JBMO/md/en-official/en-jbmo_2020_solutions.md

Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:

$$\{\begin{array}{l}a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\ a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{array}\backslash left\backslash \{\backslash begin\{array\}\{l\}\; a+b+c=\backslash frac\{1\}\{a\}+\backslash frac\{1\}\{b\}+\backslash frac\{1\}\{c\}\; \backslash \backslash \; a^\{2\}+b^\{2\}+c^\{2\}=\backslash frac\{1\}\{a^\{2\}\}+\backslash frac\{1\}\{b^\{2\}\}+\backslash frac\{1\}\{c^\{2\}\}\; \backslash end\{array\}\backslash right.$$

Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have

$$a+b+c=\frac{ab+bc+ca}{abc}a+b+c=\backslash frac\{a\; b+b\; c+c\; a\}\{a\; b\; c\}$$

Now, from the first condition and the second condition we get

$$(a+b+c{)}^2-(a^2+b^2+c^2)={(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^2-(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})(a+b+c)^\{2\}-\backslash left(a^\{2\}+b^\{2\}+c^\{2\}\backslash right)=\backslash left(\backslash frac\{1\}\{a\}+\backslash frac\{1\}\{b\}+\backslash frac\{1\}\{c\}\backslash right)^\{2\}-\backslash left(\backslash frac\{1\}\{a^\{2\}\}+\backslash frac\{1\}\{b^\{2\}\}+\backslash frac\{1\}\{c^\{2\}\}\backslash right)$$

The last one simplifies to

$$ab+bc+ca=\frac{a+b+c}{abc}a\; b+b\; c+c\; a=\backslash frac\{a+b+c\}\{a\; b\; c\}$$

First we show that $a+b+c$ and $a b+b c+c a$ are different from 0 . Suppose on contrary then from relation (1) or (2) we have $a+b+c=a b+b c+c a=0$. But then we would have

$$a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)=0a^\{2\}+b^\{2\}+c^\{2\}=(a+b+c)^\{2\}-2(a\; b+b\; c+c\; a)=0$$

which means that $a=b=c=0$. This is not possible since $a, b, c$ should be different from 0 . Now multiplying (1) and (2) we have

$$(a+b+c)(ab+bc+ca)=\frac{(a+b+c)(ab+bc+ca)}{(abc{)}^2}(a+b+c)(a\; b+b\; c+c\; a)=\backslash frac\{(a+b+c)(a\; b+b\; c+c\; a)\}\{(a\; b\; c)^\{2\}\}$$

Since $a+b+c$ and $a b+b c+c a$ are different from 0 , we get $(a b c)^{2}=1$ and using the fact that $a b c>0$ we obtain that $a b c=1$. So relations (1) and (2) transform to

$$a+b+c=ab+bc+ca\mathrm{.}a+b+c=a\; b+b\; c+c\; a\; .$$

Therefore,

$$(a-1)(b-1)(c-1)=abc-ab-bc-ca+a+b+c-1=0\text{. }(a-1)(b-1)(c-1)=a\; b\; c-a\; b-b\; c-c\; a+a+b+c-1=0\; \backslash text\; \{.\; \}$$

This means that at least one of the numbers $a, b, c$ is equal to 1 . Suppose that $c=1$ then relations (1) and (2) transform to $a+b+1=a b+a+b \Rightarrow a b=1$. Taking $a=t$ then we have $b=\frac{1}{t}$. We can now verify that any triple $(a, b, c)=\left(t, \frac{1}{t}, 1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash{0}$. From the initial observation any triple $(a, b, c)=\left(t, \frac{1}{t},-1\right)$ satisfies both conditions. $t \in \mathbb{R} \backslash{0}$. So, all triples that satisfy both conditions are $(a, b, c)=\left(t, \frac{1}{t}, 1\right),\left(t, \frac{1}{t},-1\right)$ and all permutations for any $t \in \mathbb{R} \backslash{0}$.

Comment by PSC. After finding that $a b c=1$ and

$$a+b+c=ab+bc+caa+b+c=a\; b+b\; c+c\; a$$

we can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial

$$P(x)=x^3-s{x}^2+sx-1P(x)=x^\{3\}-s\; x^\{2\}+s\; x-1$$

which has one root equal to 1 . Then, we can conclude as in the above solution.

Problem 2. Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersection of $(c)$ with $Z C$ and let $F$ be the antidiametric point of $D$ with respect to (c). Let $P$ be the point of intersection of the lines $F E$ and $C Z$. If the tangent to (c) at $Z$ meets $P A$ at $T$, prove that the points $T, E, B, Z$ are concyclic.

Solution. We will first show that $P A$ is tangent to $(c)$ at $A$.

Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.

Since $\angle F E D=90^{\circ}$, then $\angle P E D=90^{\circ}$ and so

$$\mathrm{\angle }EPD={90}^{\circ }-\mathrm{\angle }EDC={90}^{\circ }-\mathrm{\angle }BCA=\mathrm{\angle }EAC\backslash angle\; E\; P\; D=90^\{\backslash circ\}-\backslash angle\; E\; D\; C=90^\{\backslash circ\}-\backslash angle\; B\; C\; A=\backslash angle\; E\; A\; C$$

Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle C P A=90^{\circ}$ and therefore the triangle $\angle P A Z$ is right-angled. Since also $B$ is the midpoint of $A Z$, then $P B=A B=B Z$ and so $\angle Z P B=$ $\angle P Z B$.

Furthermore, $\angle E P D=\angle E A C=\angle C B A=\angle E B A$ from which it follows that the points $P, E, B, Z$ are also concyclic.

Now observe that

$$\mathrm{\angle }PAE=\mathrm{\angle }PCE=\mathrm{\angle }ZPB-\mathrm{\angle }PBE=\mathrm{\angle }PZB-\mathrm{\angle }PZE=\mathrm{\angle }EZB\backslash angle\; P\; A\; E=\backslash angle\; P\; C\; E=\backslash angle\; Z\; P\; B-\backslash angle\; P\; B\; E=\backslash angle\; P\; Z\; B-\backslash angle\; P\; Z\; E=\backslash angle\; E\; Z\; B$$

Therefore $P A$ is tangent to $(c)$ at $A$ as claimed.

It now follows that $T A=T Z$. Therefore

Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.

Problem 3. Alice and Bob play the following game: Alice picks a set $A={1,2, \ldots, n}$ for some natural number $n \geqslant 2$. Then starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers, and should differ by 1 from an already chosen number. The game ends when all numbers from the set $A$ are chosen. Alice wins if the sum of all of the numbers that she has chosen is composite. Otherwise Bob wins. Decide which player has a winning strategy.

Solution. To say that Alice has a winning strategy means that she can find a number $n$ to form the set A, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such $n$ does not exist, this would mean that Bob has a winning strategy instead.

Alice can try first to check the small values of $n$. Indeed, this gives the following winning strategy for her: she initially picks $n=8$ and responds to all possible choices made by Bob as in the list below (in each row the choices of Bob and Alice are given alternatively, starting with Bob):

$\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}2 & 3 & 1 & 4 & 5 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}2 & 3 & 4 & 1 & 5 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}3 & 2 & 1 & 4 & 5 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}3 & 2 & 4 & 5 & 1 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}3 & 2 & 4 & 5 & 6 & 1 & 7 & 8\end{array}$

$\begin{array}{llllllll}4 & 5 & 3 & 6 & 2 & 1 & 7 & 8\end{array}$

$\begin{array}{llllllll}4 & 5 & 3 & 6 & 7 & 8 & 2 & 1\end{array}$

$\begin{array}{llllllll}4 & 5 & 6 & 7 & 3 & 2 & 1 & 8\end{array}$

$\begin{array}{llllllll}4 & 5 & 6 & 7 & 3 & 2 & 8 & 1\end{array}$

$\begin{array}{llllllll}4 & 5 & 6 & 7 & 8 & 3 & 2 & 1\end{array}$

$\begin{array}{llllllll}5 & 4 & 3 & 2 & 1 & 6 & 7 & 8\end{array}$

$\begin{array}{llllllll}5 & 4 & 3 & 2 & 6 & 7 & 1 & 8\end{array}$

$\begin{array}{llllllll}5 & 4 & 3 & 2 & 6 & 7 & 8 & 1\end{array}$

$\begin{array}{llllllll}5 & 4 & 6 & 3 & 2 & 1 & 7 & 8\end{array}$

$\begin{array}{llllllll}5 & 4 & 6 & 3 & 7 & 8 & 2 & 1\end{array}$

$\begin{array}{llllllll}6 & 7 & 5 & 4 & 3 & 8 & 2 & 1\end{array}$

$\begin{array}{llllllll}6 & 7 & 5 & 4 & 8 & 3 & 2 & 1\end{array}$

$\begin{array}{llllllll}6 & 7 & 8 & 5 & 4 & 3 & 2 & 1\end{array}$

$\begin{array}{llllllll}7 & 6 & 8 & 5 & 4 & 3 & 2 & 1\end{array}$

$\begin{array}{llllllll}7 & 6 & 5 & 8 & 4 & 3 & 2 & 1\end{array}$

$\begin{array}{llllllll}8 & 7 & 6 & 5 & 4 & 3 & 2 & 1\end{array}$

In all cases, Alice's sum is either an even number greater than 2 , or else 15 or 21 , thus Alice always wins.

Problem 4. Find all pairs $(p, q)$ of prime numbers such that

$$1+\frac{p^q-q^p}{p+q}1+\backslash frac\{p^\{q\}-q^\{p\}\}\{p+q\}$$

is a prime number.

Solution. It is clear that $p \neq q$. We set

$$1+\frac{p^q-q^p}{p+q}=r1+\backslash frac\{p^\{q\}-q^\{p\}\}\{p+q\}=r$$

and we have that

$$p^q-q^p=(r-1)(p+q)p^\{q\}-q^\{p\}=(r-1)(p+q)$$

From Fermat's Little Theorem we have

$$p^q-q^p\equiv -q(p)p^\{q\}-q^\{p\}\; \backslash equiv-q\; \backslash quad(\backslash bmod\; p)$$

Since we also have that

$$(r-1)(p+q)\equiv -rq-q(p)(r-1)(p+q)\; \backslash equiv-r\; q-q\; \backslash quad(\backslash bmod\; p)$$

from (3) we get that

$$rq\equiv 0(p)\Rightarrow p\mid qrr\; q\; \backslash equiv\; 0\; \backslash quad(\backslash bmod\; p)\; \backslash Rightarrow\; p\; \backslash mid\; q\; r$$

hence $p \mid r$, which means that $p=r$. Therefore, (3) takes the form

$$p^q-q^p=(p-1)(p+q)p^\{q\}-q^\{p\}=(p-1)(p+q)$$

We will prove that $p=2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have

$$p^q-q^p\equiv p(q)p^\{q\}-q^\{p\}\; \backslash equiv\; p\; \backslash quad(\backslash bmod\; q)$$

and since

$$(p-1)(p+q)\equiv p(p-1)(q)(p-1)(p+q)\; \backslash equiv\; p(p-1)\; \backslash quad(\backslash bmod\; q)$$

we have

Now, from (4) we have

$$p^q-q^p\equiv 0(p-1)\Rightarrow 1-q^p\equiv 0(p-1)\Rightarrow q^p\equiv 1(p-1)p^\{q\}-q^\{p\}\; \backslash equiv\; 0\; \backslash quad(\backslash bmod\; p-1)\; \backslash Rightarrow\; 1-q^\{p\}\; \backslash equiv\; 0\; \backslash quad(\backslash bmod\; p-1)\; \backslash Rightarrow\; q^\{p\}\; \backslash equiv\; 1\; \backslash quad(\backslash bmod\; p-1)$$

Clearly $\operatorname{gcd}(q, p-1)=1$ and if we set $k=\operatorname{ord}_{p-1}(q)$, it is well-known that $k \mid p$ and $k<p$, therefore $k=1$. It follows that

$$q\equiv 1(p-1)\Rightarrow p-1\mid q-1\Rightarrow p-1\le q-1\Rightarrow p\le qq\; \backslash equiv\; 1\; \backslash quad(\backslash bmod\; p-1)\; \backslash Rightarrow\; p-1\; \backslash mid\; q-1\; \backslash Rightarrow\; p-1\; \backslash leq\; q-1\; \backslash Rightarrow\; p\; \backslash leq\; q$$

a contradiction.

Therefore, $p=2$ and (4) transforms to

$$2^q=q^2+q+22^\{q\}=q^\{2\}+q+2$$

We can easily check by induction that for every positive integer $n \geq 6$ we have $2^{n}>n^{2}+n+2$. This means that $q \leq 5$ and the only solution is for $q=5$. Hence the only pair which satisfy the condition is $(p, q)=(2,5)$.

Comment by the PSC. From the problem condition, we get that $p^{q}$ should be bigger than $q^{p}$, which gives

$$q\ln p>p\ln q⟺\frac{\ln p}{p}>\frac{\ln q}{q}q\; \backslash ln\; p>p\; \backslash ln\; q\; \backslash Longleftrightarrow\; \backslash frac\{\backslash ln\; p\}\{p\}>\backslash frac\{\backslash ln\; q\}\{q\}$$

The function $\frac{\ln x}{x}$ is decreasing for $x>e$, thus if $p$ and $q$ are odd primes, we obtain $q>p$.