APMO/md/en-apmo1997_sol.md

Solutions

Note: The points to be awarded for each part of the solution are indicated on the right side. Problem 1.

which is easily shown by induction. (up to 3 points) Now 5 is the sum of the reciprocals of these numbers where the last, $1993006=$ $1996 \times 1997$ $\frac{1996 \times 1997}{2}=8996$. Thus we have

$$\begin{array}{c}{\displaystyle S=\frac{1}{2}(\frac{2}{1}+\frac{3}{2}+\dots +\frac{1997}{1996})}\\ {\displaystyle =\frac{1}{2}(1996+(1+\frac{1}{2}+\dots +\frac{1}{1996}))}\end{array}\backslash begin\{gathered\}\; S=\backslash frac\{1\}\{2\}\backslash left(\backslash frac\{2\}\{1\}+\backslash frac\{3\}\{2\}+\backslash ldots+\backslash frac\{1997\}\{1996\}\backslash right)\; \backslash \backslash \; =\backslash frac\{1\}\{2\}\backslash left(1996+\backslash left(1+\backslash frac\{1\}\{2\}+\backslash ldots+\backslash frac\{1\}\{1996\}\backslash right)\backslash right)\; \backslash end\{gathered\}$$

(ap to 3 points)

$$\therefore \frac{1}{2}(1996+6)\backslash therefore\; \backslash frac\{1\}\{2\}(1996+6)$$

$$=1001=1001$$

P'roblem 2. Note that $2^{n+2}=2\left(2^{n-1}+1\right)$ so that $n$ is of the form 2 r with r odd. We will consider two cases. i) $n=2 p$ with $p$ prime. $2 p \mid 2^{2 p}+2$, implies that $p!2^{2 n-1}+1$ and hence, hence $p \mid 2^{40-2}-1$. On the follows that $p \mid 2^{d}-1$. But $d \mid p-1$ and $d \mid 4 p-2=4(p-1)+2$. Fence $d \mid 2$ and since $p-1,4 p-2$ are even $d=2$. Then $p=3$ and $n=6<100$. (up to 2 points) ii) $n=2 p q$ where $p, q$ are odd primes, $p<q$ and $p q<\frac{1997}{2}$. Now $n!2 r+2$ impites that $p i 2^{n-1}$ +1 and therefore that $\mathrm{p}^{\prime} 2^{2 \mathrm{m}-2} . .1=2^{4 \mathrm{m}-2}-1$. Once again by Femat's theorern we have pl $2^{n-1}$ 1 which implies that $\mathrm{p}-1 / 4 \mathrm{pq}$-2. The same holds tole for q so that

$$q-1\mid 4pq-2q-1\; \backslash mid\; 4\; p\; q-2$$

Both $p-1$ and $q-1$ are thus multiples of 2 but not of 4 su that $p \equiv q \equiv 3$ (mod 4). ( points) laking $p=3$, we huve $4 p q-2=12 q-2$. Now from (1) we have

if $q \geq 11$, and clearly $\frac{12 q-?}{q-1}=13$ if $q=11$. But this gives $n=2(3)(11)=66<100$. Furthermore $(p, \varphi)=:(3,7)$ does not salisty (1).

Taking $p=7$ we observe that $4 \mathrm{pq}-2=28 \mathrm{q}-2$, and from (1) we have

if $q \geq 27$ and clearly $\frac{28 q-2}{q-1}=29$ if $q=27$. But 27 is not prime and the cases $(p, q)=(7$. 11), (7,19) and (7,23) do not satisfy (1).

Taking $p=11$, then $4 p q-2=44 q-2$, and

Now clearly $\frac{44 q-2}{q-1}=45$ when $q=43$. In this case we have $n:=2 p q=2(11)(43)=946$. Furthermore, $\frac{2^{44 x}+2}{946}$ is incieed an integer. The cases $(p, q)=(11,19),(11,23)$ and 111,31$)$ do not satisfy (1). (2 points) [Aditionally for completeness, if $p=19$ then $4 p y-2=76 q-2$ and $76<\frac{76 q-2}{4-1}=77$ if $q$ $\geqq 75$. Now 75 is not prime and for the cases $(p, 4)=(19,23),(19,31),(19,43)$ and $(19,47), q$ -1 is not a divisor of $74=2 \times 37$.

Similarly, if $p=23$ then $4 \mathrm{pq}-2=92 \mathrm{q}-2$ and $92<\frac{92 q-2}{q-\Gamma} \leq 93$ if $q \geq 91$ and $\frac{92 q-2}{q-1}=93$ if $q=91$. But 91 is not prime and of the cases $(p, q)=(23,31),(23,43)$, when q $=31$ all of the conditions are satisfied. But, $n=2 p q=1426$ is not a solution because $\frac{2^{1426}+2}{1+20}$ is not an integer.

No other pairs of F, q yield numbers within the required range.] (1) point)

Problem 3.

$\angle A L D=\frac{1}{2}(\overparen{M C}+\widehat{A B})$ $\angle \triangle N M$

It is known (see Geometry Revisited) or easily derivable that

$$m_a^2=(AL{)}^2=bc(1-{\left(\frac{a}{b+c}\right)}^2)m\_\{a\}^\{2\}=(A\; L)^\{2\}=b\; c\backslash left(1-\backslash left(\backslash frac\{a\}\{b+c\}\backslash right)^\{2\}\backslash right)$$

(1 point) From $\triangle$ ADL a MAN we have

$$\begin{array}{c}{\displaystyle \frac{AD}{AL}=\frac{4M}{MN}\Rightarrow }\\ {\displaystyle h_a\cdot 2R=AL\cdot AM=m_a\cdot Ma}\\ {\displaystyle h_a=\frac{2(ABC)}{a}}\end{array}\backslash begin\{gathered\}\; \backslash frac\{A\; D\}\{A\; L\}=\backslash frac\{4\; M\}\{M\; N\}\; \backslash Rightarrow\; \backslash \backslash \; h\_\{a\}\; \backslash cdot\; 2\; R=A\; L\; \backslash cdot\; A\; M=m\_\{a\}\; \backslash cdot\; M\; a\; \backslash \backslash \; h\_\{a\}=\backslash frac\{2(A\; B\; C)\}\{a\}\; \backslash end\{gathered\}$$

$$\begin{array}{c}{\displaystyle \frac{2(ABC)}{a}\cdot 2R=m_a{M}_a}\\ {\displaystyle \frac{\frac{ubc}{d}\cdot 4R}{a}=m_a{M}_a}\\ {\displaystyle bc=m_a{M}_u}\end{array}\backslash begin\{gathered\}\; \backslash frac\{2(A\; B\; C)\}\{a\}\; \backslash cdot\; 2\; R=m\_\{a\}\; M\_\{a\}\; \backslash \backslash \; \backslash frac\{\backslash frac\{u\; b\; c\}\{d\}\; \backslash cdot\; 4\; R\}\{a\}=m\_\{a\}\; M\_\{a\}\; \backslash \backslash \; b\; c=m\_\{a\}\; M\_\{u\}\; \backslash end\{gathered\}$$

So that

$$l_a=\frac{m_a^2}{m_a{I}_a}=1-(b+c{)}^{2}l\_\{a\}=\backslash frac\{m\_\{a\}^\{2\}\}\{m\_\{a\}\; I\_\{a\}\}=1-(b+c)^\{2\}$$

with similar expressions for $l_{b}$ and $l_{c}$. (2 points) $\therefore$ Given that $\sin A=\frac{t^{2}}{2 k}$, etc. the expression we are working with becomes

But $a b c=f R(.4 B C)$ so that this last expression becomes

$$\frac{3R(a+b+c)}{4(ABC)}=\frac{3R\cdot 2s}{4sr}=3\cdot \frac{\beta }{2r}=3\backslash frac\{3\; R(a+b+c)\}\{4(A\; B\; C)\}=\backslash frac\{3\; R\; \backslash cdot\; 2\; s\}\{4\; s\; r\}=3\; \backslash cdot\; \backslash frac\{\backslash beta\}\{2\; r\}=3$$

since $R \geq 2 r$. All of the inequalitics are equalities iff $a=b=c$. (1 point)

Problem 4.

• (a) Consider the gequence of triangles on the plane 4 . $42,4, A_{2}, 4,4,4,4,4, \ldots$ It is easy to see that any pair of them are similar. Let's prove that triangles $A_{1} A_{3} A_{5}$ and $A_{3} A$; are similar. Triangles $A_{2} A_{3} A_{4}$ and $A_{4} A_{4} A_{6}$ are similar and their altitudes are $A_{4} A_{5}$ and $A_{6}, A_{0}$, then

$$\frac{A_2{A}_3}{A_4}=\frac{A_4{A}_5}{A_6{A}_7}\mathrm{.}\backslash frac\{A\_\{2\}\; A\_\{3\}\}\{A\_\{4\}\}=\backslash frac\{A\_\{4\}\; A\_\{5\}\}\{A\_\{6\}\; A\_\{7\}\}\; .$$

Triangles $A A_{} A_{s}$ and $A_{s} A_{} A$ - are similar. then

$$\frac{A+A_5}{A_0{A}_7}=\frac{A3As}{A_5{A}_7}\backslash frac\{A+A\_\{5\}\}\{A\_\{0\}\; A\_\{7\}\}=\backslash frac\{A\; 3\; A\; s\}\{A\_\{5\}\; A\_\{7\}\}$$

Now we can conclude that

$$\frac{A_1{A}_3}{A_3{A}_s}=\frac{A_3{A}_s}{A_4}\backslash frac\{A\_\{1\}\; A\_\{3\}\}\{A\_\{3\}\; A\_\{s\}\}=\backslash frac\{A\_\{3\}\; A\_\{s\}\}\{A\_\{4\}\}$$

and triangles $A, A: A$ and $A ; A: A-$ are similar. $\triangle \quad A \cdot A_{s} A_{1}=\triangle \quad A-A_{3} P=\triangle \quad A_{A} A_{1}$, so triangle $A-A, P$ has a righr angle at $P$ and lines $A_{1} A_{5}$ and $A A^{A}$ - are perpendicular. In the same way lines $A A_{2}$. and $A$ are perpendicular and lines $A, A y$ and $A_{-1} A_{11}$ are perpendicular, hence $A_{1}, A_{3}, A_{9}$ are collinear and $A_{-}, A_{1 i} . A_{j}$ are collincar. It follows that triangle $A_{l A} A_{2} A_{;}$and $A_{0} A_{10} A_{l \mid}$ are homothetic and the center of homothety is P . Moreover, all an interior point to any of these triangles and there is no other point distinct from $P$ that is interior to any of these triangles. So this is the point we are looking for.

$$\text{ (up to }+\text{ points) }\backslash text\; \{\; (up\; to\; \}+\; \backslash text\; \{\; points)\; \}$$

(b) Since $\triangle \quad A_{1} P A_{3} 90^{\prime \prime}$ then $P$ lies on the circle with diameter $A_{i}, A_{3}$. Let $A_{i} A_{3}=1, A_{2} A_{2}=s, A_{3} A_{3}$ $=r$ and let $A_{i} A_{2} A_{3}$ be clockwise. Triangles $A_{1} A_{2} A_{3}$ and $A_{3} A_{+} A_{5}$ are similar, thus $A_{2} A_{5}: r=s: 1$, and so $A_{2} A_{5}=r s$. Besides $A_{3} A_{y}=r \sqrt{1+s^{2}}$ (Pythagoras), and area of triangle $A_{1} A_{2} A_{3}=$ $\frac{1}{2} r \sqrt{1+s^{2}} \cdot \sqrt{1+s^{2}}=\frac{1}{2} s \cdot 1$. Thus $r=\frac{s}{1+s^{2}}$. By the arithmetic-geometric mean $\frac{s}{1+s^{2}} \leq \frac{1}{2}$, thus $r \leq \frac{1}{4}$ and the set of all possible values of $r$ consists of two real intervals $\left[-\frac{1}{2}, 0\right)$ and $\left(0, \frac{1}{2}\right]$. $\triangle A_{2} A_{1} P$ takes the maximum value when $r= \pm$ thus the locus of $P$ consists of two continuous arcs from the circle with dianeter $4, A_{i}$ with two extreme positions corresponding to $r=-\frac{1}{2}$ and $r=\frac{1}{2}$. (up to 3 points)

Problem 5.

A redistribution can be written as $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ where $x_{1}$ denotes the number of objects transferred from $A_{i}$ to $A_{i+1}$. Our objective is to minimize the function $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n}\left|x_{1}\right|$ After redistribution we should have at each $A_{i}, a_{i}-x_{1}+x_{i-1}=N$ for $i \in{1,2, \ldots, n}$ where $x_{0}$ means $x_{n}$. (1 point) Solving this system of linear equations we obtain: $x_{i}=x_{1}-\left[(i-1) N-a_{2}-a_{3}-\ldots-a_{i}\right]$ for $i \in{1,2, \ldots, n}$. Hence

Basically the problem reduces to find the minimum of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|$ where $\alpha_{i}=(i-1) N-\sum_{j=2}^{i} a_{j}$. (up to. 3 points)

First rearrange $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ in non decreasing order. Collecting terms which are equal to one another we write the ordered sequence $\beta_{1}<\beta_{2}<\cdots<\beta_{m}$, each $\beta_{i}$ occurs $k_{i}$ times in the family $\left{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right}$. Thus $k_{1}+k_{2}+\cdots+k_{m}=n$.

Consider the intervals $\left(-\infty, \beta_{1}\right],\left[\begin{array}{l}\beta_{1}, \beta_{2}\end{array}\right], \cdots,\left[\begin{array}{cc}\beta_{m-1}, \beta_{m}\end{array}\right],\left[\begin{array}{l}\left.\beta_{m}, \infty\right) \ \end{array}\right.$ the graph of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|=\sum_{i=1}^{m} k_{i}\left|x-\beta_{i}\right|$ is a continuos piece wise linear graph define in the following way:

$$F(x)=\{\begin{array}{c}{k}_1({\beta }_1-x)+k_2({\beta }_2-x)+\cdots +k_m({\beta }_m-x)\text{ if }\mathrm{x}\in (-\mathrm{\infty },{\beta }_1]\\ k_1(x-{\beta }_1)+k_2({\beta }_2-x)+\cdots +k_m({\beta }_m-x)\text{ if }\mathrm{x}\in [{\beta }_1,{\beta }_2]\\ \mathrm{⋮}\\ k_1(x-{\beta }_1)+k_2(x-{\beta }_2)+\cdots +k_m)(x-{\beta }_m)\text{ if }\mathrm{x}\in [{\beta }_m,\mathrm{\infty })\end{array}F(x)=\backslash left\backslash \{\backslash begin\{array\}\{c\}\; k\_\{1\}\backslash left(\backslash beta\_\{1\}-x\backslash right)+k\_\{2\}\backslash left(\backslash beta\_\{2\}-x\backslash right)+\backslash cdots+k\_\{m\}\backslash left(\backslash beta\_\{m\}-x\backslash right)\; \backslash text\; \{\; if\; \}\; \backslash mathrm\{x\}\; \backslash in\backslash left(-\backslash infty,\; \backslash beta\_\{1\}\backslash right]\; \backslash \backslash \; k\_\{1\}\backslash left(x-\backslash beta\_\{1\}\backslash right)+k\_\{2\}\backslash left(\backslash beta\_\{2\}-x\backslash right)+\backslash cdots+k\_\{m\}\backslash left(\backslash beta\_\{m\}-x\backslash right)\; \backslash text\; \{\; if\; \}\; \backslash mathrm\{x\}\; \backslash in\backslash left[\backslash beta\_\{1\},\; \backslash beta\_\{2\}\backslash right]\; \backslash \backslash \; \backslash vdots\; \backslash \backslash \; \backslash left.k\_\{1\}\backslash left(x-\backslash beta\_\{1\}\backslash right)+k\_\{2\}\backslash left(x-\backslash beta\_\{2\}\backslash right)+\backslash cdots+k\_\{m\}\backslash right)\backslash left(x-\backslash beta\_\{m\}\backslash right)\; \backslash text\; \{\; if\; \}\; \backslash mathrm\{x\}\; \backslash in\backslash left[\backslash beta\_\{m\},\; \backslash infty\backslash right)\; \backslash end\{array\}\backslash right.$$

The slopes of each line segment on each interval are respectively: $S_{0}=-k_{1}-k_{2}-k_{3}-\cdots-k_{m}$ $S_{1}=k_{1}-k_{2}-k_{3}-\cdots-k_{m}$ $S_{2}=k_{1}+k_{2}-k_{3}-\cdots-k_{m}$ $S_{m}=k_{1}+k_{2}+k_{3}+\cdots+k_{m}$ Note that this sequence of increasing numbers goes from a negative to a positive number, hence for some $t \geq 1$ there is an

In the first case the minimum occurs at $x=\beta_{t}$ or $\beta_{t+1}$ and in the second case the minimum occurs at $x=\beta_{t}$ (Up to 7 points) We can rephrase the computations above in terms of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ rather than $\beta_{1}, \beta_{2}, \cdots, \beta_{m}$. After rearranging the $\alpha^{\prime} s$ in non decreasing order, pick $x=\alpha \quad$ if $n$ is odd and take $x=\alpha$ or $\alpha \quad$ if $n$ is even.

$$\frac{n+1}{2}\frac{n}{2}\frac{n}{2}+1\backslash frac\{n+1\}\{2\}\; \backslash quad\; \backslash frac\{n\}\{2\}\; \backslash quad\; \backslash frac\{n\}\{2\}+1$$

If no justification is given for the choice of $x$, give up to 4 points.