=1135&top_left_x=1254)

Solution: Let $A B=c, B C=a$ and $C A=b$. Then it follows from the angle bisector theorem that $C D=$ $a b /(b+c)$. Similarly, $C E=a b /(a+c)$, so $C D / C E=(a+c) /(b+c)$. The angles $\angle A B G, \angle A F G$ and $\angle E D C$ are equal, and so are $\angle A G B$ and $\angle A C B$, and consequently, the triangles $C E D$ and $G A B$ are similar. The conclusion follows.


The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares.

Solution: Assume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2 m+1)^{2}=(2 n+$ $1)(2 n-1)$. Since $2 n+1$ and $2 n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is 2 , but they are both odd), one of them is a square (of an odd integer, since it is odd) and the other divided by 3 is a square. An odd number squared minus 1 is divisible by 4 since $(2 t+1)^{2}-1=4\left(t^{2}+t\right)$. From the first equation we see that $n$ is odd, say $n=2 k+1$. Then $2 n+1=4 k+3$, so the square must be $2 n-1$, say $2 n-1=(2 t+1)^{2}$. Rearrangement yields $n=t^{2}+(t+1)^{2}$. (An example: $8^{3}-7^{3}=\left(2^{2}+3^{2}\right)^{2}$. $)$


A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller
triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the given triangle. Show that $f \geq \frac{1}{3}$ and determine those points $P$ for which $f=\frac{1}{3}$.

Solution:
Let $A B C$ be the triangle and let the lines through $P$ parallel to its sides intersect the sides in the points $D, E ; F, G$ and $H, I$. The triangles $A B C$, $D E P, P F G$ and $I P H$ are similar and $B D=I P$, $E C=P F$. If $B C=a, I P=a_{1}, D E=a_{2}$ ja $P F=a_{3}$, then $a_{1}+a_{2}+a_{3}=a$. There is a posi-

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-63.jpg?height=271&width=485&top_left_y=521&top_left_x=1254)
tive $k$ such that the areas of the triangles are $k a^{2}, k a_{1}^{2}$, $k a_{2}^{2}$ and $k a_{3}^{2}$. But then

$$
f=\frac{k a_{1}^{2}+k a_{2}^{2}+k a_{3}^{2}}{k a^{2}}=\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{\left(a_{1}+a_{2}+a_{3}\right)^{2}}
$$

By the arithmetic-quadratic inequality,

$$
\frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{9} \leq \frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{3}
$$

where equality holds if and only if $a_{1}=a_{2}=a_{3}$. It is easy to see that $a_{1}=a_{2}=a_{3}$ implies that $P$ is the centroid of $A B C$. So $f \geq \frac{1}{3}$, and $f=\frac{1}{3}$ if and only if $P$ is the centroid of $A B C$.


On a faded piece of paper it is possible, with some effort, to discern the following:

$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$

Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the question: What is the value of the constant term a? We assume that all polynomials in the statement above have only integer coefficients.

Solution: We denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$. For $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \cdot 3 \cdot 3 \cdot 5$. For $x=-1$ we get $P_{a}(-1)=-184$, so $a$ is also a divisor of $184=2 \cdot 2 \cdot 2 \cdot 23$. But the only prime factor in common is 2 . So the only possibilities for $a$ are $\pm 2$ and $\pm 1$. If $a=1$, we get for $x=1$ that $P_{1}(1)=3$, while $R(1)=4-180=-176$, which cannot be divided by 3 . If $a=-2$ we get for $x=1$ that $P_{2}(1)=0$, i.e. the left side is equal to 0 , while the right side is equal to $R(1)=-176$, which is different from 0 . Neither $a=1$ nor $a=-2$ will thus work. It remains to check $a=2$ and $a=-1$. Before we use the procedure above again, we need a factorization of $R(x)$. We observe that $x^{4}+1$ is a divisor of $R(x)$, since the right side may be written as $\left(x^{4}+1\right)\left(x^{13}+x-90\right)$. If $a=-1$ we get for $x=2$ that $P_{1}(2)=5$, while $x^{4}+1=17$ and $x^{13}+x-90=8104$. So the right hand side is not divisible by 5 . Now, the only remaining possibility is $a=2$, i.e. $x^{2}+x+2$ is a divisor of $R(x)$.


The integers 1, 2, 3, 4 and 5 are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a+b$ and $a b$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are 2009?

Solution: The answer is no. First notice that in each move two integers will be replaced with two greater integers (except in the case where the number 1 is wiped out). Notice also that from the start there are three odd integers. If one chooses to replace two odd integers on the blackboard, the number of odd integers on the blackboard decreases. If one chooses to replace two integers, which are not both odd, the number of odd integers on the blackboard is unchanged. To end up in a situation, where three of the integers on the blackboard are 2009, then it is not allowed in any move to replace two odd integers. Hence the number 2009 can only be obtained as a sum $a+b$. In the first move that gives the integer 2009 on the blackboard, one has to choose $a$ and $b$ such that $a+b=2009$. In this case either $a b>2009$ or $a b=2008$. In the case $a b=2008$, one of the factors is equal to 1 , and hence 1 does no longer appear on the blackboard. The two integers $a+b=2009$ and $a b$ that appears in the creation of the first 2009 cannot be used any more in creation of the remaining two integers of 2009. Also the next 2009 can only be obtained if one chooses c and d such that $c+d=2009$ and $c d>2009$ or $c d=2008$, and in the last case 1 does not appear any longer on the blackboard. The numbers $c+d=2009$ and $c d$ cannot be used in obtaining the last integer 2009. Hence the last integer 2009 cannot be obtained.


There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches.

Solution: To determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether. Now the silver medal winner has at some point lost to number 1 ; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be the candidate who lost to the gold medalist in round $i$. Now let $C_{l}$ and $C_{2}$ play, the winner then play with $C_{3}$ etc. After 4 matches we know the silver medalist; assume she was $C_{k}$. Now the bronze medalist must have lost against the gold medalist or against $C_{k}$ or both. (If she lost to someone else, this someone else was below the second place). Now the silver medalist $C_{k}$ won $k-1$ times in the first rounds and the $5-k$ players $C_{k+1}, \ldots, C_{5}$, and if $\mathrm{k} i 1$ one player $C_{j}$ with $j<k$. So there are either $k-1+5-k=4$ or 5 candidates for the third place. At most 4 matches are again needed to determine the bronze winner.


A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$is the set of positive integers, is non-decreasing and satisfies $f(m n)=f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq(f(10))^{2}$.

Solution: Since $\mathrm{f}$ is non-decreasing, $f(91) \geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \geq f(9) f(10)$. Also $f(72) \geq f(70)$, and therefore $f(8) f(9) \geq f(7) f(10)$. Since all values of $\mathrm{f}$ are positive, we get $f(8) f(9) \cdot f(13) f(7) \geq$ $f(7) f(10) \cdot f(9) f(10)$, and dividing both sides by $f(7) f(9)>0, f(8) f(13) \geq f(l 0) f(10)=$ $(f(10))^{2}$.


Three circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The
line $A O$ intersects the circle $\Gamma_{C}$ in the poin $X \neq O$. Similarly, the line $B O$ intersects the circle $\Gamma_{B}$ in the point $Y \neq O$, and the line $C O$ intersects the circle $\Gamma_{C}$ in the point $Z \neq O$. Show that

$$
\frac{|A Y||B Z||C X|}{|A Z||B X||C Y|}=1
$$

Solution:
Let $\angle A O Y=\alpha, \angle A O Z=\beta$ and $\angle Z O B=\gamma$. So $\alpha+\beta+\gamma=180^{\circ}$. Also $\angle B O X=\alpha$ (vertical angles) and $\angle A C Y=\alpha=\angle B C X$ (angles subtending equal arcs); similarly $\angle C O X=\beta$, $\angle A B Z=\beta=\angle C B X ; \angle C O Y=\gamma ; \angle B A Z=\gamma=$ $\angle C A Y$. Each of the triangles $C Y A, C B X$ and $Z B A$ have two angles from the set $\{\alpha, \beta, \gamma\}$. All triangles are then similar.

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-65.jpg?height=439&width=443&top_left_y=583&top_left_x=1275)

Similarity implies

$$
\frac{A Y}{C Y}=\frac{A B}{B Z}, \quad \frac{C X}{B X}=\frac{A Z}{A B}
$$

Consequently

$$
\frac{A Y}{A Z} \cdot \frac{B Z}{B X} \cdot \frac{C X}{C Y}=\frac{A B}{B Z} \cdot \frac{A Z}{A B} \cdot \frac{B Z}{A Z}=1
$$


Solution: Define $\alpha, \beta$ and $\gamma$ as in Solution 1. Let the radii of the circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ be $R_{A}, R_{B}$ and $R_{C}$, respectively. The sine theorem implies $A Y=2 R_{B} \sin \alpha$, $B Z=2 R_{C} \sin \gamma, C X=2 R_{A} \sin \beta, A Z=2 R_{C} \sin \beta, B X=2 R_{A} \sin \alpha$ and $C Y=2 R_{B} \sin \gamma$. Inserting these in the expression given in the problem one immediately sees that the expression has the value 1 .


Laura has 2010 lamps connected with 2010 buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons simultaneously, so the lamps are lit simultaneously, too.

a) If Richard chooses the buttons to be pressed, what is the maximum number of different combinations of buttons he can press until Laura can assign the buttons to the lamps correctly?

b) Supposing that Laura will choose the combinations of buttons to be pressed, what is the minimum number of attempts she has to do until she is able to associate the buttons with the lamps in a correct way?

Solution:
a) Let us say that two lamps are separated, if one of the lamps is turned on while the other lamp remains off. Laura can find out which lamps belong to the buttons if every two lamps are separated. Let Richard choose two arbitrary lamps. To begin with, he turns both lamps on and then varies all the other lamps in all possible ways. There are $2^{2008}$ different combinations for the remaining $2010-2=2008$ lamps. Then Richard turns
the two chosen lamps off. Also, at this time there are $2^{2008}$ combinations for the remaining lamps. Consequently, for the $2^{2009}$ combinations in all, it is not possible to separate the two lamps of the first pair. However, we cannot avoid the separation if we add one more combination. Indeed, for every pair of lamps, we see that if we turn on a combination of lamps $2^{2009}+1$ times, there must be at least one setup where exactly one of the lamps is turned on and the other is turned off. Thus, the answer is $2^{2009}+1$.

b) For every new step with a combination of lamps turned on, we get a partition of the set of lamps into smaller and smaller subsets where elements belonging to the same subset cannot be separated. In each step every subset is either unchanged or divided into two smaller parts, i.e. the total number of subsets after $\mathrm{k}$ steps will be at most $2^{k}$. We are finished when the number of subsets is equal to 2010 , so the answer is at least $\left\lceil\log _{2} 2010\right\rceil=11$. But it is easy to see that Laura certainly can choose buttons in every step in such a way that there are at most $2^{11-k}$ lamps in every part of the partition after $k$ steps. Thus, the answer is 11 .


A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n=a_{1} \pm a_{2} \pm a_{3} \pm \cdots \pm a_{k}$, where $a_{1}, \ldots, a_{k}$ are simple.

Solution: We can always write $n=a_{l}+a_{2}+\cdots+a_{9}$ where $a_{j}$ has 1 's in the places where $n$ has digits greater or equal to $j$ and 0 's in the other places. So $k \leq 9$. To show that $k \geq 9$, consider $n=10203040506070809$. Suppose $n=a_{l}+a_{2}+\cdots+a_{j}-a_{j+l}-a_{j+2}-\cdots-a_{k}$, where $a_{l}, \ldots, a_{k}$ are simple, and $k<9$. Then all digits of $b_{l}=a_{l}+\cdots+a_{j}$ are not greater than $j$ and all digits of $b_{2}=a_{j+l}+\cdots+a_{k}$ are not greater than $k-j$. We have $n+b_{2}=b_{l}$. We perform column addition of $n$ and $b_{2}$ and consider digit $j+1$ in the number $n$. There will be no carry digit coming from lower decimal places, since the sum there is less that $10 \ldots 0+88 \ldots 8=98 \ldots 8$. So in the column of $j+1$ we get the sum of $j+1$ and the corresponding digit in $b_{2}$. The resulting digit should be less than $j+1$. Thus in the corresponding place in $b_{2}$ we have at least $9-j$. But $9-j \leq k-j$, implying $k \geq 9$. Hence, we have proved that the maximal $k$ is 9 .


When $a_{0}, a_{1}, \ldots, a_{1000}$ denote digits, can the sum of the 1001-digit numbers $a_{0} a_{1} \ldots a_{1000}$ and $a_{1000} a_{999} \ldots a_{0}$ have odd digits only?

Solution:
The answer is no. The following diagram illustrates the calculation of the sum digit by digit.

| $a_{0}$ | $a_{1}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{999}$ | $a_{1000}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{1000}$ | $a_{999}$ | $\ldots$ | $a_{1000-i}$ | $\ldots$ | $a_{500}$ | $\ldots$ | $a_{i}$ | $\ldots$ | $a_{1}$ | $a_{0}$ |
| $s_{1000}$ | $s_{999}$ | $\ldots$ | $s_{1000-i}$ | $\ldots$ | $s_{500}$ | $\ldots$ | $s_{i}$ | $\ldots$ | $s_{1}$ | $s_{0}$ |

Thus $s_{i}$ are the digits of the sum. The digit $s_{\text {lool }}$ may be absent. We call column $i$ the column in the diagram with the digit $s_{i}$. Assume that $s_{i}$ is odd for $i=0,1, \ldots, 1000$. By induction on $i$ we prove that $a_{2 i}+a_{\text {looo }-2 i}$ is odd for $i=0,1, \ldots, 250$. This implies that $a_{2.250}+a_{l 000-2.250}=2 a_{500}$ is odd, which is a contradiction. Here is the proof: Since $s_{0}$ is odd, $a_{0}+a_{1000}$ is odd, so the statement is true for $\mathrm{i}=0$. Assume that $a_{2 i}+a_{1000-2 i}$ is odd for some $i \in\{0,1, \ldots, 224\}$. Since $s_{1000-2 i}$ is odd and $a_{2 i}+a_{1000-2 i}$ is odd, there is no
carry in column $1000-2 i$, so $a_{2 i+1}+a_{1000-(2 i+1)} \leq 9$. But then because $s_{2 i+1}$ is odd and $a_{2 i+1}+a_{1000-(2 i+1)} \leq 9$, there is no carry in column $2 i+2$. Hence $a_{2 i+2}+a_{1000-(2 i+2)}$ is odd because $s_{2 i+2}$ is odd. This completes the induction step.


In a triangle $A B C$ assume $A B=A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle ABC. When does equality hold?

Solution:
Because $A E \| D C$, the triangles $A B E$ and $D B C$ are similar. So

$$
C D=\frac{B C}{B E} \cdot A E
$$

$\mathrm{ja}$

$$
C D=\frac{A E \cdot B C}{B E \cdot C E} \cdot C E
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_f01d327b68eb688a4496g-67.jpg?height=340&width=508&top_left_y=732&top_left_x=1248)

Let $A F$ be an altitude of $A B C$. Then $A E \geq A F=h$, and equality holds if and only if $E=F$. Because $A B C$ is isosceles, $F$ is the midpoint of $B C$. The arithmetic-geometric mean inequality yields

$$
B E \cdot C E \leq\left(\frac{B E+E C}{2}\right)^{2}=\left(\frac{B C}{2}\right)^{2}
$$

and equality holds if and only if $E$ is the midpoint of $B C$ i.e. $E=F$. The conclusion folows when these estimates are inserted in (1); furthermore, equality is equivalent to $E=F$


Find all functions $f$ such that

$$
f(f(x)+y)=f\left(x^{2}-y\right)+4 y f(x)
$$

for all real numbers $x$ and $y$.

Solution: Substituting $y=x^{2}$ yields $f\left(f(x)+x^{2}\right)=f(0)+4 x^{2} f(x)$ for all real $x$. And $y=-f(x)$ gives $f(0)=f\left(x^{2}+f(x)\right)-4 f(x)^{2}$ for all $x$. Combining these two equations gives $4 f(x)^{2}=4 x^{4} f(x)$, so for each $x$ either $f(x)=0$ or $f(x)=x^{2}$. In particular $f(0)=0$. Now suppose there exists a real number $a \neq 0$ such that $f(a) \neq 0$. Then $f(a)=a^{2}$ and $f\left(a^{2}+y\right)=f\left(a^{2}-y\right)+4 y a^{2}$ for all $y$. If $f\left(a^{2}-y\right)=0$ for some $y \neq 0$, then $f\left(a^{2}+y\right)=4 y a^{2} \neq 0$, so $f\left(a^{2}+y\right)=\left(a^{2}+y\right)^{2}$ must hold and then $\left(a^{2}+y\right)^{2}=4 y a^{2}$, which yields $\left(a^{2}-y\right)^{2}=0$, or $a^{2}-y=0$. This shows that if $f(a) \neq 0$ for some $a$ then $f(x)=0$ only when $x=0$. So either $f(x)=0$ for all $x$ or $f(x)=x^{2}$ for all $x$. It can easily be verified that these two functions are indeed solutions to the given equation.


Show that for any integer $n \geq 2$ the sum of the fractions $\frac{1}{a b}$, where $a$ and $b$ are relatively prime positive integers such that $a<b \leq n$ and $a+b>n$, equals $\frac{1}{2}$.

Solution:
We prove this by induction. First observe that the statement holds for $n=2$, because $a=1$ and $b=2$ are the only numbers which satisfy the conditions. Next we show that increasing $n$ by 1 does not change the sum, so it remains equal to $\frac{1}{2}$. To that end it suffices to show that the sum of the terms removed from the sum equals the sum of the new terms added. All the terms in the sum for $n-1$ remain in the sum for $n$ except the fractions $\frac{1}{a b}$ with $a$ and $b$ relatively prime, $0<a<b \leq_{n}$ and $a+b=n$. On the other hand the new fractions added to the sum for $n$ have the form $\frac{1}{a n}$ with $0<a<n$. So it suffices to show

$$
\sum_{\substack{0<a<n / 2 \\(a, n-a)=1}} \frac{1}{n(a-n)}=\sum_{\substack{0<a<n \\(a, n)=1}} \frac{1}{a n}
$$

(We denote the greatest common factor of $x$ and $y$ by $(x, y)$.) Since $(a, n)=(n-a, n)$, the terms in the sum on the right hand side can be grouped into pairs

$$
\frac{1}{a n}+\frac{1}{(n-a) n}=\frac{(n-a)+a}{a(n-a) n}=\frac{1}{a(n-a)}
$$

No term is left out because if $\mathrm{n}$ is even and greater than 2 then $(n / 2, n)=\frac{n}{2}>1$. So the right hand side is given by

$$
\sum_{\substack{0<a<n \\(a, n)=1}} \frac{1}{a n}=\sum_{\substack{0<a<n / 2 \\(a, n)=1}} \frac{1}{a(n-a)}=\sum_{\substack{0<a<n / 2 \\(a, n-a)=1}} \frac{1}{a(n-a)}
$$

which is what we had to prove.