$\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ are points in the plane and $\mathrm{r}_{1}, \mathrm{r}_{2}, \ldots, \mathrm{r}_{\mathrm{n}}$ are real numbers such that the distance between $P_{i}$ and $P_{j}$ is $r_{i}+r_{j}$ (for $i$ not equal to $j$ ). Find the largest $n$ for which this is possible.

Solution:
Answer: $\mathrm{n}=4$.

Draw a circle radius $r_{i}$ at $P_{i}$. Then each pair of circles must touch. But that is possible iff $n \leq 4$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-34.jpg?height=477&width=557&top_left_y=2126&top_left_x=247)


$\mathrm{k}$ is the positive root of the equation $\mathrm{x}^{2}-1998 \mathrm{x}-1=0$. Define the sequence $\mathrm{x}_{0}, \mathrm{x}_{1}, \mathrm{x}_{2}, \ldots$ by $\mathrm{x}_{0}$ $=1, \mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{k} \mathrm{x}_{\mathrm{n}}\right]$. Find the remainder when $\mathrm{x}_{1998}$ is divided by 1998.

Solution:
Put $\mathrm{p}(\mathrm{x})=\mathrm{x}^{2}-1998 \mathrm{x}-1$. Then $\mathrm{p}(1998)=-1, \mathrm{p}(1999)=1998$, so $1998<\mathrm{k}<1999$. Also $\mathrm{k}$ is irrational (using the formula for the root of a quadratic). We have $x_{n}=\left[k x_{n-1}\right]$, so $x_{n}<k x_{n-1}$ and $>\mathrm{kx}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}} / \mathrm{k}<\mathrm{x}_{\mathrm{n}-1}<\mathrm{x}_{\mathrm{n}} / \mathrm{k}+1 / \mathrm{k}$, so $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=\mathrm{x}_{\mathrm{n}-1}-1$.

$\mathrm{k}=(1998 \mathrm{k}+1) / \mathrm{k}=1998+1 / \mathrm{k}$. Hence $\mathrm{kx}_{\mathrm{n}}=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x} / \mathrm{k}$. Hence $\mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{kx}_{\mathrm{n}}\right]=1998 \mathrm{x}_{\mathrm{n}}+$ $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}+1}=\mathrm{x}_{\mathrm{n}-1}-1 \bmod 1998$. So $\mathrm{x}_{1998}=1-999=1000 \bmod$ 1998.


Find all positive integers $n<1000$ such that the cube of the sum of the digits of $n$ equals $n^{2}$.

Solution: $\mathrm{n}<1000$, so the sum of the digits is at most 27 , so $\mathrm{n}^{2}$ is a cube not exceeding $27^{3}$. So we are looking for $\mathrm{m}^{3}$ which is also a square. That implies $\mathrm{m}$ is a square. So the only possibilities are $\mathrm{m}=1,4,9,16,25$. Giving $\mathrm{n}=1,8,27,64,125$. The corresponding cubes of the digit sums are $1,512,729,1000,512$, whereas the corresponding squares are $1,64,729,4096,15625$. Thus the only solutions are $\mathrm{n}=1,27$.


Given two circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ we say that $\mathrm{C}$ bisects $\mathrm{C}^{\prime}$ if their common chord is a diameter of $\mathrm{C}^{\prime}$. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of the bisecting circles.

Solution:
Let $\mathrm{C}, \mathrm{C}^{\prime}$ have center $\mathrm{O}, \mathrm{O}^{\prime}$ respectively and radius $\mathrm{r}, \mathrm{r}^{\prime}$ respectively. Let a circle center $\mathrm{P}$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $A B$ is perpendicular to $O P$ and is a diameter of C. Hence $\mathrm{PA}^{2}=\mathrm{OP}^{2}+\mathrm{r}^{2}$. Conversely, the circle center $\mathrm{P}$, radius $\sqrt{ }\left(\mathrm{OP}^{2}+\mathrm{r}^{2}\right)$ bisects $\mathrm{C}$. So $\mathrm{P}$ will bisect $\mathrm{C}$ and $\mathrm{C}^{\prime}$ iff $\mathrm{OP}^{2}+\mathrm{r}^{2}=\mathrm{OP}^{\prime 2}+\mathrm{r}^{\prime 2}$.

It is well-known that the locus of points $\mathrm{P}^{\prime}$ with equal tangents to $\mathrm{C}$ and $\mathrm{C}^{\prime}$ is the radical axis. Call the radical axis $\mathrm{R}$. For a point $\mathrm{P}^{\prime}$ on the radical axis we have $\mathrm{P}^{\prime} \mathrm{O}^{2}-\mathrm{r}^{2}=\mathrm{P}^{\prime} \mathrm{O}^{\prime 2}-\mathrm{r}^{\prime 2}$. If we reflect $\mathrm{P}^{\prime}$ in the perpendicular bisector of $\mathrm{OO}^{\prime}$ to get $\mathrm{P}$, then $\mathrm{PO}=\mathrm{P}^{\prime} \mathrm{O}^{\prime}$ and $\mathrm{PO}^{\prime}=\mathrm{P}^{\prime} \mathrm{O}$, so $\mathrm{PO}^{\prime 2}$ $-r^{2}=\mathrm{PO}^{2-r^{2}}$ and hence $\mathrm{PO}^{2}+\mathrm{r}^{2}$. Call the reflection of the $\mathrm{R}$ in the perpendicular bisector of OO' the line $\mathrm{R}^{\prime}$. We have established that points on $\mathrm{R}^{\prime}$ form part of the locus. Conversely, if $\mathrm{P}^{\prime}$ is such that there is a circle center $\mathrm{P}^{\prime}$ bisecting both circles, then $\mathrm{OP}^{\prime 2}+\mathrm{r}^{2}=\mathrm{O}^{\prime} \mathrm{P}^{\prime 2}+\mathrm{r}^{\prime 2}$, so if $\mathrm{P}$ is the reflection of $\mathrm{P}^{\prime}$ then $\mathrm{OP}^{2}-\mathrm{r}^{2}=\mathrm{O}^{2}-\mathrm{r}^{\mathrm{r}^{2}}$ and hence $\mathrm{P}$ lies on the radical axis R. Hence $\mathrm{P}^{\prime}$ must lie on $\mathrm{R}^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-36.jpg?height=460&width=534&top_left_y=267&top_left_x=247)

We have $\mathrm{PT}^{2}=\mathrm{PO}^{2}-\mathrm{r}^{2}=\mathrm{PX}^{2}+\mathrm{OX}^{2}-\mathrm{r}^{2}$, and similarly $\mathrm{PT}^{\prime 2}=\mathrm{PX}^{2}+\mathrm{O}^{\prime} \mathrm{X}^{2}-\mathrm{r}^{\prime 2}$. So $\mathrm{PT}=\mathrm{PT}^{\prime}$ iff $O X^{2}-r^{2}=O^{\prime} X^{2}-r^{\prime 2}$. There is evidently a unique point $X$ for which that is true, so the locus of such $\mathrm{P}$ is the line through $\mathrm{X}$ perpendicular to $\mathrm{OO}^{\prime}$

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-36.jpg?height=503&width=505&top_left_y=1002&top_left_x=250)

If the circles intersect, then the point $\mathrm{X}$ evidently lies on the line joining the two common points, because $O X^{2}-r^{2}=-X Y^{2}=O^{\prime} X^{2}-r^{\prime 2}$. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it.


Given points $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ on a line we construct a circle on diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ for each pair $\mathrm{i}, \mathrm{j}$ and we color the circle with one of $\mathrm{k}$ colors. For each $\mathrm{k}$, find all $\mathrm{n}$ for which we can always find two circles of the same color with a common external tangent.

Solution:
Answer: $\mathrm{n}>\mathrm{k}+1$.

There are $\mathrm{n}-1$ circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}+1}$. Obviously, each pair has a common tangent. If $\mathrm{n}-1$ $>\mathrm{k}$, then two of them must have the same color.

If $\mathrm{n}-1 \leq \mathrm{k}$, then color all circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ and $\mathrm{i}<\mathrm{j}$ with color $\mathrm{i}$. Then if two circles have the same color, then both have a tangent at one of the points. Hence one lies inside the other and they do not have a common external tangent.


Show that any integer greater than 10 whose digits are all members of $\{1,3,7,9\}$ has a prime factor $\geq 11$.

Solution:
Such a number cannot be divisible by 2 (or its last digit would be even) or by 5 (or its last digit would be 0 or 5). So if the result is false then the number must be of the form $3^{\mathrm{m}} 7^{\mathrm{n}}$ for non-negative integers $\mathrm{m}, \mathrm{n}$. But we claim that a number of this form must have even 10s digit.

It is easy to prove the claim by induction. It is true for 3 and 7 (the digit is 0 in both cases). But if we multiply such a number by 3 or 7 , then the new 10 s digit has the same parity as the carry from the units digit. But multiplying 1,3,7, 9 by 3 gives a carry of $0,0,2,6$ respectively, which is always even, and multiplying by 7 gives a carry of $0,2,4,6$, which is also always even. So the new number also has an even 10 s digit.


$\mathrm{O}$ is the circumcenter of the acute-angled triangle $\mathrm{ABC}$. The altitudes are $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$. The line $\mathrm{EF}$ cuts the circumcircle at $\mathrm{P}$ and $\mathrm{Q}$. Show that $\mathrm{OA}$ is perpendicular to $\mathrm{PQ}$. If $\mathrm{M}$ is the midpoint of $\mathrm{BC}$, show that $\mathrm{AP}^{2}=2 \mathrm{AD} \cdot \mathrm{OM}$.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-37.jpg?height=576&width=597&top_left_y=1371&top_left_x=250)

Let $\mathrm{OA}$ and $\mathrm{PQ}$ meet at $\mathrm{T}$. $\angle \mathrm{AEH}=\angle \mathrm{AFH}=90^{\circ}$, so $\mathrm{AEHF}$ is cyclic, so $\angle \mathrm{AFT}=\angle \mathrm{AFE}$ $($ same angle $)=\angle \mathrm{AHE}=90^{\circ}-\angle \mathrm{HAE}=90^{\circ}-\angle \mathrm{DAC}($ same angle $)=\angle \mathrm{C}$. But $\angle \mathrm{TAF}=\angle \mathrm{OAF}$ (same angle) $=90^{\circ}-(1 / 2) \angle \mathrm{AOB}=90^{\circ}-\angle \mathrm{C}$. Hence $\angle \mathrm{AFT}=90^{\circ}$, which establishes that OA and $P Q$ are perpendicular.

Let the circumradius be $\mathrm{R}$ and let $\mathrm{AA}^{\prime}$ be a diameter. We have $\mathrm{AF}=\mathrm{AC} \cos \mathrm{A}=2 \mathrm{R} \sin \mathrm{B} \cos$ A. Hence $\mathrm{AT}=\mathrm{AF} \cos \mathrm{OAB}=\mathrm{AF} \sin \mathrm{C}=2 \mathrm{R} \cos \mathrm{A} \sin \mathrm{B} \sin \mathrm{C} . \mathrm{Now} \mathrm{PT}^{2}=\mathrm{PT} \cdot \mathrm{TQ}=$ $A T \cdot A^{\prime} T=A T(2 R-A T)$. Hence $A P^{2}=2 R \cdot A T=4 R^{2} \cos A \sin B \sin C$.

We have $\mathrm{AD}=\mathrm{AC} \sin \mathrm{C}=2 \mathrm{R} \sin \mathrm{B} \sin \mathrm{C}$, and $\mathrm{OM}=\mathrm{OC} \cos \mathrm{COM}=\mathrm{R} \cos \mathrm{A}$. Hence 2 $\mathrm{AD} \cdot \mathrm{OM}=\mathrm{AP}^{2}$.


Given two points $\mathrm{A}$ and $\mathrm{B}$, take $\mathrm{C}$ on the perpendicular bisector of $\mathrm{AB}$. Define the sequence $C_{1}, C_{2}, C_{3}, \ldots$ as follows. $C_{1}=C$. If $C_{n}$ is not on $A B$, then $C_{n+1}$ is the circumcenter of the triangle $\mathrm{ABC}_{\mathrm{n}}$. If $\mathrm{C}_{\mathrm{n}}$ lies on $\mathrm{AB}$, then $\mathrm{C}_{\mathrm{n}+1}$ is not defined and the sequence terminates. Find all points $\mathrm{C}$ such that the sequence is periodic from some point on.

Solution:
Answer: any $\mathrm{C}$ such that $\angle \mathrm{ACB}=180^{\circ} \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime integers and $\mathrm{s}$ not a power of 2 .

Let $\angle \mathrm{AC}_{\mathrm{n}} \mathrm{B}=\mathrm{x}_{\mathrm{n}}$, where the angle is measured clockwise, so that $\mathrm{x}_{\mathrm{n}}$ is positive on one side of $A B$ and negative on the other side. Then $x_{n}$ uniquely identifies $C_{n}$ on the perpendicular bisector.

We have $\mathrm{x}_{\mathrm{n}+1}=2 \mathrm{x}_{\mathrm{n}}$. To make this work in all cases we have to take it mod $180^{\circ}$ (so that if $A C_{n} B$ is obtuse, then $C_{n+1}$ lies on the other side of $A B$ ). If $x_{n}$ is eventually periodic then $x_{m+1}$ $=\mathrm{x}_{\mathrm{n}+1}$, for some $\mathrm{n}>\mathrm{m}$, so $\left(2^{\mathrm{n}}-2^{\mathrm{m}}\right) \mathrm{x}_{1}=0 \bmod 180$. Hence $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$ for some relatively prime integers $\mathrm{r}, \mathrm{s}$. Also $\mathrm{s}$ cannot be a power of 2 for then we would have $\mathrm{x}_{\mathrm{k}}=180 \mathrm{r}$ for some $\mathrm{k}$, in which case the sequence would terminate rather than be periodic.

Conversely, suppose $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime and $\mathrm{s}$ not a power of 2 . Then $\mathrm{x}_{\mathrm{n}+1}=1802^{\mathrm{n}} \mathrm{r} / \mathrm{s}$ cannot be $0 \bmod 180$, so the sequence does not terminate. Put $\mathrm{s}=2^{\mathrm{b}} \mathrm{c}$, with $\mathrm{c}$ odd. Let $\mathrm{d}=\varphi(\mathrm{c})$, where $\varphi(\mathrm{m})$ is Euler's phi function, so that $2^{\mathrm{d}}=1 \mathrm{mod} \mathrm{c}$. Then $\mathrm{x}_{\mathrm{b}+1}=180 \mathrm{r} / \mathrm{c}$ $\bmod 180$ and $2^{\mathrm{b}+\mathrm{d}}=2^{\mathrm{b}} \bmod \mathrm{c}$, so $\mathrm{x}_{\mathrm{b}+\mathrm{d}+1}=180 \mathrm{r} / \mathrm{c}$ mod 180 . Hence the sequence is periodic.


Label the vertices of a regular $n$-gon from 1 to $n>3$. Draw all the diagonals. Show that if $n$ is odd then we can label each side and diagonal with a number from 1 to $n$ different from the labels of its endpoints so that at each vertex the sides and diagonals all have different labels.

Solution: Labeling the diagonal/side between $\mathrm{i}$ and $\mathrm{j}$ as $\mathrm{i}+\mathrm{j}$ (reduced if necessary mod $\mathrm{n}$ ) almost works. The labels for all the lines at a given vertex will be different. But the line between i and $n$ will have label i, the same as one endpoint. However, we are not using the label 2i for the lines from vertex i. So for the line between $i$ and $n$ we use $2 i$ instead of $i+n$. The only points that need checking are (1) whether a line from $\mathrm{i}$ to $\mathrm{n}$ has a label different from $\mathrm{n}$, and (2) whether all the lines at $\mathrm{n}$ have different labels. Both points are ok because $\mathrm{n}$ is odd.


Two circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ have centers $\mathrm{O}$ and $\mathrm{O}^{\prime}$ and meet at $\mathrm{M}$ and $\mathrm{N}$. The common tangent closer to $\mathrm{M}$ touches $\mathrm{C}$ at $\mathrm{A}$ and $\mathrm{C}^{\prime}$ at $\mathrm{B}$. The line through $\mathrm{B}$ perpendicular to $\mathrm{AM}$ meets the line $\mathrm{OO}^{\prime}$ at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-39.jpg?height=425&width=625&top_left_y=256&top_left_x=247)

A neat coordinate solution by Massaki Yamamoto (a competitor) is as follows.

Take $A B$ as the $x$-axis and the perpendicular line through $M$ as the $y$-axis. Choose the unit of length so that $\mathrm{M}$ has coordinates $(0,1)$. Let $\mathrm{A}$ be $(-\mathrm{m}, 0)$ and $\mathrm{B}$ be $(\mathrm{n}, 0)$. Then considering the right-angled triangle $O^{\prime} M K$, where $K$ is $(n, 1)$ we find that $O^{\prime}$ is $\left(n,\left(n^{2}+1\right) / 2\right)$. Similarly, $O$ is $\left.\left(-\mathrm{m},\left(\mathrm{m}^{2}+1\right) / 2\right)\right)$.

The gradient of the lie AM is $1 / \mathrm{m}$, so the gradient of the line $\mathrm{BD}$ is $-\mathrm{m}$ and hence its equation is $m x+y=m n$. The gradient of the line OO' is $(n-m) / 2$, so its equation is $2 y-x(n-m)=m n+1$. These intersect at $\left((m n-1) /(m+n),\left(m n^{2}+m\right) /(m+n)\right)$. B' is $\left(n, n^{2}+1\right)$. It is now easy to check that the lines MB' and MD both have gradient $\mathrm{n}$, so $\mathrm{M}, \mathrm{D}, \mathrm{B}^{\prime}$ are collinear.


Find all solutions to $(\mathrm{m}+1)^{\mathrm{a}}=\mathrm{m}^{\mathrm{b}}+1$ in integers greater than 1 .

## Answer

$(\mathrm{m}, \mathrm{a}, \mathrm{b})=(2,2,3)$.

Solution:
Taking equation mod $\mathrm{m}+1$ we get $(-1)^{\mathrm{b}}=-1$, so $\mathrm{b}$ is odd. Hence we can divide the rhs by $\mathrm{m}+1$ to get $m^{b-1}-m^{b-2}+\ldots-m+1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(\mathrm{m}+1)^{\mathrm{a}-1}$ is even (note that $\left.\mathrm{a}>1\right)$. Contradicton, so $\mathrm{m}$ is even.

We have $\mathrm{m}^{\mathrm{b}}=(\mathrm{m}+1)^{\mathrm{a}}-1$. Expanding the rhs by the binomial theorem, and using $\mathrm{b}>1$, we see that $\mathrm{m}$ must divide $\mathrm{a}$. So a is even also. Put $\mathrm{a}=2 \mathrm{~A}, \mathrm{~m}=2 \mathrm{M}$. We can factorise $(\mathrm{m}+1)^{\mathrm{a}}-1$ as ( $\left.(m+1)^{A}+1\right)\left((m+1)^{A}-1\right)$. The two factors have difference 2 , so their gcd divides 2 , but both factors are even, so their gcd is exactly 2.

If $\mathrm{M}=1$ or a power of 2 , then the smaller factor $3^{\mathrm{A}}-1$ must be 2 , so $\mathrm{A}=1$ and we have $3^{\mathrm{A}}+$ $1=4$, so $(2 M)^{b}=8$. Hence $M=1$ and $b=3$ and we have the solution $(m, a, b)=(2,2,3)$.

If $\mathrm{M}$ is not a power of 2 , then $\mathrm{M}^{\mathrm{b}}>2^{\mathrm{b}}$, so we must have the larger factor $2 \cdot \mathrm{M}^{\mathrm{b}}$ and the smaller factor $2^{b-1}$. But the larger factor is now $>2^{b+1}$, so the difference between the factors is at least 3. $2^{\mathrm{b}-1}>2$. Contradiction.


Some terms are deleted from an infinite arithmetic progression $1, \mathrm{x}, \mathrm{y}, \ldots$ of real numbers to leave an infinite geometric progression $1, \mathrm{a}, \mathrm{b}, \ldots$. Find all possible values of a.

Solution:
Answer: the positive integers.

If a is negative, then the terms in the GP are alternately positive and negative, whereas either all terms in the AP from a certain point on are positive or all terms from a certain point on are negative. So a cannot be negative. If a is zero, then all terms in the GP except the first are zero, but at most one term of the AP is zero, so a cannot be zero. Thus a must be positive, so the AP must have infinitely many positive terms and hence $\mathrm{x} \geq 1$.

Let $\mathrm{d}=\mathrm{x}-1$, so all terms of the AP have the form 1+nd for some positive integer $\mathrm{n}$. Suppose $a=1+m d, a^{2}=1+n d$, then $(1+m d)^{2}=1+n d$, so $d=(n-2 m) / m^{2}$, which is rational. Hence $\mathrm{a}$ is rational. Suppose $\mathrm{a}=\mathrm{b} / \mathrm{c}$, where $\mathrm{b}$ and $\mathrm{c}$ are relatively prime positive integers and $\mathrm{c}>1$. Then the denominator of the nth term of the GP is $\mathrm{c}^{\mathrm{n}}$, which becomes arbitrarily large as $\mathrm{n}$ increases. But if $\mathrm{d}=\mathrm{h} / \mathrm{k}$, then all terms of the AP have denominator at most $\mathrm{k}$. So we cannot have $\mathrm{c}>1$. So a must be a positive integer.

On the other hand, it is easy to see that any positive integer works. Take $\mathrm{x}=2$, then the AP includes all positive integers and hence includes any GP with positive integer terms.


Given a pile of 2000 stones, two players take turns in taking stones from the pile. Each player must remove $1,2,3,4$, or 5 stones from the pile at each turn, but may not take the same number as his opponent took on his last move. The player who takes the last stone wins. Does the first or second player have a winning strategy?

Solution:
The first player has a winning strategy. He takes 4 on his first move leaving 7 mod $13(2000=$ $153.13+7+4)$. Now we claim that the first player can always leave: (1) $0 \bmod 13$, (2) $3 \bmod$ 13 by taking away 3 , (3) $5 \bmod 13$ by taking away 5 , or (4) $7 \bmod 13$, and that the second player can never leave 0 mod 13.

Let us look at each of these in turn. If the first player leaves $0 \bmod 13$, then the second player can take 3 and leave 10. In that case the first player takes 5 (a type (3) move). If the second player takes $1,2,4$ or 5 , leaving $12,11,9$ or $8 \bmod 13$, then the first player takes $5,4,2,1$ (respectively) and leaves 7 mod 13 (a type (4) move).

If the first player leaves $3 \bmod 13$ by taking away 3 , then the second player cannot leave 0 $\bmod 13$, because he cannot take 3 stones. If he takes 1,2 leaving $2,1 \bmod 13$ respectively, then the first player takes 2,1 leaving 0 mod 13 (a type (1) move). If the second player takes 4, 5 leaving 12, 11 mod 13, then the first player takes 5,4 leaving $7 \bmod 13$ (a type (4) move).

If the first player leaves $5 \bmod 13$ by taking 5 , then the second player cannot leave $0 \bmod 13$, because he cannot take 5 stones. If he takes $1,2,3,4$ stones, leaving 4,3,2, 1 mod 13 , then the first player takes $4,3,2,1$ stones leaving 0 mod 13 (a type (1) move).

Finally, if the first player leaves $7 \bmod 13$, and the second player takes 1 stone, then the first player takes 3 stones leaving 3 mod 13 (a type (2) move). If the second player takes $2,3,4$, or 5 stones leaving $5,4,3,2 \bmod 13$, then the first player takes $5,4,3,2$ stones leaving $0 \bmod 13$ (a type (1) move).

So the second player can never leave 0 mod 13 and hence, in particular, can never take the last stone. But we have shown that the first player can always make a move of one of the four types, so can always move and hence must win (since after less than 2000 moves there will be no stones left).


A convex hexagon is called a unit if it has four diagonals of length 1, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area $\mathrm{k}$ for any $0<\mathrm{k} \leq 1$. What is the largest possible area for a unit?

Solution:
Answer: We can get arbitrarily close to (but not achieve) ( $3 \sqrt{3}$ )/4 (approx 1.3) by:

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=182&width=302&top_left_y=1234&top_left_x=266)

To prove the first part, consider the diagram below. Take $\mathrm{AB}=\mathrm{AC}=1$ and angle $\mathrm{BAC}=2 \theta$. Take $\mathrm{DE}=\mathrm{DF}=1$ and take the points of intersection $\mathrm{X}$ and $\mathrm{Y}$ such that $\mathrm{AX}=\mathrm{DX}=\mathrm{AY}=$ $\mathrm{DY}=2 / 3$. It is easy to check that the area of the hexagon is $\sin 2 \theta$. So by taking $\theta$ in the interval $(0, \pi / 4]$ we can get any area $0<\mathrm{k} \leq 1$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=351&width=526&top_left_y=1755&top_left_x=248)

It is easy to check that there are six possible configurations for the unit diagonals, as shown in the diagram below.
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=340&width=684&top_left_y=2331&top_left_x=250)

Consider case 1.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=211&width=260&top_left_y=360&top_left_x=264)

The area of the hexagon is area AEDC + area AFE + area BAC. The part of the segment BF that lies inside AEDC is wasted. The rest goes to provide height for the triangles on bases AE and $\mathrm{AC}$. So area AFE + area BAC can be maximised by taking $\mathrm{F}$ close to $\mathrm{A}$ and $\angle \mathrm{BAC}$ as close to a right angle as possible, so that the height of the triangle BAC (on the base $\mathrm{AC}$ ) is as large as possible. We can then get arbitrarily close to the area of:

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=215&width=303&top_left_y=949&top_left_x=251)

We obviously make AEB a straight line. Now area $\mathrm{ADE}+$ area $\mathrm{ADC}=$ area $\mathrm{ACE}+$ area CDE. So if we regard every point except $\mathrm{D}$ as fixed, then we maximise the area by taking $\angle \mathrm{EAD}=\angle \mathrm{CAD}$, so that $\mathrm{D}$ is the maximum distance from $\mathrm{CE}$. Thus a maximal configuration must have $\angle \mathrm{AED}=\angle \mathrm{CAD}$. Similarly, it must have $\angle \mathrm{CAD}=\angle \mathrm{CAB}$, so all three angles must be equal. That disposes of case 1 .

In cases 2 and 6 we find by a similar (but more tedious argument) the same maximum, although in one case we have to use the argument at the end for the final optimisation. In the other cases the maximum is smaller.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=169&width=651&top_left_y=1723&top_left_x=246)

4

5

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=160&width=303&top_left_y=1942&top_left_x=842)

6

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=143&width=297&top_left_y=2116&top_left_x=845)

However, all these details would take an already long solution way over length. Does anyone have a better approach?

No. 6 (second case) can be made arbitrarily close to the figure below (with $\mathrm{AB}=\mathrm{AC}=\mathrm{BD}=$ 1). To optimise it, suppose $\angle A C B=\theta$. Area $A B D C=$ area $A B C+$ area $B C D$. If we fix $\theta$, then $\mathrm{BC}$ is fixed, so to maximise area BCD we must take $\angle \mathrm{CBD}=90^{\circ}$. But $\theta$ cannot be optimal unless also $\angle \mathrm{CAD}=90^{\circ}$. We have $\mathrm{BA}=\mathrm{BD}$ and hence $\angle \mathrm{BAD}=\angle \mathrm{BDA}=45^{\circ}-\theta / 2$. Hence
$90^{\circ}=\angle \mathrm{CAD}=\angle \mathrm{BAC}-\angle \mathrm{BAD}=\left(180^{\circ}-2 \theta\right)-\left(45^{\circ}-\theta / 2\right)$. Hence $\theta=30^{\circ}$. So $\angle \mathrm{ACD}=$ $\angle \mathrm{BDC}=60^{\circ}$ and $\angle \mathrm{CAB}=\angle \mathrm{ABD}=120^{\circ}$. It is easy to check that this has area $(3 \sqrt{3}) / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-43.jpg?height=286&width=532&top_left_y=411&top_left_x=248)


Show that there are arbitrarily large numbers $n$ such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides $n$.

Solution: $3232=16 \times 202$ and $10000=16 \times 625$. So any number with 3232 as its last 4 digits is divisible by 16 . So consider $\mathrm{N}=22223232$. Its sum of digits is 18 , so it is divisible by 9 . Hence it is divisible by $9 \cdot 16=144$. But any four digits have at most four $2 \mathrm{~s}$ and at most two 3s, so the product of any four digits divides 144 and hence $\mathrm{N}$. But now we can extend $\mathrm{N}$ by inserting an additional $9 \mathrm{~m} 2 \mathrm{~s}$ at the front. Its digit sum is increased by $18 \mathrm{~m}$, so it remains divisible by 144 and it is still divisible by the product of any four digits.


Solution: The number 111111111 with nine 1 s is divisible by 9 . Hence the number with twenty-seven $1 \mathrm{~s}$ which equals $111111111 \times 1000000001000000001$ is divisible by 27 . So $\mathrm{N}$, the number with twenty-seven $3 \mathrm{~s}$, is divisible by $3^{4}$. Now the number with $27 \mathrm{n} 3 \mathrm{~s}$ is divisible by $\mathrm{N}$ and hence by $3^{4}$.


Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of $n$ distinct real numbers?

## Answer
 $(\mathrm{m}-1) \mathrm{m}$ for $\mathrm{n}=2 \mathrm{~m}$
 $\mathrm{m}^{2}$ for $\mathrm{n}=2 \mathrm{~m}+1$

Solution:
Let $\mathrm{X}$ be one of the elements. What is the largest number of trios that can have $\mathrm{X}$ as middle element? Obviously, at most $\max (\mathrm{b}, \mathrm{a})$, where $\mathrm{b}$ is the number of elements smaller than $\mathrm{X}$ and a is the number larger. Thus if $n=2 m$, the no. of trios is at most $0+1+2+\ldots+m-1+m-1+$ $\mathrm{m}-2+\ldots+1+0=(\mathrm{m}-1) \mathrm{m}$. If $\mathrm{n}=2 \mathrm{~m}+1$, then the no. is at most $0+1+2+\ldots+\mathrm{m}-1+\mathrm{m}+\mathrm{m}-$ $1+\ldots+1+0=\mathrm{m}^{2}$

These maxima can be achieved by taking the numbers $1,2,3, \ldots, \mathrm{n}$.


The numbers $1,2, \ldots, 2002$ are written in order on a blackboard. Then the 1 st, 4 th, 7 th, $\ldots$, $3 \mathrm{k}+1$ th, $\ldots$ numbers in the list are erased. Then the 1 st, $4 \mathrm{th}, 7 \mathrm{th}, \ldots 3 \mathrm{k}+1$ th numbers in the remaining list are erased (leaving $3,5,8,9,12, \ldots$ ). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased?

Solution:
Answer: 1598.

Let $\mathrm{a}_{\mathrm{n}}$ be the first number remaining after $\mathrm{n}$ iterations, so $\mathrm{a}_{0}=1, \mathrm{a}_{1}=2, \mathrm{a}_{3}=3, \mathrm{a}_{4}=5$ etc. We claim that:

```
\(a_{n+1}=3 / 2 a_{n} \quad\) if \(a_{n}\) is even, and
    \(3 / 2\left(a_{n}+1\right)-1\) if \(a_{n}\) is odd.
```

We use induction on $n$. Suppose $a_{n}=2 N$. Consider the number $3 N$. There are initially $\mathrm{N}$ smaller numbers $=1 \mathrm{mod} 3$. So after the first iteration, it will lie in 2Nth place. Hence, it will lie in first place after $n+1$ iterations. Similarly, suppose $\mathrm{a}_{\mathrm{n}}=2 \mathrm{~N}+1$. Consider $3 N+2$. There are initially $\mathrm{N}+1$ smaller numbers $=1 \bmod 3$. So after the first iteration, it will lie in $2 \mathrm{~N}+1$ st place. Hence, it will lie in first place after $\mathrm{n}+1$ iterations. That completes the induction.

We may now calculate successively the members of the sequence: $1,2,3,5,8,12,18,27,41$, $62,93,140,210,315,473,710,1065,1598,2397$. Hence 1598 is the last surviving number from $1,2, \ldots, 2002$.


Given a set of 9 points in the plane, no three collinear, show that for each point $\mathrm{P}$ in the set, the number of triangles containing $\mathrm{P}$ formed from the other 8 points in the set must be even.

Solution: Join each pair of points, thus dividing the plane into polygonal regions. If a point $\mathrm{P}$ moves around within one of the regions then the number of triangles it belongs to does not change. But if it crosses one of the lines then it leaves some triangles and enters others. Suppose the line is part of the segment joining the points $\mathrm{Q}$ and $\mathrm{R}$ of the set. Then it can only enter or leave a triangle QRX for some $\mathrm{X}$ in the set. Suppose $\mathrm{x}$ points in the set lie on the same side of the line $\mathrm{QR}$ as $\mathrm{P}$. Then there are 6 - $\mathrm{x}$ points on the other side of the line $\mathrm{QR}$. So $\mathrm{P}$ leaves $\mathrm{x}$ triangles and enters $6-\mathrm{x}$. Thus the net change is even. Thus if we move $\mathrm{P}$ until it is in the outer infinite region (outside the convex hull of the other 8 points), then we change the number of triangles by an even number. But in the outside region it belongs to no triangles.


$\mathrm{ABC}$ is an equilateral triangle. $\mathrm{P}$ is a variable interior point such that $\square \mathrm{APC}=120^{\circ}$. The ray $\mathrm{CP}$ meets $\mathrm{AB}$ at $\mathrm{M}$, and the ray $\mathrm{AP}$ meets $\mathrm{BC}$ at $\mathrm{N}$. What is the locus of the circumcenter of the triangle MBN as $\mathrm{P}$ varies?

Solution:
Answer: the segment of the perpendicular bisector of $\mathrm{BG}$ (where $\mathrm{G}$ is the center of the triangle) which forms a rectangle with $\mathrm{AC}$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-46.jpg?height=580&width=646&top_left_y=498&top_left_x=251)

$\angle \mathrm{MPN}=\angle \mathrm{APC}=120^{\circ}$ and $\angle \mathrm{MBN}=60^{\circ}$, so $\mathrm{MBNP}$ is cyclic, in other words, $\mathrm{P}$ lies on the circumcircle of BMN.

$\mathrm{P}$ also lies on the circle $\mathrm{AGC}$, so $\angle \mathrm{CPG}=\angle \mathrm{CAG}$ (if $\mathrm{P}$ is on the same side of $\mathrm{AG}$ as $\mathrm{A})=30^{\circ}$ $=\angle$ MBG. So PMBG is cyclic. In other words, $\mathrm{G}$ also lies on the circumcircle of BMN. If P lies on the other side, the same conclusion follows from considering $\angle$ APG.

Since B and G lie on the circumcircle, the center O must lie on the perpendicular bisector of BG. But it is clear that the extreme positions of $\mathrm{O}$ occur when $\mathrm{P}$ is at $\mathrm{A}$ and $\mathrm{B}$ and that these are the feet of the perpendiculars from A and B to the perpendicular bisector.


$\mathrm{ABC}$ is a triangle. $\mathrm{BD}$ is the an angle bisector. $\mathrm{E}, \mathrm{F}$ are the feet of the perpendiculars from $\mathrm{A}$, $C$ respectively to the line $B D . M$ is the foot of the perpendicular from $D$ to the line $B C$. Show that $\angle \mathrm{DME}=\angle \mathrm{DMF}$.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-46.jpg?height=435&width=659&top_left_y=2104&top_left_x=253)

Let $\mathrm{H}$ be the foot of the perpendicular from $\mathrm{D}$ to $\mathrm{AB} . \angle \mathrm{AHD}=\angle \mathrm{AED}=90^{\circ}$, so $\mathrm{AHED}$ is cyclic. Hence $\angle \mathrm{DAE}=\angle \mathrm{DHE}$. But $\mathrm{M}$ is the reflection of $\mathrm{H}$ is the line $\mathrm{BD}$, so $\angle \mathrm{DME}=$ $\angle \mathrm{DAE}$.

$\mathrm{AE}$ is parallel to $\mathrm{CD}$, so $\angle \mathrm{DAE}=\angle \mathrm{DCF}$. $\angle \mathrm{DFC}=\angle \mathrm{DMC}$, so $\mathrm{DMCF}$ is cyclic. Hence $\angle \mathrm{DCF}$ $=\angle \mathrm{DMF}$. Hence $\angle \mathrm{DME}=\angle \mathrm{DMF}$.


The sequence $a_{n}$ is defined as follows: $a_{1}=56, a_{n+1}=a_{n}-1 / a_{n}$. Show that $a_{n}<0$ for some $n$ such that $0<\mathrm{n}<2002$.

Solution:
Note that whilst $a_{n}$ remains positive we have $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$. Hence if $a_{m}$ and $a_{m+n}$ are in this part of the sequence, then $a_{m+1}=a_{m}-1 / a_{m}, a_{m+2}=a_{m+1}-1 / a_{m+1}<a_{m+1}-1 / a_{m}=a_{m}-2 / a_{m}$. By a trivial induction $\mathrm{a}_{\mathrm{m}+\mathrm{n}}<\mathrm{a}_{\mathrm{m}}-\mathrm{n} / \mathrm{a}_{\mathrm{m}}$.

If we use one step then we need $56^{2}=3136$ terms to get $a_{1+3136}<56-56^{2} / 56=0$, which is not good enough. So we try several steps.

Thus suppose that $\mathrm{a}_{\mathrm{n}}>0$ for all $\mathrm{n}<=2002$. Then we get successively:

$a_{337}<56-336 / 56=50$

$a_{837}<50-500 / 50=40$

$a_{1237}<40-400 / 40=30$

$a_{1537}<30-300 / 30=20$

$\mathrm{a}_{1737}<20-200 / 20=10$

$\mathrm{a}_{1837}<10-100 / 10=0$.

Contradiction. So we must have $\mathrm{a}_{\mathrm{n}}<0$ for some $\mathrm{n}<2002$.


A game is played on a 2001 x 2001 board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the top left square in a single move. The policeman starts in the central square, and the robber starts one square diagonally northeast of the policeman. If the policeman moves onto the same square as the robber, then the robber is captured and the first player wins. However, the robber may move onto the same square as the policeman without being captured (and play continues). Show that the robber can avoid capture for at least 10000 moves, but that the policeman can ultimately capture the robber.

Solution:
Color the squares with three colors as follows:

```
0<ccccccccoc
1cccccccccoc
2cccccccccccollo
```

```
\(0 \begin{array}{llllllllll}0 & 1 & 2 & 0 & 1 & 2 & 0 & \ldots & 2\end{array}\)
\(\begin{array}{llllllllll}1 & 2 & 0 & 1 & 2 & 0 & 1 & \cdots & 0\end{array}\)
. .
\(\begin{array}{lllllllll}2 & 0 & 1 & 2 & 0 & 1 & 2 & \cdots & 1\end{array}\)
```

The middle square is color 2 (moving 999+1 squares E from the top left increases the color by 1, then moving $999+1 \mathrm{~S}$ increases it by another 1) and the square immediately $\mathrm{NE}$ of it is also 2. So both $\mathrm{P}$ and $\mathrm{R}$ start on color 2 . Note that any move increases the color by 1 mod 3 , except for P's special move which changes the color from 1 to 0 .

Until $\mathrm{P}$ has made this move, after each move of $\mathrm{P}, \mathrm{P}$ 's color is always 1 more than R's color $(\bmod 3)$, so $\mathrm{P}$ cannot win (irrespective of the moves made by either player). Immediately after he makes the special move for the first time, $\mathrm{P}$ is on color 0 and $\mathrm{R}$ is on color 1 , so immediately after his move P's color is now 1 less than R's color mod 3. Again P cannot win. But after P has made the special move for the second time, P's color is the same as R's (mod 3 ) immediately after P's move.

Note that it takes P at least 2001 moves to complete his special move for the first time and at least 6002 moves (in total) to complete his special move for the second time. This solves the first part of the question. Suppose R just moves down to the bottom right and then moves in small circles (one move NW, one move S, one move E) waiting for P. It takes P at least 6002 +3999 (moving from top left to the capture square, one square short of the bottom right) $=$ 10001 to capture him, so R makes at least 10000 moves before being captured.

We claim that $\mathrm{P}$ wins if he can get into any of the positions shown below relative to $\mathrm{R}$, with $\mathrm{R}$ to move $(*)$ :

```
X P X X X
P \(\quad \mathrm{x} \quad \mathrm{X} \quad \mathrm{P} \quad \mathrm{x}\)
\(x \quad x \quad R \quad x \quad x\)
X \(\quad \mathrm{P} \quad \mathrm{x} \quad \mathrm{X} \quad \mathrm{P}\)
x \(x\) x \(\quad \mathrm{x}\)
```

If follows that $\mathrm{P}$ can also win from the four positions below $(* *)$ :

```
X X X P X X X
x \(\quad x \quad x \quad x \quad x \quad x \quad x\)
X \(x\) x \(x\) x \(x\) x
P X X \(\quad\) R \(\quad \mathrm{x}\) X \(\quad\) P
x \(\quad \mathrm{x} \quad \mathrm{x} \quad \mathrm{x}\) x \(\quad \mathrm{x} \quad \mathrm{x}\)
x \(\quad x \quad x \quad x \quad x \quad x \quad x\)
x \(\quad x \quad x \quad P \quad x \quad x \quad x\)
```

For in each case at least one of R's possible moves allow $\mathrm{P}$ to move immediately into one of the winning positions at $(*)$. But $\mathrm{R}$ can only make the other moves a limited number of times before running into the border. [That is obvious if the other two moves are $\mathrm{E}$ and $\mathrm{S}$. If they are NW and E, then every NW move takes R closer to the top border, but his total number of E moves can never exceed his total number of NW moves by more than 2000 because of the right border. Similarly, for NW and S.]

Now let $\mathrm{d}$ be the number of rows plus the number of columns that $\mathrm{R}$ and $\mathrm{P}$ are apart. It is easy to check that the positions in $(*)$ and $(* *)$ represent the only possibilities for $\mathrm{d}=2$ and 3 . We show that $\mathrm{P}$ can always get to $\mathrm{d}=2$ or 3 . For $\mathrm{P}$ can always copy R's move, so he can certainly move so that $\mathrm{d}$ never increases. But one of R's moves will always allow $\mathrm{P}$ to decrease $\mathrm{d}$ by 1 or 2 . There are three cases to consider:

Case 1. If $\mathrm{P}$ is east of $\mathrm{R}$ and $\mathrm{R}$ moves $\mathrm{E}$, then $\mathrm{P}$ moving $\mathrm{NW}$ will decrease $\mathrm{d}$ by 1 or 2 . That is not possible if $\mathrm{P}$ is in the top row, but then moving $\mathrm{S}$ will decrease $\mathrm{d}$ by 2 unless $\mathrm{R}$ is also in the top row. If both are in the top row, then P moves S. Now after R's next move, P moves NW which reduces $\mathrm{d}$ by 2 .

Case 2. If $\mathrm{P}$ is south of $\mathrm{R}$ and $\mathrm{R}$ moves $\mathrm{S}$, then a similar argument, shows that $\mathrm{P}$ can always decrease $\mathrm{d}$ by 1 or 2 in one or two moves.

Case 3. If $\mathrm{P}$ is not south or east or $\mathrm{R}$, and $\mathrm{R}$ moves $\mathrm{NW}$, then $\mathrm{P}$ can always decrease $\mathrm{d}$ by 1 or 2 by moving $\mathrm{S}$ or $\mathrm{E}$.

But repeated decreases by 1 or 2 must bring $\mathrm{d}$ ultimately to 2 or 3 and hence to one of $(*)$ or $(* *)$. So $\mathrm{P}$ can always win.

It remains to prove the claim that (*) are winning positions. The reason is that in each case $\mathrm{R}$ has one move blocked off, so must make one of the other two. P then copies R's move, so next turn $\mathrm{R}$ has the same move blocked off. Repeated use of the other two moves will bring him ultimately to one of the sides.

We start with the easiest case: in the two following positions. $\mathrm{R}$ cannot move to $\mathrm{z}$, so he must move east or south on each move. Hence he will (after at most 4000 moves) reach the bottom right corner. He then loses moving out of it.

```
X P X
P z x
x X R
```

The other cases of $(*)$ are slightly more complicated. Starting from either of the two positions below, we show that $\mathrm{R}$ must eventually reach the extreme left column.

```
w \(\quad x \quad P \quad x\)
\(x \quad R \quad z \quad x\)
x y x P
```

$\mathrm{R}$ cannot move to $\mathrm{z}$, so he can only make $\mathrm{NW}$ and $\mathrm{S}$ moves. But his total number of $\mathrm{S}$ moves can never exceed his total number of NW moves by more than 2000 because he cannot move off the bottom of the board, so he must eventually reach the extreme left column. [If he reaches the bottom row at $\mathrm{y}$, then $\mathrm{P}$ can always move to $\mathrm{z}$ to preserve the configuration. If $\mathrm{R}$ reaches the top row by moving to $\mathrm{w}$, then $\mathrm{P}$ can always move to $\mathrm{z}$ to preserve the configuration.]

Having reached the extreme left column he is forced to move south. Eventually moving to y will take him to the corner. $\mathrm{P}$ then moves to $\mathrm{z}$ and $\mathrm{R}$ is captured on his next move.

The final case to consider is the two positions below. $\mathrm{R}$ cannot move to $\mathrm{z}$, so must move $\mathrm{E}$ or $\mathrm{NW}$. A similar argument to the previous case shows that he must eventually reach the top row. Having reached it at w, P moves to z. So R is forced to move right along the top row. When he reaches the corner at $\mathrm{y}, \mathrm{P}$ moves to $\mathrm{z}$ and $\mathrm{R}$ is captured when he moves out of the corner.

```
W x x
x R y
P z x
X X P
```


Let A, B be two sets of $\mathrm{N}$ consecutive integers. If $\mathrm{N}=2003$, can we form $\mathrm{N}$ pairs (a,b) with a $\angle \mathrm{A}, \mathrm{b} \angle \mathrm{B}$ such that the sums of the pairs are $\mathrm{N}$ consecutive integers? What about $\mathrm{N}=2004$ ?

## Answer

Yes, no.

Solution:
wlog $A=B=\{1,2, \ldots, N\}$ - if we have a solution for $A=\{a+1, a+2, \ldots, a+N\}$ and $B=\{b+1$, $b+2, \ldots, b+N\}$, then subtracting a from every element of A and $b$ from every element of $b$ gives a solution for $A=B=\{1,2, \ldots, N\}$. Suppose the sum set is $(m+1),(m+2), \ldots,(m+N)$. It has sum $N(2 m+N+1) / 2$ and $A$ and $B$ each have sum $N(N+1) / 2$, so we must have $2 m=N+1$, hence $\mathrm{N}$ must be odd. So we cannot do it for $\mathrm{N}=2004$.

Suppose $\mathrm{N}=2 \mathrm{M}+1$, take the pairs $(1, \mathrm{M}+1),(3, \mathrm{M}),(5, \mathrm{M}-1), \ldots,(2 \mathrm{M}+1,1),(2,2 \mathrm{M}+1),(4$, $2 \mathrm{M}), \ldots,(2 \mathrm{M}, \mathrm{M}+2)$.


$\mathrm{C}$ is a point on the semicircle with diameter $\mathrm{AB}$. D is a point on the $\operatorname{arc} \mathrm{BC} . \mathrm{M}, \mathrm{P}, \mathrm{N}$ are the midpoints of AC, CD and BD. The circumcenters of ACP and BDP are O, O'. Show that MN and $\mathrm{OO}^{\prime}$ are parallel.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-51.jpg?height=257&width=526&top_left_y=266&top_left_x=248)

Let the center of the circle be $X$ and the radius $r$. Let $\angle A X M=\theta, \angle B X N=\varphi$. Note that $O$ is the intersection of $\mathrm{XM}$ and the perpendicular to $\mathrm{CD}$ at $\mathrm{Q}$, the midpoint of $\mathrm{CP}$. We have $\mathrm{XM}=$ r $\cos \theta$. Let $\mathrm{CD}$ and $\mathrm{XM}$ meet at $\mathrm{Y}$. Then $\angle \mathrm{PYX}=90^{\circ}-\angle \mathrm{PXY}=90^{\circ}-\angle \mathrm{PXC}-\angle \mathrm{CXM}=\theta+$ $\varphi-\varphi=\theta$. Hence $O X=P Q \sec \varphi$, so $O X / X M=P Q /(r \cos \theta \cos \varphi)$. Similarly, $O^{\prime} X / O N=P Q /(r$ $\cos \theta \cos \varphi)$, so OO' and MN are parallel.


Pablo was trying to solve the following problem: find the sequence $\mathrm{x}_{0}, \mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{2003}$ which satisfies $\mathrm{x}_{0}=1,0 \leq \mathrm{x}_{\mathrm{i}} \leq 2 \mathrm{x}_{\mathrm{i}-1}$ for $1 \leq \mathrm{i} \leq 2003$ and which maximises $\mathrm{S}$. Unfortunately he could not remember the expression for $\mathrm{S}$, but he knew that it had the form $\mathrm{S}= \pm \mathrm{x}_{1} \pm \mathrm{x}_{2} \pm \ldots \pm \mathrm{x}_{2002}$ $+\mathrm{x}_{2003}$. Show that he can still solve the problem.

Solution: For any combination of signs the maximum is obtained by taking all $\mathrm{x}_{\mathrm{i}}$ as large as possible. Suppose we have a different set of $\mathrm{x}_{\mathrm{i}}$. Then for some $\mathrm{k}$ we must have $\mathrm{x}_{\mathrm{k}}<2 \mathrm{x}_{\mathrm{k}-1}$ and $\mathrm{x}_{\mathrm{i}}=2 \mathrm{x}_{\mathrm{i}-1}$ for all $\mathrm{i}>\mathrm{k}$. Suppose $2 \mathrm{x}^{\mathrm{k}-1}-\mathrm{x}^{k}=\mathrm{h}>0$. Then we can increase $\mathrm{x}_{\mathrm{k}}$ by $\mathrm{h}, \mathrm{x}_{\mathrm{k}+1}$ by $2 \mathrm{~h}, \mathrm{x}_{\mathrm{k}+2}$ by $4 \mathrm{~h}$, ... So the sum will be increased by $\mathrm{h}\left( \pm 1 \pm 2 \pm \ldots \pm 2^{\mathrm{m}-1}+2^{\mathrm{m}}\right)$ for some $\mathrm{m} \geq 0$. But $\pm 1 \pm 2 \pm$ $\ldots \pm 2^{\mathrm{m}-1} \geq-\left(1+2+\ldots+2^{\mathrm{m}-1}\right)=-2^{\mathrm{m}}+1$, so the overall sum will be increased by at least 1 . So the set of $x_{i}$ was not maximal.


A $\square\{1,2,3, \ldots, 49\}$ does not contain six consecutive integers. Find the largest possible value of |A|. How many such subsets are there (of the maximum size)?

## Answer

$\max =41$; no. ways 495

Solution:
We must exclude at least one element of each of the 8 sets $\{1,2, \ldots, 6\},\{7, \ldots, 12\},\{13, \ldots$, $18\}, \ldots,\{43, \ldots, 48\}$. So $|\mathrm{A}| \leq 41$. But a value of 41 is certainly possible, for example, exclude $2,8,14, \ldots, 44$.

The largest excluded element must be at least 44 (or we have the 6 consecutive elements 44, 45, 46, 47, 48, 49). The smallest excluded element must be at most 6 . If we exclude 2 and 44, then the difference between them is 7.6 and so the other 6 excluded elements are fixed. But if we exclude 3 and 44, for example, then there are several possible choices for the other elements.

There are 5 ways of choosing the smallest and largest excluded element to get a difference of 7. 6 between them ( 2 and 44, 3 and 45, 4 and 46, 5 and 47,6 and 48). There are 4 ways to get a difference of 7. $6-1$ ( 3 and 44, 4 and 45, 5 and 46, 6 and 47). There are 3 ways to get a difference of 7. $6-2$ ( 4 and 44, 5 and 45, 6 and 46), 2 ways to get a difference of 7. $6-3$ (5 and 44, 6 and 45), and 1 way to get a difference of 7. $6-4$ (6 and 44).

If the difference is $7.6-1$, then we can shorten any of the 7 gaps, so there are 7 possibilities. For example, with 3 and 44, we could shorten the first gap, so excluding 3, 8, 14, 20, 26, 32, 38 and 44 , or the second gap, so excluding $3,9,14,20,26,32,38$ and 44 , and so on.

If the difference is $7 \cdot 6-2$, then we can shorten one gap by two ( 7 possibilities) or two gaps by one (21 possibilities), total 28 . If the difference is $7 \cdot 6-3$, then we can shorten on gap by three (7), one by two and one by one (42) or three by one (35), total 84. Finally, if the difference is 7. 6 - 4, we can shorten one by four (7), one by three and one by 1 (42), two by two (21), one by two and two by one (105), or four by one (35), total 210.

So the total number of possibilities is $5 \cdot 1+4 \cdot 7+3 \cdot 28+2 \cdot 84+1 \cdot 210=495$.


$\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side lengths $B X, X Y, Y D$.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-52.jpg?height=439&width=328&top_left_y=1508&top_left_x=247)

We have $\mathrm{DY}<\mathrm{BY} \leq \mathrm{BX}+\mathrm{XY}$ (this is almost obvious, but to prove formally use the cosine formula for BAY and DAY and notice that $\angle \mathrm{BAY}>\angle \mathrm{DAY}$ ). Similarly, $\mathrm{BX}<\mathrm{DX} \leq \mathrm{DY}+$ YX. So it remains to show that $X Y<B X+D Y$.

Take $\mathrm{Q}^{\prime}$ on the extension of $\mathrm{BC}$ so that $\mathrm{BQ}^{\prime}=\mathrm{DQ}$, as shown in the diagram. Take $\mathrm{Y}^{\prime}$ on $\mathrm{AQ}^{\prime}$ so that $\mathrm{AY}^{\prime}=\mathrm{AY}$. Then $\mathrm{XY}^{\prime} \leq \mathrm{BX}+\mathrm{BY}^{\prime}=\mathrm{BX}+\mathrm{DY}$. Now we claim that $\angle \mathrm{PAQ}^{\prime}>\angle \mathrm{PAQ}$, so it follows by the same observation as above that $\mathrm{XY}^{\prime}>\mathrm{XY}$. But the claim is almost obvious. Note that $\mathrm{PQ}^{\prime}=\mathrm{AB}$

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-53.jpg?height=334&width=371&top_left_y=256&top_left_x=266)

So take $\mathrm{P}^{\prime}$ on $\mathrm{AD}$ with $\angle \mathrm{P}^{\prime} \mathrm{PQ} \mathrm{P}^{\prime}=90^{\circ}$. Then $\mathrm{A}$ lies inside the circle $\mathrm{P}^{\prime} \mathrm{PQ}^{\prime}$, so extend $\mathrm{PA}$ to meet it again at $\mathrm{A}^{\prime}$. Then $\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}=\angle \mathrm{PP}^{\prime} \mathrm{Q}^{\prime}=45^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}=\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}+\angle \mathrm{AQ}^{\prime} \mathrm{Q}^{\prime}>45^{\circ}$. But $\angle \mathrm{PAQ}^{\prime}+\angle \mathrm{PAQ}=90^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}>\angle \mathrm{PAQ}$ as claimed.


The sequences $\mathrm{a}_{0}, \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots$ and $\mathrm{b}_{0}, \mathrm{~b}_{1}, \mathrm{~b}_{2}, \ldots$ are defined by $\mathrm{a}_{0}=1, \mathrm{~b}_{0}=4, \mathrm{a}_{\mathrm{n}+1}=\mathrm{a}_{\mathrm{n}}{ }^{2001}+\mathrm{b}_{\mathrm{n}}$, $b_{n+1}=b_{n}^{2001}+a_{n}$. Show that no member of either sequence is divisible by 2003.

Solution:
2003 is prime, so $\mathrm{a}^{2002}=1 \bmod 2003$ for any a not divisible by 2003. Thus $a_{n+1}=a_{n}^{-1}+b_{n}$ $\bmod 2003, b_{n+1}=b_{n}^{-1}+a_{n} \bmod 2003$. Put $c_{n}=a_{n} b_{n}$. Then $c_{n+1}=c_{n}+1 / c_{n}+2=\left(c_{n}+1\right)_{2} / c_{n}$ $\bmod 2003$. So if $\mathrm{c}_{\mathrm{n}} \neq 0 \bmod 2003$, then $\mathrm{c}_{\mathrm{n}+1} \neq 0 \bmod 2003$ unless $\mathrm{c}_{\mathrm{n}}=-1 \bmod 2003$. Then if $c_{n+1}=-1 \bmod 2003$, we must have $\left(c_{n}{ }^{2}+3 c_{n}+1\right) / c_{n}=0 \bmod 2003$, so $c_{n}{ }^{2}+3 c_{n}+1=0 \bmod$ 2003. Note that $c_{0}=4$. So it is sufficient to show that there are no solutions to $x^{2}+3 x+1=0$ $\bmod 2003$, or equivalently to $(x-1000)^{2}=1000^{2}-1=502 \bmod 2003$. In other words, we have to show that 502 is a quadratic non-residue mod 2003.

The easiest way to do that is to use the law of quadratic reciprocity, but that is almost certainly outside the syllabus. We note that 4. $502=5 \bmod 2003$, so 502 is a square iff 5 is a square. It is sufficient to show that $5^{1001}=-1 \bmod 2003$, for then if we had $x^{2}=5$, we would have $\mathrm{x}^{2002}=-1 \bmod 2003$, whereas we know that $\mathrm{x}^{2002}=1 \bmod 2003$. We note that $1001=$ 7. 11. 13. We start by showing that $5^{7}=8 \bmod 2003$. We have $5^{5}=3125=1122 \bmod 2003$, so $5^{6}=5610=1604 \bmod 2003$, so $5^{7}=8020=8 \bmod 2003$.

We calculate successively $2^{11}=2048=45 \bmod 2003$, so $2^{22}=2025=22 \bmod 2003$. Multiplying by 22 is relatively easy, so $2^{44}=484,2^{66}=10648=633,2^{88}=13926=-95,2^{110}$ $=-2090=-87,2^{132}=-1914=89,2^{143}=4005=-1$ all $\bmod 2003$. Hence $8^{11 \cdot 13}=-1 \bmod 2003$, so $5^{1001}=-1 \bmod 2003$, as required, and we are done.