RoJBMO
Collection
1 item • Updated
id int64 1 508 | problem_uid stringlengths 12 21 | problem_source stringclasses 3
values | year int64 2.01k 2.03k | type stringclasses 4
values | statement stringlengths 64 921 | solution stringlengths 182 6.82k | stmt_len int64 64 921 | sol_len int64 182 6.82k | type_label stringclasses 4
values | problem_num int64 1 27 |
|---|---|---|---|---|---|---|---|---|---|---|
219 | jbmo_2010_p1 | jbmo | 2,010 | a | The real numbers $a, b, c, d$ satisfy simultaneously the equations
\[
abc - d = 1,\quad bcd - a = 2,\quad cda - b = 3,\quad dab - c = -6.
\]
Prove that $a + b + c + d
eq 0$. | Suppose that $a + b + c + d = 0$. Then
\[
abc + bcd + cda + dab = 0. \tag{1}
\]
If $abcd = 0$, then one of the numbers, say $d$, must be $0$. In this case $abc = 0$, and so at least two of the numbers $a, b, c, d$ will be equal to $0$, making one of the given equations impossible. Hence $abcd
eq 0$ and, from (1),
\[
\... | 174 | 664 | Algebra | 1 |
220 | jbmo_2010_p2 | jbmo | 2,010 | n | Find all integers $n$, $n \geqslant 1$, such that $n \cdot 2^{n+1} + 1$ is a perfect square. | Answer: $n = 0$ and $n = 3$.
Clearly $n \cdot 2^{n+1} + 1$ is odd, so, if this number is a perfect square, then $n \cdot 2^{n+1} + 1 = (2x+1)^2$, $x \in \mathbb{N}$, whence $n \cdot 2^{n-1} = x(x+1)$.
The integers $x$ and $x+1$ are coprime, so one of them must be divisible by $2^{n-1}$, which means that the other mus... | 92 | 575 | Number Theory | 2 |
221 | jbmo_2010_p3 | jbmo | 2,010 | g | Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$. | The point $M$ lies on the circumcircle of $\triangle ABK$ (since both $AL$ and the perpendicular bisector of $BK$ bisect the arc $BK$ of this circle). Then $\angle CBK = \angle ABK = \angle AMK = \angle NLA$. Thus $ABLN$ is cyclic, whence $\angle NAL = \angle NBL = \angle CBK = \angle NLA$. Now it follows that $LN = NA... | 292 | 322 | Geometry | 3 |
222 | jbmo_2010_p4 | jbmo | 2,010 | c | A $9 \times 7$ rectangle is tiled with tiles of the two types shown in the picture below (the tiles are composed by three, respectively four unit squares and the L-shaped tiles can be rotated repeatedly with $90^\circ$).
Let $n \geqslant 0$ be the number of the $2 \times 2$ tiles which can be used in such a tiling. Fi... | Answer: $0$ or $3$.
Denote by $x$ the number of pieces of the L-shaped type and by $y$ the number of pieces of the type $2 \times 2$. Mark $20$ squares of the rectangle as in the figure.
Obviously, each piece covers at most one marked square. Thus,
\[
x + y \geq 20 \tag{1}
\]
and consequently $3x + 3y \geq 60$ (2). O... | 345 | 514 | Combinatorics | 4 |
223 | jbmo_2012_p1 | jbmo | 2,012 | a | Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that
\[
\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c} + 6 \geq 2\sqrt{2}\!\left(\frac{1-a}{a} + \frac{1-b}{b} + \frac{1-c}{c}
ight).
\]
When does equality hold? | Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
\[
\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} + 6 \geq 2\sqrt{2}\!\left(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}
ight)
\]
and equivalently
\[
\left(\frac{b+c}{a} - 2\sqrt{2}\cdot\frac{b+c}{a} + 2
igh... | 275 | 823 | Algebra | 1 |
224 | jbmo_2012_p2 | jbmo | 2,012 | g | Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$, that touches $k_1$ and $k_2$ at $M$ and $N$, respectively. If $t \perp AM$ and $MN = 2AM$, evaluate $\angle NMB$. | \textit{Solution 1.} Let $P$ be the symmetric of $A$ with respect to $M$. Then $AM = MP$ and $t \perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\angle NAP = \angle NPA$. We have $\angle BAP = \angle BAM = \angle BMN$ and $\angle BAN = \angle BNM$.
Thus we have
\[
180^\circ - \angle NBM = \a... | 243 | 1,216 | Geometry | 2 |
225 | jbmo_2012_p3 | jbmo | 2,012 | c | On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
\begin{enumerate}[label=\alph*)]
\item $6$?
\item $7$?
\end{enumerate} | \textit{Solution.} (a) The answer is \textbf{no}.
Suppose it is possible. Consider some color, say blue. Each blue string is the side of $4$ triangles formed with vertices on the given points. As there exist $\binom{5}{2} = \tfrac{5 \cdot 4}{2} = 10$ pairs of colors other than blue, and for any such pair of colors tog... | 306 | 1,825 | Combinatorics | 3 |
226 | jbmo_2012_p4 | jbmo | 2,012 | n | Find all positive integers $x, y, z$ and $t$ such that
\[
2^x \cdot 3^y + 5^z = 7^t.
\] | \textit{Solution.} Reducing modulo $3$ we get $5^z \equiv 1$, therefore $z$ is even, $z = 2c$, $c \in \mathbb{N}$.
Next we prove that $t$ is even. Obviously $t \geq 2$. Let us suppose that $t$ is odd, say $t = 2d+1$, $d \in \mathbb{N}$. The equation becomes $2^x \cdot 3^y + 25^c = 7 \cdot 49^d$. If $x \geq 2$, reducin... | 87 | 2,772 | Number Theory | 4 |
197 | shl_jbmo_2012_a1 | shl_jbmo | 2,012 | a | \textit{Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove the inequality}
\[
\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c} + 6 \geq 2\sqrt{2}\left(\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}
ight)
\]
\textit{When does equality hold?... | \textbf{Solution.} Replacing $1-a, 1-b, 1-c$ with $b+c, a+c, a+b$ respectively on the right hand side, the given inequality becomes
\[
\frac{a+c}{b} + \frac{b+c}{a} + \frac{a+b}{c} + 6 \geq 2\sqrt{2}\left(\sqrt{\frac{b+c}{a}} + \sqrt{\frac{a+c}{b}} + \sqrt{\frac{a+b}{c}}
ight)
\]
and equivalently
\[
\left(\frac{a+c}{b}... | 321 | 878 | Algebra | 1 |
198 | shl_jbmo_2012_a2 | shl_jbmo | 2,012 | a | \textit{Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that}
\[
\frac{1}{a^3 + bc} + \frac{1}{b^3 + ca} + \frac{1}{c^3 + ab} \leq \frac{(ab + bc + ca)^2}{6}.
\] | \textbf{Solution.} By the AM-GM inequality we have $a^3 + bc \geq 2\sqrt{a^3 bc} = 2\sqrt{a^2(abc)} = 2a$ and so
\[
\frac{1}{a^3 + bc} \leq \frac{1}{2a}.
\]
Similarly, $\dfrac{1}{b^3 + ca} \leq \dfrac{1}{2b}$, $\dfrac{1}{c^3 + ab} \leq \dfrac{1}{2c}$ and then
\[
\frac{1}{a^3 + bc} + \frac{1}{b^3 + ca} + \frac{1}{c^3 + ... | 182 | 722 | Algebra | 2 |
199 | shl_jbmo_2012_a3 | shl_jbmo | 2,012 | a | \textit{Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = a^2 + b^2 + c^2$. Show that}
\[
\frac{a^2}{a^2 + ab} + \frac{b^2}{b^2 + bc} + \frac{c^2}{c^2 + ca} \geq \frac{a+b+c}{2}.
\] | \textbf{Solution.} By the Cauchy-Schwarz inequality it is
\[
\left(\frac{a^2}{a^2+ab} + \frac{b^2}{b^2+bc} + \frac{c^2}{c^2+ca}
ight)\left((a^2+ab)+(b^2+bc)+(c^2+ca)
ight) \geq (a+b+c)^2
\]
\[
\Rightarrow \quad \frac{a^2}{a^2+ab} + \frac{b^2}{b^2+bc} + \frac{c^2}{c^2+ca} \geq \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}
\]
S... | 197 | 761 | Algebra | 3 |
200 | shl_jbmo_2012_a4 | shl_jbmo | 2,012 | a | \textit{Solve the following equation for $x, y, z \in \mathbb{N}$}
\[
\left(1 + \frac{x}{y+z}
ight)^2 + \left(1 + \frac{y}{z+x}
ight)^2 + \left(1 + \frac{z}{x+y}
ight)^2 = \frac{27}{4}
\] | \textbf{Solution 1.} Call $a = 1 + \dfrac{x}{y+z}$, $b = 1 + \dfrac{y}{z+x}$, $c = 1 + \dfrac{z}{x+y}$ to get
\[
a^2 + b^2 + c^2 = \frac{27}{4}.
\]
Since it is also true that
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2,
\]
the quadratic-harmonic means inequality implies
\[
\frac{3}{2} = \sqrt{\frac{a^2+b^2+c^2}{3}} ... | 187 | 2,162 | Algebra | 4 |
201 | shl_jbmo_2012_a5 | shl_jbmo | 2,012 | a | \textit{Find the largest positive integer $n$ for which the inequality}
\begin{equation}
\frac{a+b+c}{abc+1} + \sqrt[n]{abc} \leq \frac{5}{2} \tag{1}
\end{equation}
\textit{holds for all $a, b, c \in [0,1]$. Here $\sqrt[n]{abc} = abc$.} | \textbf{Solution.} Let $n_{max}$ be the sought largest value of $n$, and let $E_{a,b,c}(n) = \dfrac{a+b+c}{abc+1} + \sqrt[n]{abc}$. Then $E_{a,b,c}(m) - E_{a,b,c}(n) = \sqrt[m]{abc} - \sqrt[n]{abc}$ and since $abc \leq 1$ we clearly have $E_{a,b,c}(m) \geq E_{a,b,c}(n)$ for $m \geq n$. So if $E_{a,b,c}(n) \geq \frac{5}... | 236 | 2,636 | Algebra | 5 |
202 | shl_jbmo_2012_g1 | shl_jbmo | 2,012 | g | \textit{Let $ABC$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $ABC$ and distinct from $A$, $B$ and $C$. If the lines through $P$ and parallel to $BC$, $CA$, $AB$ intersect the lines $CA$, $AB$, $BC$ at $M$, $N$ and $Q$ respectively, prove that $M$, $N$ and $Q$ are collinear.} | \textbf{Solution.} Without any loss of generality, let $P$ be in the minor arc of the chord $AC$ as in Figure 1. Since $\angle PNA = \angle NPM = 60^\circ$ and $\angle NAM = \angle PMA = 120^\circ$, it follows that the points $A$, $M$, $P$ and $N$ are concyclic. This yields
\begin{equation}
\angle NMP = \angle NAP. \ta... | 312 | 786 | Geometry | 6 |
203 | shl_jbmo_2012_g2 | shl_jbmo | 2,012 | g | \textit{Let $ABC$ be an isosceles triangle with $AB = AC$. Let also $c(K, KC)$ be a circle tangent to the line $AC$ at point $C$ which it intersects the segment $BC$ again at an interior point $H$. Prove that $HK \perp AB$.} | \textbf{Solution 1.} Let lines $KH$, $AB$ intersect at $M$ (Figure 5a). From the quadrilateral $KMAC$ we have
$\angle KMA = 360^\circ - \angle A - \angle ACK - \angle CKM = 360^\circ - \angle A - 90^\circ - (180^\circ - 2\angle KCH) = 90 - \angle A + 2\angle KCH = 90 - \angle A + 2(90^\circ - \angle ACB) = 270^\circ -... | 224 | 1,761 | Geometry | 7 |
204 | shl_jbmo_2012_g3 | shl_jbmo | 2,012 | g | \textit{Let $AB$ and $CD$ be chords in a circle of center $O$ with $A$, $B$, $C$, $D$ distinct, and let the lines $AB$ and $CD$ meet at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $AC$ and $BD$ respectively. If $MN \perp OE$, prove that $AD \| BC$.} | \textbf{Solution.} $E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A$, $B$, $C$, $D$ are distinct. Let lines $AC$ and $NE$ meet at point $P$. Then $EN = DN = BN$ (median in a right triangle), so $\angle PEC = \angle NED = \angle ND... | 274 | 795 | Geometry | 8 |
205 | shl_jbmo_2012_g4 | shl_jbmo | 2,012 | g | \textit{Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O$, $H$ be the triangle's circumcenter and orthocenter respectively. Let also $A'$ be the point where the angle bisector of angle $BAC$ meets $\Gamma$. If $A'H = AH$, find the measure of angle $BAC$.} | \textbf{Solution.} The segment $AA'$ bisects $\angle OAH$: if $\angle BCA = y$ (Figure 4), then $\angle BOA = 2y$, and since $OA = OB$, it is $\angle OAB = \angle OBA = 90^\circ - y$. Also since $AH \perp BC$, it is $\angle HAC = 90^\circ - y = \angle OAB$ and the claim follows.
Since $AA'$ bisects $\angle OAH$ and $A... | 283 | 1,162 | Geometry | 9 |
206 | shl_jbmo_2012_g5 | shl_jbmo | 2,012 | g | \textit{Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$ that touches them at $M$ and $N$ respectively. If $t \perp AM$ and $MN = 2AM$, evaluate $\angle NMB$.} | \textbf{Solution.} Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 5). Then $AM = MP$ and $t \perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\angle NAP = \angle NPA$. We have $\angle BAP = \angle BAM = \angle BMN$ and $\angle BAN = \angle BNM$. Thus
\[
180^\circ - \angle NBM = \... | 239 | 952 | Geometry | 10 |
207 | shl_jbmo_2012_g6 | shl_jbmo | 2,012 | g | \textit{Let $O_1$ be a point in the exterior of the circle $c(O, R)$ and let $O_1N$, $O_1D$ be the tangent segments from $O_1$ to the circle. On the segment $O_1N$ consider the point $B$ such that $BN = R$. Let the line from $B$ parallel to $ON$ intersect the segment $O_1D$ at $C$. If $A$ is a point on the segment $O_1... | \textbf{Solution.} Obviously, the segment $BC$ is tangent to the circle $c$. Let $M$ be the point of tangency (Figure 6). Call $Q$, $M$ the tangency points of $BA$, $BC$ with $c'$ and $c$ respectively, and call $H$ the midpoint of segment $AC$. It is well known that
\[
AQ = \frac{1}{2}(AO_1 + AB - BO_1) \quad \text{and... | 470 | 1,588 | Geometry | 11 |
208 | shl_jbmo_2012_g7 | shl_jbmo | 2,012 | g | \textit{Let $MNPQ$ be a square of side length 1, and $A$, $B$, $C$, $D$ points on the sides $MN$, $NP$, $PQ$, and $QM$ respectively such that $AC \cdot BD = \dfrac{5}{4}$. Can the set $\{AB, BC, CD, DA\}$ be partitioned into two subsets $S_1$ and $S_2$ of two elements each, so that each one of the sums of the elements ... | \textbf{Solution.} The answer is negative.
Suppose such a partitioning was possible (Figure 7). Then $AB + BC + CD + DA \in \mathbb{N}$. But $(AB + BC) + (CD + DA) > AC + AC \geq 2$, hence $AB + BC + CD + DA > 2$. On the other hand, $AB + BC + CD + DA < (AN + NB) + (BP + PC) + (CQ + QD) + (DM + MA) = 4$, hence $AB + B... | 362 | 2,324 | Geometry | 12 |
209 | shl_jbmo_2012_c1 | shl_jbmo | 2,012 | c | \textit{Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{th}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with his nam... | \textbf{Solution.} Yes it is: Rotating the table by the angles $\dfrac{360^\circ}{11}$, $2 \cdot \dfrac{360^\circ}{11}$, $3 \cdot \dfrac{360^\circ}{11}$, \ldots, $10 \cdot \dfrac{360^\circ}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee ... | 458 | 622 | Combinatorics | 13 |
210 | shl_jbmo_2012_c2 | shl_jbmo | 2,012 | c | \textit{$n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails connected by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7?} | \textbf{Solution.} (a) The answer is no:
Suppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\binom{5}{2} = \frac{5 \cdot 4}{2} = 10$ pairs of colors other than blue, and for any such pair of colors together with t... | 246 | 1,814 | Combinatorics | 14 |
211 | shl_jbmo_2012_c3 | shl_jbmo | 2,012 | c | \textit{In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less than or equal to $\dfrac{1}{72}$.} | \textbf{Solution.} \underline{Lemma}: If a triangle $ABC$ lies in a rectangle $KLMN$ with sides $KL = a$ and $LM = b$, then the area of the triangle is less than or equal to $\dfrac{ab}{2}$.
Proof of the lemma: Without any loss of generality assume that among the distance of $A$, $B$, $C$ from $KL$, that of $A$ is bet... | 201 | 2,402 | Combinatorics | 15 |
212 | shl_jbmo_2012_n1 | shl_jbmo | 2,012 | n | \textit{If $a$, $b$ are integers and $s = a^3 + b^3 - 60ab(a+b) \geq 2012$, find the least possible value of $s$.} | \textbf{Solution.} It is $s = (a+b)^3 - 63ab(a+b)$ which gives the same residue modulo 7 as $(a+b)^3$. But the residues modulo 7 of perfect cubes can only be 0, 1 or 6. So the residue of $s$ modulo 7 is 0, 1 or 6. Now for $a = 6$, $b = -1$ we get $s = 2015 \geq 2012$ and this is the least possible value of $s$ because ... | 114 | 502 | Number Theory | 16 |
213 | shl_jbmo_2012_n2 | shl_jbmo | 2,012 | n | \textit{Do there exist prime numbers $p$ and $q$ such that $p^2(p^3 - 1) = q(q+1)$?} | \textbf{Solution.} Write the given equation in the form
\begin{equation}
p^2(p-1)(p^2+p+1) = q(q+1). \tag{9}
\end{equation}
First observe that it must not be $p = q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct primes, (9) immediately yields $p... | 84 | 1,834 | Number Theory | 17 |
214 | shl_jbmo_2012_n3 | shl_jbmo | 2,012 | n | \textit{Decipher the equality}
\[
(\overline{VER} - \overline{IA}) : (\overline{GRE} + \overline{ECE}) = G^{R^E},
\]
\textit{assuming that the number $\overline{GREECE}$ has a maximum value. It is supposed that each letter corresponds to a unique digit from 0 to 9 and different letters correspond to different digits, a... | \textbf{Solution.} Denote
\[
x = \overline{VER} - \overline{IA}, \quad y = \overline{GRE} + \overline{ECE}, \quad z = G^{R^E}.
\]
Then obviously, we have
\begin{align*}
(201 + 131 \ \text{or} \ 231 + 101) &\leq y \leq (879 + 969 \ \text{or} \ 869 + 979 \ \text{or} \ 769 + 989) \\
\Rightarrow \quad 332 &\leq y \leq 1848... | 512 | 4,044 | Number Theory | 18 |
215 | shl_jbmo_2012_n4 | shl_jbmo | 2,012 | n | \textit{Determine all triples $(m, n, p)$ satisfying}
\begin{equation}
n^{2p} = m^2 + n^2 + p + 1 \tag{27}
\end{equation}
\textit{where $m$ and $n$ are integers and $p$ is a prime number.} | \textbf{Solution.} By Fermat's theorem $n^{2p} \equiv n^2 \pmod{p}$, therefore $m^2 + n^2 + p + 1 \equiv n^2 \pmod{p} \Rightarrow m^2 \equiv -1 \pmod{p}$.
\textit{Case 1}: $p = 4k + 3$. We have $(m^2)^{2k+1} \equiv (-1)^{2k+1} \pmod{p}$. Therefore,
\begin{equation}
m^{p-1} \equiv -1 \pmod{p} \tag{28}
\end{equation}
an... | 188 | 1,247 | Number Theory | 19 |
216 | shl_jbmo_2012_n5 | shl_jbmo | 2,012 | n | \textit{Find all the positive integers $x$, $y$, $z$, $t$ such that $2^x \cdot 3^y + 5^z = 7^t$.} | \textbf{Solution.} Reducing modulo 3 we get $5^z \equiv 1$, therefore $z$ is even, $z = 2c$, $c \in \mathbb{N}$.
Next we prove that $t$ is even. Obviously, $t \geq 2$. Let us suppose that $t$ is odd, $t = 2d + 1$, $d \in \mathbb{N}$. The equation becomes $2^x \cdot 3^y + 25^c = 7 \cdot 49^d$.
If $x \geq 2$, reducing ... | 97 | 2,965 | Number Theory | 20 |
217 | shl_jbmo_2012_n6 | shl_jbmo | 2,012 | n | \textit{If $a$, $b$, $c$, $d$ are integers and $A = 2(a - 2b + c)^4 + 2(b - 2c + a)^4 + 2(c - 2a + b)^4$, $B = d(d+1)(d+2)(d+3) + 1$, prove that $\left(\sqrt{A} + 1
ight)^2 + B$ cannot be a perfect square.} | \textbf{Solution.} First we prove the following Lemma
\underline{Lemma}: \textit{If $x$, $y$, $z$ real numbers such that $x + y + z = 0$, then $2(x^4 + y^4 + z^4) = (x^2 + y^2 + z^2)^2$.}
Proof of the Lemma:
\begin{align*}
x^4 + y^4 + z^4 &= x^2 x^2 + y^2 y^2 + z^2 z^2 = x^2(y+z)^2 + y^2(z+x)^2 + z^2(x+y)^2 \\
&= 2(x... | 206 | 1,420 | Number Theory | 21 |
218 | shl_jbmo_2012_n7 | shl_jbmo | 2,012 | n | \textit{Find all natural numbers $a$, $b$, $c$ for which $1997^a + 15^b = 2012^c$.} | \textbf{Solution.} $1997^a + 15^b = 2012^c \Rightarrow 1 + (-1)^b \equiv 0 \pmod{4}$, so $b$ is an odd number.
$1997^a + 15^b = 2012^c \Rightarrow 1 + 0 \equiv 2^c \pmod{3}$, so $c$ is even, say $c = 2c_1$.
We intend to consider the given equation modulo 8 and for this reason we discern two cases:
(1): $c = 1$. Clea... | 83 | 2,261 | Number Theory | 22 |
227 | jbmo_2013_p1 | jbmo | 2,013 | n | Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\dfrac{a^3b - 1}{a+1}$ and $\dfrac{b^3a + 1}{b - 1}$ are both positive integers. | \textbf{Solution.} As $a^3b - 1 = b(a^3 + 1) - (b + 1)$ and $a + 1 \mid a^3 + 1$, we have $a + 1 \mid b + 1$.
As $b^3a + 1 = a(b^3 - 1) + (a + 1)$ and $b - 1 \mid b^3 - 1$, we have $b - 1 \mid a + 1$.
So $b - 1 \mid b + 1$ and hence $b - 1 \mid 2$.
\begin{itemize}
\item If $b = 2$, then $a + 1 \mid b + 1 = 3$ give... | 156 | 631 | Number Theory | 1 |
228 | jbmo_2013_p2 | jbmo | 2,013 | g | Let $ABC$ be an acute triangle with $AB < AC$ and let $O$ be the center of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE... | \textbf{Solution.} We will show that $MOPD$ is a parallelogram. From this it follows that $M$, $N$, $P$ are collinear.
Since $\angle BAD = \angle CAO = 90^\circ - \angle ABC$, $D$ is the foot of the perpendicular from $A$ to side $BC$. Since $M$ is the midpoint of the line segment $BE$, we have $BM = ME = MD$ and henc... | 404 | 1,082 | Geometry | 2 |
229 | jbmo_2013_p3 | jbmo | 2,013 | a | Show that
\[
\left(a + 2b + \frac{2}{a+1}
ight)\!\left(b + 2a + \frac{2}{b+1}
ight) \geq 16
\]
for all positive real numbers $a$ and $b$ such that $ab \geq 1$. | \textbf{Solution 1.} By the AM-GM Inequality we have:
\[
\frac{a+1}{2} + \frac{2}{a+1} \ge 2
\]
Therefore
\[
a + 2b + \frac{2}{a+1} \ge \frac{a+3}{2} + 2b,
\]
and, similarly,
\[
b + 2a + \frac{2}{b+1} \ge 2a + \frac{b+3}{2}.
\]
On the other hand,
\[
(a + 4b + 3)(b + 4a + 3) \ge \left(\sqrt{ab} + 4\sqrt{ab} + 3
... | 159 | 1,702 | Algebra | 3 |
230 | jbmo_2013_p4 | jbmo | 2,013 | c | Let $n$ be a positive integer. Two players, Alice and Bob, are playing the following game:
\begin{itemize}
\item Alice chooses $n$ real numbers, not necessarily distinct.
\item Alice writes all pairwise sums on a sheet of paper and gives it to Bob (there are $\tfrac{n(n-1)}{2}$ such sums, not necessarily distinct).
\it... | \textbf{Solution.}
\textbf{a.} Yes. Let $a \le b \le c \le d \le e$ be the numbers chosen by Alice. As each number appears in a pairwise sum $4$ times, by adding all $10$ pairwise sums and dividing the result by $4$, Bob obtains $a + b + c + d + e$. Subtracting the smallest and the largest pairwise sums $a + b$ and $d... | 773 | 1,621 | Combinatorics | 4 |
179 | shl_jbmo_2013_a1 | shl_jbmo | 2,013 | a | Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:
\[
x^3 = \frac{z}{y} - 2\frac{y}{z}
\]
\[
y^3 = \frac{x}{z} - 2\frac{z}{x}
\]
\[
z^3 = \frac{y}{x} - 2\frac{x}{y}
\] | \textbf{Solution.} We have
\[
x^3yz = z^2 - 2y^2
\]
\[
y^3zx = x^2 - 2z^2
\]
\[
z^3xy = y^2 - 2x^2
\]
with $xyz
eq 0$.
Adding these up we obtain $(x^2 + y^2 + z^2)(xyz + 1) = 0$. Hence $xyz = -1$. Now the system of equations becomes:
\[
x^2 = 2y^2 - z^2
\]
\[
y^2 = 2z^2 - x^2
\]
\[
z^2 = 2x^2 - y^2
\]
Then the first... | 215 | 486 | Algebra | 1 |
180 | shl_jbmo_2013_a2 | shl_jbmo | 2,013 | a | Find the largest possible value of the expression $\left|\sqrt{x^2 + 4x + 8} - \sqrt{x^2 + 8x + 17}
ight|$ where $x$ is a real number. | \textbf{Solution.} We observe that
\[
\left|\sqrt{x^2 + 4x + 8} - \sqrt{x^2 + 8x + 17}
ight| = \left|\sqrt{(x-(-2))^2 + (0-2)^2} - \sqrt{(x-(-4))^2 + (0-1)^2}
ight|
\]
is the absolute difference of the distances from the point $P(x, 0)$ in the $xy$-plane to the points $A(-2, 2)$ and $B(-4, 1)$.
By the Triangle Inequal... | 134 | 751 | Algebra | 2 |
181 | shl_jbmo_2013_a3 | shl_jbmo | 2,013 | a | Show that
\[
\left(a + 2b + \frac{2}{a+1}
ight)\left(b + 2a + \frac{2}{b+1}
ight) \geq 16
\]
for all positive real numbers $a$, $b$ satisfying $ab \geq 1$. | \textbf{Solution 1.} By the AM-GM Inequality we have:
\[
\frac{a+1}{2} + \frac{2}{a+1} \geq 2
\]
Therefore
\[
a + 2b + \frac{2}{a+1} \geq \frac{a+3}{2} + 2b\,.
\]
and, similarly,
\[
b + 2a + \frac{2}{b+1} \geq 2a + \frac{b+3}{2}\,.
\]
On the other hand,
\[
(a + 4b + 3)(b + 4a + 3) \geq (\sqrt{ab} + 4\sqrt{ab} + 3)^2 \g... | 155 | 1,707 | Algebra | 3 |
182 | shl_jbmo_2013_c1 | shl_jbmo | 2,013 | c | Find the largest number of distinct integers that can be chosen from the set $\{1, 2, \ldots, 2013\}$ so that the difference of no two of them is equal to 17. | \textbf{Solution.} Consider the sets $A_{mn} = \{34m + n - 34, 34m + n - 17\}$ for $1 \leq m \leq 59$ and $1 \leq n \leq 17$, and $B_k = \{2006 + k\}$ for $1 \leq k \leq 7$. As we cannot choose more than one number from each of these sets, we can choose at most $59 \cdot 17 + 7 = 1010$ numbers. On the other hand, choos... | 158 | 512 | Combinatorics | 4 |
183 | shl_jbmo_2013_c2 | shl_jbmo | 2,013 | c | On a billiards table in the shape of a rectangle $ABCD$ with $AB = 2013$ and $AD = 1000$, a billiard ball is shot along the bisector of the angle $\angle BAD$. Assuming that the ball is reflected from the sides at the same angle it comes in, determine whether it will ever go to the corner $B$. | \textbf{Solution 1.} The ball travels a horizontal distance of 1000 units between two bounces from the sides $AB$ and $CD$ as it always moves on a line making a $45^\circ$ angle with the sides. Hence it is always at a distance of even number of units to the line $AD$ when it hits $AB$ or $CD$. Hence it can never hit $A... | 294 | 1,285 | Combinatorics | 5 |
184 | shl_jbmo_2013_c3 | shl_jbmo | 2,013 | c | All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if \textbf{a.} $n = 4$, \textbf{b.} $n = 5$, \textbf{c.} $n = 6$? | \textbf{Solution. a.} No. Four apples with weights $1, 5, 7, 9$ and with weights $2, 4, 6, 10$ both give the results $6, 8, 10, 12, 14, 16$ when weighed in pairs.
\textbf{b.} Yes. Let $a \leq b \leq c \leq d \leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and d... | 208 | 1,381 | Combinatorics | 6 |
185 | shl_jbmo_2013_g1 | shl_jbmo | 2,013 | g | Let $AB$ be a diameter of a circle $\omega$ with center $O$ and $OC$ be a radius of $\omega$ which is perpendicular to $AB$. Let $M$ be a point on the line segment $OC$. Let $N$ be the second point of intersection of the line $AM$ with $\omega$, and let $P$ be the point of intersection of the lines tangent to $\omega$ ... | \textbf{Solution.} Since the lines $PN$ and $BP$ are tangent to $\omega$, $NP = PB$ and $OP$ is the bisector of $\angle NOB$. Therefore the lines $OP$ and $NB$ are perpendicular. Since $\angle ANB = 90^\circ$, it follows that the lines $AN$ and $OP$ are parallel. As $MO$ and $PB$ are also parallel and $AO = OB$, the tr... | 393 | 505 | Geometry | 7 |
186 | shl_jbmo_2013_g2 | shl_jbmo | 2,013 | g | $\omega_1$ and $\omega_2$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\omega_3$ at the points $K$ and $L$, respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\omega_1$ and $\omega_2$ intersects $\omega_3$. Show that... | \textbf{Solution.} Let $C$ be the intersection point of the tangent lines to the circles $\omega_1$ at $K$ and $\omega_2$ at $L$. Point $C$ lies on the radical axis of circles $\omega_1$ and $\omega_3$, and also on the radical axis of the circles $\omega_2$ and $\omega_3$. Therefore $C$ lies on the radical axis of the ... | 405 | 925 | Geometry | 8 |
187 | shl_jbmo_2013_g3 | shl_jbmo | 2,013 | g | \textbf{G3.} Let $D$ be a point on the side $BC$ of an acute triangle $ABC$ such that $\angle BAD = \angle CAO$ where $O$ is the center of the circumcircle $\omega$ of the triangle $ABC$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. Let $M$, $N$, $P$ be the midpoints of the line segments $... | \textbf{Solution.} We will show that $MOPD$ is a parallelogram. From this it follows that $M$, $N$, $P$ are collinear.
Since $\angle BAD = \angle CAO = 90^\circ - \angle ABC$, $D$ is the foot of the perpendicular from $A$ to side $BC$. Since $M$ is the midpoint of the line segment $BE$, we have $BM = ME = MD$ and henc... | 389 | 1,082 | Geometry | 9 |
188 | shl_jbmo_2013_g4 | shl_jbmo | 2,013 | g | \textbf{G4.} Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, ... | \textbf{Solution.} First we will show that points $P$ and $Q$ are not on the line segment $AB$.
Assume that $Q$ is on the line segment $AB$. Since $CI = QI$ and $\angle IBQ = \angle IBC$, either the triangles $CBI$ and $QBI$ are congruent or $\angle ICB + \angle IQB = 180^\circ$. In the first case, we have $BC = BQ$ w... | 439 | 1,166 | Geometry | 10 |
189 | shl_jbmo_2013_g5 | shl_jbmo | 2,013 | g | \textbf{G5.} A circle passing through the midpoint $M$ of the side $BC$ and the vertex $A$ of a triangle $ABC$ intersects the sides $AB$ and $AC$ for the second time at the points $P$ and $Q$, respectively. Show that if $\angle BAC = 60^\circ$ then
\[
AP + AQ + PQ < AB + AC + \tfrac{1}{2}\,BC\,.
\] | \textbf{Solution.} Since the quadrilateral $APMQ$ is cyclic, we have $\angle PMQ = 180^\circ - \angle PAQ = 180^\circ - \angle BAC = 120^\circ$. Therefore $\angle PMB + \angle QMC = 180^\circ - \angle PMQ = 60^\circ$.
Let the point $B'$ be the symmetric of the point $B$ with respect to the line $PM$ and the point $C'$... | 299 | 1,169 | Geometry | 11 |
190 | shl_jbmo_2013_g6 | shl_jbmo | 2,013 | g | Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively, of a rectangle $ABCD$. Let $K$ and $M$ be the points of intersection of the line $PD$ with $QB$ and $QA$, respectively, and let $N$ be the point of intersection of the lines $PA$ and $QB$.
Let $X$, $Y$, $Z$ be the midpoints of the line segments... | \textbf{Solution.} Let $R$ be the midpoint of the side $AD$. Then the lines $BR$ and $PD$ are parallel. Since $\angle MAN = \angle QAP = \angle QBR = \angle QKM$, the points $A$, $N$, $K$, $M$ are concyclic.
Let $\ell_4$ be the line passing through the midpoint $W$ of the line segment $MK$ and perpendicular to the lin... | 615 | 1,731 | Geometry | 12 |
191 | shl_jbmo_2013_n1 | shl_jbmo | 2,013 | n | Find all positive integers $n$ for which $1^3 + 2^3 + \cdots + 16^3 + 17^n$ is a perfect square. | \textbf{Solution.} We have $1^3 + 2^3 + \cdots + 16^3 = (1 + 2 + \cdots + 16)^2 = 8^2 \cdot 17^2$. Hence, if $1^3 + 2^3 + \cdots + 16^3 + 17^n = m^2$ for a positive integer $m$, then $17 \mid m$. If $m = 17k$ for some positive integer $k$, then $17^{n-2} = (k+8)(k-8)$. As $(k+8) - (k-8) = 16$, this can only happen when... | 96 | 405 | Number Theory | 13 |
192 | shl_jbmo_2013_n2 | shl_jbmo | 2,013 | n | Find all ordered triples $(x, y, z)$ of integers satisfying $20^x + 13^y = 2013^z$. | \textbf{Solution.} We have $1^3 + 2^3 + \cdots + 16^3 = (1 + 2 + \cdots + 16)^2 = 8^2 \cdot 17^2$. Hence, if $1^3 + 2^3 + \cdots + 16^3 + 17^n = m^2$ for a positive integer $m$, then $17 \mid m$. If $m = 17k$ for some positive integer $k$, then $17^{n-2} = (k+8)(k-8)$. As $(k+8) - (k-8) = 16$, this can only happen when... | 83 | 921 | Number Theory | 14 |
193 | shl_jbmo_2013_n3 | shl_jbmo | 2,013 | n | Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\dfrac{a^3b - 1}{a + 1}$ and $\dfrac{b^3a + 1}{b - 1}$ are positive integers. | \textbf{Solution.} As $a^3b - 1 = b(a^3 + 1) - (b + 1)$ and $a + 1 \mid a^3 + 1$, we have $a + 1 \mid b + 1$.
As $b^3a + 1 = a(b^3 - 1) + (a + 1)$ and $b - 1 \mid b^3 - 1$, we have $b - 1 \mid a + 1$.
So $b - 1 \mid b + 1$ and hence $b - 1 \mid 2$.
\begin{itemize}
\item If $b = 2$, then $a + 1 \mid b + 1 = 3$ gi... | 153 | 635 | Number Theory | 15 |
194 | shl_jbmo_2013_n4 | shl_jbmo | 2,013 | n | A rectangle in the $xy$-plane is called \textit{latticed} if all its vertices have integer coordinates.
\textbf{a.} Find a latticed rectangle with area 2013 whose sides are not parallel to the axes.
\textbf{b.} Show that if a latticed rectangle has area 2011, then its sides are parallel to the axes. | \textbf{Solution. a.} The rectangle $PQRS$ with $P(0, 0)$, $Q(165, 198)$, $R(159, 203)$, $S(-6, 5)$ has area 2013.
\textbf{b.} Suppose that the latticed rectangle $PQRS$ has area 2011 and its sides are not parallel to the axes. Without loss of generality we may assume that its vertices are $P(0, 0)$, $Q(a, b)$, $S(c, ... | 302 | 882 | Number Theory | 16 |
195 | shl_jbmo_2013_n5 | shl_jbmo | 2,013 | n | Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation
\[
\frac{1}{x^2} + \frac{y}{xz} + \frac{1}{z^2} = \frac{1}{2013}\,.
\] | \textbf{Solution.} We have $x^2z^2 = 2013(x^2 + xyz + z^2)$. Let $d = \gcd(x, z)$ and $x = da$, $z = db$. Then $a^2b^2d^2 = 2013(a^2 + aby + b^2)$.
As $\gcd(a, b) = 1$, we also have $\gcd(a^2, a^2 + aby + b^2) = 1$ and $\gcd(b^2, a^2 + aby + b^2) = 1$. Therefore $a^2 \mid 2013$ and $b^2 \mid 2013$. But $2013 = 3 \cdot... | 152 | 604 | Number Theory | 17 |
196 | shl_jbmo_2013_n6 | shl_jbmo | 2,013 | n | Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:
\[
x^2 - y^2 = z
\]
\[
3xy + (x - y)z = z^2
\] | \textbf{Solution.} If $z = 0$, then $x = 0$ and $y = 0$, and $(x, y, z) = (0, 0, 0)$.
Let us assume that $z
eq 0$, and $x + y = a$ and $x - y = b$ where $a$ and $b$ are nonzero integers such that $z = ab$. Then $x = (a + b)/2$ and $y = (a - b)/2$, and the second equation gives $3a^2 - 3b^2 + 4ab^2 = 4a^2b^2$.
Hence
... | 141 | 1,121 | Number Theory | 18 |
231 | jbmo_2014_p1 | jbmo | 2,014 | n | Find all distinct prime numbers $p$, $q$ and $r$ such that
\[
3p^4 - 5q^4 - 4r^2 = 26.
\] | \subsection*{Solution}
First notice that if both primes $q$ and $r$ differ from $3$, then
$q^2 \equiv r^2 \equiv 1 \pmod{3}$,
hence the left hand side of the given equation is congruent to zero modulo $3$, which is
impossible since $26$ is not divisible by $3$. Thus, $q = 3$ or $r = 3$. We consider two cases.
\textbf... | 89 | 1,084 | Number Theory | 1 |
232 | jbmo_2014_p2 | jbmo | 2,014 | g | Consider an acute triangle $ABC$ with area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocenters of the triangles $MNC$ and $MND$ respectively. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$. | \subsection*{Solution 1}
Let $O$, $P$, $K$, $R$ and $T$ be the mid-points of the segments $CD$, $MN$, $CN$, $CH_1$
and $MH_1$, respectively. From $\triangle MNC$ we have that $PK = \tfrac{1}{2}MC$ and
$PK \parallel MC$. Analogously, from $\triangle MH_1C$ we have that $TR = \tfrac{1}{2}MC$ and
$TR \parallel MC$. Conse... | 291 | 1,615 | Geometry | 2 |
233 | jbmo_2014_p3 | jbmo | 2,014 | a | Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
\[
\left(a + \frac{1}{b}
ight)^2 + \left(b + \frac{1}{c}
ight)^2 + \left(c + \frac{1}{a}
ight)^2 \geq 3(a + b + c + 1).
\]
When does equality hold? | \subsection*{Solution 1}
By using AM-GM ($x^2 + y^2 + z^2 \geq xy + yz + zx$) we have
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq
\left(a+\frac{1}{b}
ight)\!\left(b+\frac{1}{c}
ight)
+\left(b+\frac{1}{c}
ight)\!\left(c+\frac{1}{a}
ight)
+\left(c+\frac{1}{a}
ight)\!\left(... | 218 | 4,413 | Algebra | 3 |
234 | jbmo_2014_p4 | jbmo | 2,014 | c | For a positive integer $n$, two players $A$ and $B$ play the following game: Given a pile of $s$ stones, the players take turns alternatively with $A$ going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a positive multiple of $n$ stones. The winner is the one who ta... | Denote by $k$ the sought number and let $\{s_1, s_2, \ldots, s_k\}$ be the corresponding values
for $s$. We call each $s_i$ a \emph{losing number} and every other nonnegative integer a
\emph{winning number}.
Clearly every multiple of $n$ is a winning number.
Suppose there are two different losing numbers $s_i > s_j$,... | 435 | 2,624 | Combinatorics | 4 |
154 | shl_jbmo_2014_a1 | shl_jbmo | 2,014 | a | \textit{For any real number $a$, let $\lfloor a
floor$ denote the greatest integer not exceeding $a$. In positive real numbers solve the following equation}
\[
n + \left\lfloor \sqrt{n}
ight
floor + \left\lfloor \sqrt[3]{n}
ight
floor = 2014.
\] |
oindent\textbf{Solution 1.} Obviously $n$ must be a positive integer. Now note that $44^2 = 1936 < 2014 < 2025 = 45^2$ and $12^3 < 1900 < 2014 < 13^3$.
If $n < 1950$ then $2014 = n + \lfloor\sqrt{n}
floor + \lfloor\sqrt[3]{n}
floor < 1950 + 44 + 12 = 2006$, a contradiction!
So $n \geq 1950$. Also if $n > 2000$ then ... | 248 | 1,152 | Algebra | 1 |
155 | shl_jbmo_2014_a2 | shl_jbmo | 2,014 | a | \textit{Let $a$, $b$ and $c$ be positive real numbers such that $abc = \dfrac{1}{8}$. Prove the inequality}
\[
a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 \geq \frac{15}{16}.
\]
\textit{When does equality hold?} |
oindent\textbf{Solution 1.} By using the Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
\begin{align*}
&a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 \\
&= \frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{c^2}{4}+\frac{c^2}{4... | 210 | 1,018 | Algebra | 2 |
156 | shl_jbmo_2014_a3 | shl_jbmo | 2,014 | a | \textit{Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that:}
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 \geq 3(a+b+c+1).
\]
\textit{When does equality hold?} |
oindent\textbf{Solution 1.} By using AM-GM ($x^2+y^2+z^2 \geq xy+yz+zx$) we have
\begin{align*}
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 &\geq \left(a+\frac{1}{b}
ight)\left(b+\frac{1}{c}
ight)+\left(b+\frac{1}{c}
ight)\left(c+\frac{1}{a}
ight)+\left(c+\frac{1}{a}
ight)\left(... | 217 | 1,873 | Algebra | 3 |
157 | shl_jbmo_2014_a4 | shl_jbmo | 2,014 | a | \textit{Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that}
\[
\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} \geq \frac{69}{4}.
\]
\textit{When does equality hold?} |
oindent\textbf{Solution 1.} The inequality can be written as:
\[
\frac{5+2(1+b)}{1+a}+\frac{5+2(1+c)}{1+b}+\frac{5+2(1+a)}{1+c} \geq \frac{69}{4}.
\]
We substitute $1+a=x, 1+b=y, 1+c=z$.
So, we have to prove the inequality
\[
\frac{5+2y}{x}+\frac{5+2z}{y}+\frac{5+2x}{z} \geq \frac{69}{4} \Longleftrightarrow 5\left(\f... | 187 | 1,176 | Algebra | 4 |
158 | shl_jbmo_2014_a5 | shl_jbmo | 2,014 | a | \textit{Let $x,y,z$ be non-negative real numbers satisfying $x+y+z=xyz$. Prove that}
\[
2\left(x^2+y^2+z^2
ight) \geq 3(x+y+z),
\]
\textit{and determine when equality occurs.} |
oindent\textbf{Solution.} Equality holds when $x=y=z=0$.
Apply AM-GM to $x+y+z=xyz$,
\begin{align*}
xyz = x+y+z &\geq 3\sqrt[3]{xyz} \Rightarrow (xyz)^3 \geq \left(3\sqrt[3]{xyz}
ight)^3\\
&\Rightarrow x^3y^3z^3 \geq 27xyz\\
&\Rightarrow x^2y^2z^2 \geq 27\\
&\Rightarrow \sqrt[4]{x^2y^2z^2} \geq 3
\end{align*}
Also by... | 175 | 1,802 | Algebra | 5 |
159 | shl_jbmo_2014_a6 | shl_jbmo | 2,014 | a | \textit{Let $a,b,c$ be positive real numbers. Prove that}
\[
\left((3a^2+1)^2+2\left(1+\frac{3}{b}
ight)^2
ight)\left((3b^2+1)^2+2\left(1+\frac{3}{c}
ight)^2
ight)\left((3c^2+1)^2+2\left(1+\frac{3}{a}
ight)^2
ight) \geq 48^3.
\]
\textit{When does equality hold?} |
oindent\textbf{Solution.} Let $x$ be a positive real number. By AM-GM we have $\dfrac{1+x+x+x}{4} \geq x^{3/4}$, or equivalently $1+3x \geq 4x^{3/4}$. Using this inequality we obtain:
\[
(3a^2+1)^2 \geq 16a^3 \quad \text{and} \quad 2\left(1+\frac{3}{b}
ight)^2 \geq 32b^{-3/2}.
\]
Moreover, by the inequality of arithme... | 262 | 648 | Algebra | 6 |
160 | shl_jbmo_2014_a7 | shl_jbmo | 2,014 | a | \textit{Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=48$. Prove}
\[
a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16} \leq 24^2.
\]
\textit{When does equality hold?} |
oindent\textbf{Solution.} Observe that $2x^3+16 = 2(x^3+8) = 2(x+2)(x^2-2x+4)$. From AM-GM:
\[
\sqrt{2x^3+16} = \sqrt{(2x+4)(x^2-2x+4)} \leq \frac{2x+4+x^2-2x+4}{2} = \frac{x^2+8}{2} \tag{1}.
\]
By adding the inequality (1) obtained for $x=a$, $x=b$ and $x=c$ it suffices to prove:
\[
a^2b^2+8a^2+b^2c^2+8b^2+c^2a^2+8c^... | 184 | 495 | Algebra | 7 |
161 | shl_jbmo_2014_a8 | shl_jbmo | 2,014 | a | \textit{Let $x$, $y$ and $z$ be positive real numbers such that $xyz=1$. Prove the inequality}
\[
\frac{1}{x(ay+b)}+\frac{1}{y(az+b)}+\frac{1}{z(ax+b)} \geq 3, \quad \text{if:}
\]
\begin{enumerate}[label=\alph*)]
\item $a=0$ and $b=1$;
\item $a=1$ and $b=0$;
\item $a+b=1$ for $a,b>0$
\end{enumerate}
\textit{When ... |
oindent\textbf{Solution.}
oindent a) The inequality reduces to $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
oindent b) Here the inequality reduces to $\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx} \geq 3$, i.e.\ $x+y+z \geq 3$, ... | 650 | 1,198 | Algebra | 8 |
162 | shl_jbmo_2014_a9 | shl_jbmo | 2,014 | a | \textit{Let $n$ be a positive integer, and let $x_1,\ldots,x_n,y_1,\ldots,y_n$ be positive real numbers such that $x_1+\ldots+x_n = y_1+\ldots+y_n = 1$. Show that}
\[
|x_1-y_1|+\ldots+|x_n-y_n| \leq 2 - \min_{1\leq i\leq n}\frac{x_i}{y_i} - \min_{1\leq i\leq n}\frac{y_i}{x_i}.
\] |
oindent\textbf{Solution.} Up to reordering the real numbers $x_i$ and $y_i$, we may assume that $\dfrac{x_1}{y_1} \leq \ldots \leq \dfrac{x_n}{y_n}$. Let $A = \dfrac{x_1}{y_1}$ and $B = \dfrac{x_n}{y_n}$, and $S = |x_1-y_1|+\ldots+|x_n-y_n|$. Our aim is to prove that $S \leq 2-A-\dfrac{1}{B}$.
First, note that we can... | 280 | 1,266 | Algebra | 9 |
163 | shl_jbmo_2014_c1 | shl_jbmo | 2,014 | c | \textit{Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\dfrac{ab}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:}
\begin{enumerate}[label=\alph*)]
\item \tex... |
oindent\textbf{Solution.} a) Observe that $\dfrac{1}{\frac{ab}{a+b}} = \dfrac{1}{a}+\dfrac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_1, a_2, \ldots, a_n$, and $c$ is the length of the last remaining segment, then
\[
\frac{1}{c} = \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n},
\]
prov... | 609 | 550 | Combinatorics | 10 |
164 | shl_jbmo_2014_c2 | shl_jbmo | 2,014 | c | \textit{In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.} |
oindent\textbf{Solution.} Denote by $X_1, X_2, \ldots, X_n$ the cities in the country and let $X_i$ be connected to exactly $m_i$ other cities by direct two-way airline. Then $X_i$ is a final destination of $m_i$ direct routes and an intermediate stop of $m_i(m_i-1)$ non-direct routes. Thus $r = m_1^2+\ldots+m_n^2$. A... | 296 | 722 | Combinatorics | 11 |
165 | shl_jbmo_2014_c3 | shl_jbmo | 2,014 | c | \textit{For a given positive integer $n$, two players A and B play the following game: Given is a pile of $a$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the l... |
oindent\textbf{Solution.} Denote by $k$ the sought number and let $\{a_1, a_2, \ldots, a_k\}$ be the corresponding values for $a$. We will call each $a_i$ a losing number and every other positive integer a winning number. Clearly every multiple of $n$ is a winning number.
Suppose there are two different losing number... | 413 | 1,676 | Combinatorics | 12 |
166 | shl_jbmo_2014_c4 | shl_jbmo | 2,014 | c | \textit{Let $A = 1\cdot4\cdot7\cdots2014$ be the product of the numbers less or equal to $2014$ that give remainder $1$ when divided by $3$. Find the last non-zero digit of $A$.} |
oindent\textbf{Solution.} Grouping the elements of the product by ten we get:
\begin{align*}
&(30k+1)(30k+4)(30k+7)(30k+10)(30k+13)(30k+16)\\
&(30k+19)(30k+22)(30k+25)(30k+28) =\\
&= (30k+1)(15k+2)(30k+7)(120k+40)(30k+13)(15k+8)\\
&\quad (30k+19)(15k+11)(120k+100)(15k+14)
\end{align*}
(We divide all even numbers not d... | 178 | 1,596 | Combinatorics | 13 |
167 | shl_jbmo_2014_g1 | shl_jbmo | 2,014 | g | \textit{Let $ABC$ be a triangle with $\angle B = \angle C = 40^\circ$. The bisector of $\angle B$ meets $AC$ at the point $D$. Prove that $\overline{BD}+\overline{DA}=\overline{BC}$.} |
oindent\textbf{Solution.} Since $\angle BAC = 100^\circ$ and $\angle BDC = 120^\circ$ we have $\overline{BD} < \overline{BC}$. Let $E$ be the point on $\overline{BC}$ such that $\overline{BD}=\overline{BE}$. Then $\angle DEC = 100^\circ$ and $\angle EDC = 40^\circ$, hence $\overline{DE}=\overline{EC}$, and $\angle BAC... | 183 | 495 | Geometry | 14 |
168 | shl_jbmo_2014_g2 | shl_jbmo | 2,014 | g | \textit{Let $ABC$ be an acute triangle with $\overline{AB} < \overline{AC} < \overline{BC}$ and $c(O,R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_1\!\left(A,\overline{AE}
ight)$ intersects $\overline{AC}$ at point $K$, the c... |
oindent\textbf{Solution.} Let $M$ be the point of intersection of the line $DL$ with the circle $c(O,R)$ (we choose $M \equiv D$ if $LD$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is sufficient to prove that the points $E$, $K$ and $M$ are collinear.
We have that $\angle EAC = 90^\ci... | 481 | 1,187 | Geometry | 15 |
169 | shl_jbmo_2014_g3 | shl_jbmo | 2,014 | g | \textit{Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$) for an acute triangle $ABC$ with area $S$. If $H_1$ and $H_2$ are the orthocentres of the triangles $MNC$ and $MND$ respectively. Evaluate the area of the quadrilateral $AH_1BH_2$.} |
oindent\textbf{Solution 1.} Let $O$, $P$, $K$, $R$ and $T$ be the midpoints of the segments $CD$, $MN$, $CN$, $CH_1$ and $MH_1$, respectively. From $\triangle MNC$ we have that $\overline{PK} = \frac{1}{2}\overline{MC}$ and $PK \parallel MC$. Analogously, from $\triangle MH_1C$ we have that $\overline{TR} = \frac{1}{2... | 279 | 1,812 | Geometry | 16 |
170 | shl_jbmo_2014_g4 | shl_jbmo | 2,014 | g | \textit{Let $ABC$ be a triangle such that $\overline{AB}
eq \overline{AC}$. Let $M$ be a midpoint of $\overline{BC}$, $H$ the orthocenter of $ABC$, $O_1$ the midpoint of $\overline{AH}$ and $O_2$ the circumcenter of $BCH$. Prove that $O_1AMO_2$ is a parallelogram.} |
oindent\textbf{Solution 1.} Let $O_2'$ be the point such that $O_1AMO_2'$ is a parallelogram. Note that $\overrightarrow{MO_2'} = \overrightarrow{AO_1} = \overrightarrow{O_1H}$. Therefore, $O_1HO_2'M$ is a parallelogram and $\overrightarrow{MO_1} = \overrightarrow{O_2'H}$.
Since $M$ is the midpoint of $\overline{BC}$... | 266 | 2,120 | Geometry | 17 |
171 | shl_jbmo_2014_g5 | shl_jbmo | 2,014 | g | \textit{Let $ABC$ be a triangle with $\overline{AB}
eq \overline{BC}$, and let $BD$ be the internal bisector of $\angle ABC$ ($D \in AC$). Denote the midpoint of the arc $AC$ which contains point $B$ by $M$. The circumcircle of the triangle $BDM$ intersects the segment $AB$ at point $K
eq B$, and let $J$ be the refle... |
oindent\textbf{Solution 1.}
Let the circumcircle of the triangle $BDM$ intersect the line segment $BC$ at point $L
eq B$. From $\angle CBD = \angle DBA$ we have $\overline{DL} = \overline{DK}$. Since $\angle LCM = \angle BCM = \angle BAM = \angle KAM$, $\overline{MC} = \overline{MA}$ and
\begin{align*}
\angle LMC &=... | 447 | 2,142 | Geometry | 18 |
172 | shl_jbmo_2014_g6 | shl_jbmo | 2,014 | g | \textit{Let $ABCD$ be a quadrilateral whose sides $AB$ and $CD$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_1$ and $H_2$ the orthocenters of the triangles $OAB$ and $OCD$, respectively. If $M$ and $N$ are the midpoints of the segments $\overline{AB}$ and $\overline{CD}$, respectiv... |
oindent\textbf{Solution.}
Let $A'$ and $B'$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $AOB$, and $C'$ and $D'$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $COD$. Obviously, $A'$ and $D'$ belong to the circle $c_1$ of diameter $\overline{AD}$, while $B'$ ... | 425 | 1,491 | Geometry | 19 |
173 | shl_jbmo_2014_n1 | shl_jbmo | 2,014 | n | \textit{Each letter of the word OHRID corresponds to a different digit belonging to the set $\{1,2,3,4,5\}$. Decipher the equality $(O+H+R+I+D)^2:(O-H-R+I+D)=O^{H^{R^{I^D}}}$.} |
oindent\textbf{Solution.} Since $O$, $H$, $R$, $I$ and $D$ are distinct numbers from $\{1,2,3,4,5\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D = O+H+R+I+D-2(H+R)<15$. From this
\[
O^{H^{R^{I^D}}} = \frac{(O+H+R+I+D)^2}{O-H-R+I+D} = \frac{225}{15-2(H+R)},
\]
hence $O^{H^{R^{I^D}}} > 15$ and divides $225$, which is only po... | 176 | 519 | Number Theory | 20 |
174 | shl_jbmo_2014_n2 | shl_jbmo | 2,014 | n | \textit{Find all triples $(p,q,r)$ of distinct primes $p$, $q$ and $r$ such that}
\[
3p^4 - 5q^4 - 4r^2 = 26.
\] |
oindent\textbf{Solution.} First notice that if both primes $q$ and $r$ differ from 3, then $q^2 \equiv r^2 \equiv 1\pmod{3}$, hence the left hand side of the given equation is congruent to zero modulo 3, which is impossible since 26 is not divisible by 3. Thus, $q=3$ or $r=3$. We consider two cases.
\medskip
oinden... | 112 | 1,011 | Number Theory | 21 |
175 | shl_jbmo_2014_n3 | shl_jbmo | 2,014 | n | \textit{Find the integer solutions of the equation}
\[
x^2 = y^2(x+y^4+2y^2).
\] |
oindent\textbf{Solution.} If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x,y)=(0,0)$ is a solution of the problem. Assume $x
eq 0$ and $y
eq 0$ satisfy the equation. The equation can be transformed in the form $x^2-xy^2 = y^6+2y^4$. Then $4x^2-4xy^2+y^4 = 4y^6+9y^4$ and consequently
\[
... | 80 | 886 | Number Theory | 22 |
176 | shl_jbmo_2014_n4 | shl_jbmo | 2,014 | n | \textit{Prove there are no integers $a$ and $b$ satisfying the following conditions:}
\begin{enumerate}[label=
oman*)]
\item $16a-9b$ is a prime number
\item $ab$ is a perfect square
\item $a+b$ is a perfect square
\end{enumerate} |
oindent\textbf{Solution.} Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
\begin{align}
16a-9b &= p \tag{1}\\
ab &= n^2 \tag{2}\\
a+b &= m^2 \tag{3}
\end{align}
Moreover, let $d = \gcd(a,b)$ and $a=dx$, $b... | 236 | 1,259 | Number Theory | 23 |
177 | shl_jbmo_2014_n5 | shl_jbmo | 2,014 | n | \textit{Find all nonnegative integers $x$, $y$, $z$ such that}
\[
2013^x + 2014^y = 2015^z.
\] |
oindent\textbf{Solution.} Clearly, $y>0$, and $z>0$. If $x=0$ and $y=1$, then $z=1$ and $(x,y,z)=(0,1,1)$ is a solution. If $x=0$ and $y \geq 2$, then modulo 4 we have $1+0 \equiv (-1)^z$, hence $z$ is even ($z=2z_1$ for some integer $z_1$). Then $2^y\cdot1007^y = (2015^{z_1}-1)(2015^{z_1}+1)$, and since $\gcd(1007,20... | 94 | 1,574 | Number Theory | 24 |
178 | shl_jbmo_2014_n6 | shl_jbmo | 2,014 | n | \textit{Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:}
\medskip
\textit{Vukasin: ``Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect... |
oindent\textbf{Solution.} Let the middle number be $n$, so the numbers are $n-1$, $n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2\pmod{4}$. Furthermore, neither 3, 5 nor 7 divides $n$. Also $n+1+11 \equiv 2\pmod{4}$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3... | 921 | 2,527 | Number Theory | 25 |
235 | jbmo_2015_p1 | jbmo | 2,015 | n | Find all prime numbers $a$, $b$, $c$ and positive integers $k$ satisfying the equation
\[
a^2 + b^2 + 16c^2 + 9k^2 = 1.
\] |
oindent\textbf{Solution:}
The relation $9 \cdot k^2 + 1 \equiv 1 \pmod{3}$ implies
$$a^2 + b^2 + 16 \cdot c^2 \equiv 1 \pmod{3} \iff a^2 + b^2 + c^2 \equiv 1 \pmod{3}.$$
Since $a^2 \equiv 0, 1 \pmod{3}$, $b^2 \equiv 0, 1 \pmod{3}$, $c^2 \equiv 0, 1 \pmod{3}$, we have:
\bigskip
\begin{center}
\begin{tabular}{|c|c|c|... | 122 | 1,785 | Number Theory | 1 |
236 | jbmo_2015_p2 | jbmo | 2,015 | a | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of the expression
\[
A = \frac{2 + a^3}{2 + b^3} + \frac{2 + b^3}{2 + c^3} + \frac{2 + c^3}{2 + a^3}.
\] |
oindent\textbf{Solution:}
We can rewrite $A$ as follows:
\begin{align*}
A &= \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c} = 2\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}
ight) - a^2 - b^2 - c^2 \\
&= 2\left(\frac{ab + bc + ca}{abc}
ight) - (a^2 + b^2 + c^2) = 2\left(\frac{ab + bc + ca}{abc}
ight) - ((a+... | 197 | 1,196 | Algebra | 2 |
237 | jbmo_2015_p3 | jbmo | 2,015 | g | Let $ABC$ be an acute triangle. The lines $\ell_1$ and $\ell_2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $\ell_1$ and $\ell_2$ at the points $E$ and $F$, respectively. If $D$ is the intersection point of... |
oindent\textbf{Solution:}
Let $H$, $G$ be the points of intersection of $ME$, $MF$ with $AC$, $BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get $\dfrac{MH}{MA} = \dfrac{MA}{ME}$, thus
$$MA^2 = MH \cdot ME. \tag{1}$$
Similarly, from the similarity of triangles $\triangle MB... | 387 | 968 | Geometry | 3 |
238 | jbmo_2015_p4 | jbmo | 2,015 | c | An L-shape is one of the following four pieces, each consisting of three unit squares. A $5 \times 5$ board, consisting of $25$ unit squares, a positive integer $k \leq 25$ and an unlimited supply of L-shapes are given. Two players, $A$ and $B$, play the following game: starting with $A$ they alternatively mark a previ... |
oindent\textbf{Solution:}
We will show that player $\boldsymbol{A}$ wins if $k = 1, 2, 3$, but player $\boldsymbol{B}$ wins if $k = 4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4.
If $k = 1$, player $\boldsymbol{A}$ marks the upper left corner of the square and t... | 752 | 2,294 | Combinatorics | 4 |
134 | shl_jbmo_2015_a1 | shl_jbmo | 2,015 | a | Let $x$, $y$, $z$ be real numbers, satisfying the relations
\[
\begin{cases}
x \geq 20,\\
y \geq 40,\\
z \geq 1675,\\
x + y + z = 2015.
\end{cases}
\]
Find the greatest value of the product $P = x \cdot y \cdot z$. | \subsubsection*{Solution 1:}
By virtue of $z \geq 1675$ we have
\[
y + z < 2015 \iff y < 2015 - z \leq 2015 - 1675 < 1675.
\]
It follows that $(1675 - y)\cdot(1675 - z) \leq 0 \iff y \cdot z \leq 1675 \cdot (y + z - 1675)$.
By using the inequality $u \cdot v \leq \left(\dfrac{u+v}{2}
ight)^2$ for all real numbers $u$,... | 214 | 1,705 | Algebra | 1 |
135 | shl_jbmo_2015_a2 | shl_jbmo | 2,015 | a | 3) If $x^3 - 3\sqrt{3}\,x^2 + 9x - 3\sqrt{3} - 64 = 0$, find the value of $x^6 - 8x^5 + 13x^4 - 5x^3 + 49x^2 - 137x + 2015$. | \subsubsection*{Solution}
$x^3 - 3\sqrt{3}\,x^2 + 9x - 3\sqrt{3} - 64 = 0 \iff (x-\sqrt{3})^3 = 64 \iff (x-\sqrt{3}) = 4 \iff x - 4 = \sqrt{3} \iff x^2 - 8x + 16 = 3 \iff$
$x^2 - 8x + 13 = 0$
$x^6 - 8x^5 + 13x^4 - 5x^3 + 49x^2 - 137x + 2015 = (x^2-8x+13)(x^4-5x+9)+1898 = 0 + 1898 = 1898$ | 124 | 288 | Algebra | 2 |
136 | shl_jbmo_2015_a3 | shl_jbmo | 2,015 | a | Let $a, b, c$ be positive real numbers. Prove that
\[
\frac{a}{b} + \sqrt{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} > 2.
\] | \subsubsection*{Solution:}
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
\begin{align*}
2\frac{a}{b} + 2\sqrt{\frac{b}{c}} + 2\sqrt[3]{\frac{c}{a}}
&= \frac{a}{b} + \left(\frac{a}{b} + \sqrt{\frac{b}{c}} + \sqrt{\... | 118 | 844 | Algebra | 3 |
137 | shl_jbmo_2015_a4 | shl_jbmo | 2,015 | a | Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of
\[
A = \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c}.
\] | \subsubsection*{Solution:}
We rewrite $A$ as follows:
\begin{align*}
A &= \frac{2-a^3}{a} + \frac{2-b^3}{b} + \frac{2-c^3}{c}
= 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) - a^2 - b^2 - c^2 =\\
&2\left(\frac{ab+bc+ca}{abc}
ight) - (a^2+b^2+c^2)
= 2\left(\frac{ab+bc+ca}{abc}
ight) - ((a+b+c)^2 - 2(ab+bc+ca)) =\\
&2... | 160 | 1,374 | Algebra | 4 |
138 | shl_jbmo_2015_a5 | shl_jbmo | 2,015 | a | Let $x, y, z$ be positive real numbers that satisfy the equality $x^2+y^2+z^2=3$. Prove that
\[
\frac{x^2+yz}{x^2+yz+1} + \frac{y^2+zx}{y^2+zx+1} + \frac{z^2+xy}{z^2+xy+1} \leq 2.
\] | \subsubsection*{Solution:}
We have
\begin{align*}
&\frac{x^2+yz}{x^2+yz+1} + \frac{y^2+zx}{y^2+zx+1} + \frac{z^2+xy}{z^2+xy+1} \leq 2 \iff\\[6pt]
&\frac{x^2+yz+1}{x^2+yz+1} + \frac{y^2+zx+1}{y^2+zx+1} + \frac{z^2+xy+1}{z^2+xy+1} \leq 2 + \frac{1}{x^2+yz+1}+\frac{1}{y^2+zx+1}+\frac{1}{z^2+xy+1} \iff\\[6pt]
&3 \leq 2 + \... | 182 | 864 | Algebra | 5 |
139 | shl_jbmo_2015_c1 | shl_jbmo | 2,015 | c | A board $n\times n$ ($n\geq 3$) is divided into $n^2$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2\times2$ square of the board are different. Find all $n$ for which such boards exist. | \subsubsection*{Solution:}
The number of the $2\times2$ squares in a board $n\times n$ is equal to $(n-1)^2$. All possible sums of the numbers in such squares are $0, 1, \ldots, 4n$. A necessary condition for the existence of a board with the required property is $4n+1\geq(n-1)^2$ and consequently $n(n-6)\leq 0$. Thus ... | 286 | 1,087 | Combinatorics | 6 |
140 | shl_jbmo_2015_c2 | shl_jbmo | 2,015 | c | 2015 points are given in a plane such that from any five points we can choose two points with distance less than 1 unit. Prove that 504 of the given points lie on a unit disc. | \subsubsection*{Solution:}
Start from an arbitrary point $A$ and draw a unit disc with center $A$. If all other points belong to this disc then we are done. Otherwise, take any point $B$ outside of the disc. Draw a unit disc with center $B$. If two drawn discs cover all 2015 points, by PHP, one of the discs contains at... | 175 | 801 | Combinatorics | 7 |
141 | shl_jbmo_2015_c3 | shl_jbmo | 2,015 | c | Positive integers are put into the following table:
\[
\begin{array}{ccccccccc}
1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & \cdots\\
2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & \cdots\\
4 & 8 & 13 & 19 & 26 & 34 & 43 & 53 & \cdots\\
7 & 12 & 18 & 25 & 33 & 42 & & &\\
11 & 17 & 24 & 32 & 41 & & & &\\
16 & 23 & & & & & & &\\
\cdots & ... | \subsubsection*{Solution 1:}
We shall observe straight lines as on the next picture. We can call these lines diagonals.
On the first diagonal is number 1.\\
On the second diagonal are two numbers: 2 and 3.\\
On the 3rd diagonal are three numbers: 4, 5 and 6.\\
$\ldots$
On the $n$-th diagonal are $n$ numbers. These nu... | 443 | 2,450 | Combinatorics | 8 |
142 | shl_jbmo_2015_c4 | shl_jbmo | 2,015 | c | Let $n\geq1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^2$ squares of side length 1. Find the number of parallelograms which have vertices among the vertices of the $n^2$ squares of side length 1, with both sides smaller or equal to 2, and which have the area eq... | \subsubsection*{Solution:}
We can divide all these parallelograms into 7 classes (types I -- VII), according to Figure.
\begin{center}
[Figure: Grid showing 7 types of parallelograms labeled I through VII]
\end{center}
Type I: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogr... | 329 | 1,216 | Combinatorics | 9 |
143 | shl_jbmo_2015_c5 | shl_jbmo | 2,015 | c | We have a $5\times5$ chessboard and a supply of L-shaped triominoes, i.e. $2\times2$ squares with one corner missing. Two players $A$ and $B$ play the following game: A positive integer $k\leq 25$ is chosen. Starting with $A$, the players take alternating turns marking squares of the chessboard until they mark a total ... | \subsubsection*{Solution:}
We will show that player $A$ wins if $k=1, 2$ or 3, but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to 4.
If $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.
\begin{center}
[Figure: $5\time... | 700 | 2,295 | Combinatorics | 10 |
144 | shl_jbmo_2015_g1 | shl_jbmo | 2,015 | g | Around the triangle $ABC$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $BC$ and $AC$ at the points $D$ and $E$, respectively. Prove that the points $A, B, D, E$ belong to the same circle. | \subsubsection*{Solution:}
Let $O$ be the center of a circumscribed circle $k$ of the triangle $ABC$, and let $F$ and $G$ be the points of intersection of the line $CO$ with the line $p$ and the circle $k$, respectively (see Figure). From $p\|t$ it follows that $p\perp CO$. Furthermore, $\angle ABC = \angle AGC$, becau... | 298 | 778 | Geometry | 11 |
145 | shl_jbmo_2015_g2 | shl_jbmo | 2,015 | g | The point $P$ is outside of the circle $\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\Omega$ at the points $A$ and $B$. The median $AM$, $M\in(BP)$, intersects the circle $\Omega$ at the point $C$ and the line $PC$ intersects again the circle $\Omega$ at the point $D$. Prove that the lines ... | \subsubsection*{Solution:}
Since $\angle BAC = \angle BAM = \angle MBC$, we have $\triangle MAB \cong \triangle MBC$.
We obtain $\dfrac{MA}{MB} = \dfrac{MB}{MC} = \dfrac{AB}{BC}$. The equality $MB = MP$ implies $\dfrac{MA}{MP} = \dfrac{MP}{MC}$ and $\angle PMC = \angle PMA$ gives the relation $\triangle PMA \cong \tri... | 347 | 488 | Geometry | 12 |
146 | shl_jbmo_2015_g3 | shl_jbmo | 2,015 | g | Let $c \equiv c(O,K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $\widehat{ABO}$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $AOB$, say $c_1$ at point $K$ and the circumcircle of the triangle $AOC$... | \subsubsection*{Solution:}
The segments $OB, OC$ are equal, as radii of the circle $c$. Hence $OBC$ is an isosceles triangle and
\[
\hat{B}_1 = \hat{C}_1 = \hat{x}. \tag{1}
\]
\begin{center}
[Figure: Circle $c$ with points $A$, $B$, $C$, center $O$, circles $c_1$, $c_2$, and points $K$, $L$.]
\end{center}
The chord $... | 470 | 1,763 | Geometry | 13 |
147 | shl_jbmo_2015_g4 | shl_jbmo | 2,015 | g | Let $\triangle ABC$ be an acute triangle. The lines $(\varepsilon_1)$, $(\varepsilon_2)$ are perpendicular to $AB$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the sides of the triangle $AC$, $BC$ intersect the lines $(\varepsilon_1)$, $(\varepsilon_2)$ at the points $E... | \subsubsection*{Solution:}
Let $H$, $G$ be the points of intersection of $ME$, $MF$, with $AC$, $BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get
\[
\frac{MH}{MA} = \frac{MA}{ME}
\]
thus, $MA^2 = MH \cdot ME$ \quad (1).
Similarly, from the similarity of triangles $\triangle... | 456 | 1,245 | Geometry | 14 |
148 | shl_jbmo_2015_g5 | shl_jbmo | 2,015 | g | Let $ABC$ be an acute triangle with $AB
eq AC$. The incircle $\omega$ of the triangle touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The perpendicular line erected at $C$ onto $BC$ meets $EF$ at $M$, and similarly, the perpendicular line erected at $B$ onto $BC$ meets $EF$ at $N$. The line $D... | \begin{center}
[Figure: Triangle $ABC$ with incircle $\omega$, tangent points $D$, $E$, $F$, points $M$, $N$, $S$, $P$, $Q$.]
\end{center}
\subsubsection*{Proof 1.1.}
Let $\{T\} = EF \cap BC$. Applying Menelaus' theorem to the triangle $ABC$ and the transversal line $E-F-T$ we obtain $\dfrac{TB}{TC}\cdot\dfrac{EC}{EA}... | 420 | 1,537 | Geometry | 15 |
End of preview. Expand in Data Studio
RoJBMO: Junior Balkan Mathematical Olympiad Benchmark
RoJBMO is a benchmark of 508 competition mathematics problems drawn from the Junior Balkan Mathematical Olympiad (JBMO), its official shortlists, and the Romanian Team Selection Tests (TST). It is designed to evaluate the mathematical reasoning capabilities of large language models on problems that are underrepresented in existing benchmarks and resistant to training data contamination.
Sources
| Source | Problems | AoPS exposure |
|---|---|---|
jbmo — Official JBMO competition problems |
60 | 95.0% |
shl_jbmo — JBMO official shortlist problems |
282 | 46.5% |
tst_jbmo_ro — Romanian Team Selection Test problems |
166 | 13.9% |
The tst_jbmo_ro subset is the most contamination-resistant, with only 13.9% of problems findable via AoPS 5-gram search.
Dataset Structure
Each row contains one problem with the following fields:
| Field | Type | Description |
|---|---|---|
id |
int | Unique row index |
problem_uid |
string | Unique problem identifier (e.g. jbmo_2012_p3) |
problem_source |
string | Source category: jbmo, shl_jbmo, or tst_jbmo_ro |
year |
int | Publication year (2010–2025) |
type |
string | Domain code: a (Algebra), g (Geometry), c (Combinatorics), n (Number Theory) |
type_label |
string | Full domain name |
problem_num |
int | Problem number within the edition |
statement |
string | Problem statement (LaTeX-formatted) |
solution |
string | Official solution (LaTeX-formatted) |
stmt_len |
int | Statement length in characters |
sol_len |
int | Solution length in characters |
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- 20
Size
: 508 problems
Language
: English
Domains
: Algebra, Geometry, Combinatorics, Number Theory
Years
: 2010–2025
License
: CC BY-NC 4.0
Number of rows:
508
Total file size:
1.04 MB