id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0d7r | Given triangle $A B C$ inscribed in $(O)$. Two tangents at $B$, $C$ of $(O)$ intersect at $P$. The bisector of angle $A$ intersects $(P, P B)$ (the circle with center $P$ and radius $P B$) at point $E$ lying inside triangle $A B C$. Let $M$, $N$ be the midpoints of two arcs $B C$ of $(O)$ such that $M$ and $A$ are on d... | [
"\n\nLet $I$ be the midpoint of $B C$. We have $\\angle I C M = \\angle M A C = \\angle M C P$ then $C M$ is the bisector of $\\angle I C P$. This follows that $M$ is the insimilicenter of $(I)$ and $(P)$. However, $\\angle M C N = 90^{\\circ}$ hence $N$ is the exsimilicenter of $(I)$ and $... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
... | English | proof only | null | |
084a | Problem:
Edoardo è andato in vacanza nella città di Altanbulat. Il suo aereo, all'andata, è partito da Milano alle 13:00 ed è arrivato ad Altanbulat alle 9:00 del giorno dopo (ora locale). Il volo di ritorno invece è partito da Altanbulat alle 9:00 ed è atterrato alle 15:00 dello stesso giorno a Milano (di nuovo, tutt... | [
"Solution:\n\nLa risposta è (C). Indichiamo con $d$ la differenza di fuso orario tra l'Italia e Altanbulat. Traducendo tutte le ore nel fuso orario italiano, Edoardo parte alle 13 e arriva alle $9-d$ del giorno dopo, quindi impiega $24+(9-d)-13=20-d$ ore per il viaggio di andata. Nel ritorno invece parte alle $9-d$... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
08c8 | Problem:
$ABCD$ è un tetraedro con la seguente proprietà: detti $A'$, $B'$, $C'$, $D'$, rispettivamente, gli incentri delle facce $BCD$, $ACD$, $ABD$ ed $ABC$, si ha che le rette $AA'$, $BB'$, $CC'$ e $DD'$ hanno un punto in comune. Dimostrare che il prodotto delle lunghezze di due spigoli opposti del tetraedro è cost... | [
"Solution:\n\nSia $P$ il punto d'intersezione delle rette $AA'$, $BB'$, $CC'$, $DD'$, e si consideri il piano $\\Pi$ passante per $A$, $P$, $B$. Poiché $\\Pi$ contiene le rette $AP$ e $BP$, esso contiene i punti $A'$, $B'$, che si trovano su tali rette; dunque, detta $AK$ la bisettrice dell'angolo in $A$ nel triang... | Italy | XXXIII Olimpiade Italiana di Matematica | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
08qq | Problem:
Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$ | [
"Solution:\nFirst 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\n$$\na+b+c=\\frac{a b+b c+c a}{a b c}\n$$\nNow, from the first condition and the second condition we get\n$$\n(a+b+c)^{2}-\\left(a^{2}+b^{2}... | JBMO | JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All permutations of (t, 1/t, 1) and (t, 1/t, −1) for any nonzero real t. | |
082a | Problem:
Una gara di sci è divisa in due manches; un atleta si è piazzato al $3^{\circ}$ posto nella prima ed al $5^{\circ}$ nella seconda. Sapendo che la classifica finale è stilata sulla base della somma dei tempi ottenuti nelle singole manches, che ci sono 70 concorrenti e supponendo che non ci siano stati ex-aequo... | [
"Solution:\n\nLa risposta è (D). Infatti, poiché solo in 2 hanno fatto un tempo migliore nella prima manche e solo in quattro hanno fatto meglio di lui nella seconda manche, l'atleta in questione ha ottenuto in entrambe le manches tempi migliori dei restanti 63 atleti, quindi non si è potuto classificare più che se... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | D | |
00do | Let $ABC$ be an acute triangle with $AB < AC$. Let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$ respectively. The circumcircles of $AEF$ and $ABC$ meet again at $M$. Suppose the line $BM$ is tangent to the circumcircle of $AEF$. Prove that $M$, $F$, $D$ are collinear. | [
"First, by the tangency condition and $AMBC$ being a cyclic quadrilateral we have that\n$$\n\\angle AEM = 180^\\circ - \\angle AMB = \\angle ACB\n$$\nand hence $ME$ is parallel to $BC$. Next, we claim that $M$ and $E$ are symmetric with respect to $AD$. This is because $AH$ is a diameter of the circumcircle of $\\t... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Colline... | English | proof only | null | |
01k8 | Two cats, Bill and Tom, play the following game. They, in turn (Bill starts), eat fishes from the heap of 50 fishes. Per move it is allowed to eat exactly 1, or exactly 4, or exactly 7 fishes. The player wins if he eats the last fish of the heap.
Who of the cats wins if both of them play to win? | [
"It is easy to see that if either $1$ or $4$ fishes remain in the heap, then the player who must move wins, but if $2$ fishes remain, then this player loses. So we will solve the problem moving backward. We write all numbers from $1$ to $50$ and mark them with \"+\" or \"-\". If $k$ fishes remain in the heap before... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Tom | |
0bzi | Let $a$, $b$, $c$, $d$ be non-negative real numbers satisfying $a + b + c + d = 3$. Prove that
$$
\frac{a}{1 + 2b^3} + \frac{b}{1 + 2c^3} + \frac{c}{1 + 2d^3} + \frac{d}{1 + 2a^3} \ge \frac{a^2 + b^2 + c^2 + d^2}{3}.
$$
When does the equality hold? | [
"*First solution.* We notice the equality case $a = 3$, $b = c = d = 0$ and we try to mix the variables as follows:\n$$\nf(a, b, c, d) \\ge f(a + b + c + d, 0, 0, 0) = a + b + c + d - \\frac{(a + b + c + d)^2}{3} = 0,\n$$\nwhere\n$$\nf(a, b, c, d) = \\frac{a}{1 + 2b^3} + \\frac{b}{1 + 2c^3} + \\frac{c}{1 + 2d^3} + ... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds precisely for the cyclic permutations of the quadruples (3, 0, 0, 0) and (2, 1, 0, 0). | |
0ehb | Problem:
V pravokotni trikotnik $ABC$ s pravim kotom pri $B$ včrtamo tri kvadrate, kot to prikazuje slika. Stranici manjših dveh kvadratov sta dolgi 3 oziroma 4 enote. Izračunaj dolžino stranice $AC$ trikotnika $ABC$.
 | [
"Solution:\n\nOznačimo oglišča kvadratov, kot prikazuje slika. Trikotniki $CKL$, $LJF$, $FHI$ in $IGA$ so podobni trikotniki, zato velja\n$$\n\\frac{|CK|}{3} = \\frac{3}{|JF|} = \\frac{|FH|}{4} = \\frac{4}{|AG|}\n$$\nOznačimo dolžino stranice kvadrata $BEFD$ z $x$. Potem iz enakosti (1) sledi\n$$\n3 \\cdot 4 = |FH|... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 245/12 | |
04oj | Find all three-digit numbers such that the sum of their digits is $11$, and the number obtained by swapping the hundreds digit and the ones digit is by $594$ greater than the original number. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 137, 218 | |
05og | Problem:
Soit un quadrilatère $ABCD$ convexe. On se donne $E, F$ deux points tels que $E, B, C, F$ soient alignés dans cet ordre. On suppose de plus que $\widehat{BAE} = \widehat{CDF}$ et $\widehat{EAF} = \widehat{FDE}$. Montrer que $\widehat{FAC} = \widehat{EDB}$. | [
"Solution:\n\n\n\nC'est encore une chasse aux angles. Comme $\\widehat{FAE} = \\widehat{FDE}$ et $A, D$ sont du même côté de $(EF)$, les points $A, D, E, F$ sont cocycliques.\n\nDe plus, pour montrer que $\\widehat{FAC} = \\widehat{EDB}$, il suffit de montrer que $\\widehat{CAB} = \\widehat... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hpf | Problem:
Neville Nevermiss and Benjamin Baskethound are two players on the Simpson School basketball team. During Season I, Neville made a higher percentage of his attempted baskets than Ben. The same happened in Season II. Prove or disprove: When the statistics of the two seasons are combined, Neville necessarily mad... | [
"Solution:\n\nIt does not follow that Neville made a higher percentage over the two seasons. Consider the following (deliberately extreme) situation. In Season I:\n- Neville attempts $80$ baskets and makes $1$ of them;\n- Ben attempts $20$ baskets and misses them all.\n\nThen Neville obviously has a higher success ... | United States | Berkeley Math Circle Monthly Contest | [
"Statistics > Mathematical Statistics",
"Math Word Problems"
] | null | proof only | null | |
095v | Problem:
În triunghiul $ABC$ cu $m(\angle A) = 90^{\circ}$ construim $AD \perp BC$ ($D \in BC$) și bisectoarea $AE$ ($E \in BC$). Notăm cu $L$ și $F$ proiecțiile ortogonale ale punctului $E$ pe catetele $[AB]$ și $[AC]$. Să se demonstreze că dreptele $AD$, $BF$ și $CL$ sunt concurente.
 | [
"Solution:\n\nNotăm $AB = c$, $BC = a$ și $AC = b$. Conform teoremei bisectoarei, avem $\\frac{c}{b} = \\frac{BE}{EC}$.\n\n$EL \\parallel AC \\Rightarrow \\frac{AL}{LB} = \\frac{CE}{EB}$. (2)\n\nDin (1) și (2) rezultă $\\frac{AL}{LB} = \\frac{b}{c}$. (3)\n\n$EF \\parallel AB \\Rightarrow \\frac{CF}{FA} = \\frac{CE}... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cfi | Let $a$ and $b$ be two given natural numbers and $(x_n)_{n \ge 0}$ the sequence defined by $x_n = n^2 + a n + b$ for any natural number $n$.
a) Prove that the sequence $(x_n)_{n \ge 0}$ contains infinitely many perfect squares if and only if $a^2 = 4b$.
b) Prove that the sequence $(x_n)_{n \ge 0}$ does not contain an... | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0gnu | Let $\Gamma$ be the circumcircle of a triangle $ABC$, and let $D$ and $E$ be two points different from the vertices on the sides $AB$ and $AC$, respectively. Let $A'$ be the second point where $\Gamma$ intersects the bisector of the angle $\angle BAC$, and let $P$ and $Q$ be the second points where $\Gamma$ intersects ... | [
"Since $\\angle RPD = \\angle RAD = \\angle A'AC = \\angle A'PC = \\angle DPC$, $P, R, C$ are collinear. Then $\\angle PRD = \\angle PAD = \\angle PAB = \\angle PCB$ implies that $DR \\parallel BC$. Similarly, $SE \\parallel BC$, and consequently, $SE \\parallel DR$ and $\\frac{VD}{DA} = \\frac{SR}{RA}$.\n\nLet $l$... | Turkey | 17th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
05yg | Problem:
Trouver tous les triplets de réels positifs ou nuls $\left(a, b, c\right)$ tels que
$$
\left\{
\begin{array}{l}
a^{2}+a b=c \\
b^{2}+b c=a \\
c^{2}+c a=b
\end{array}
\right.
$$ | [
"Solution:\nSupposons d'abord qu'un des réels est nul, sans perte de généralité $a=0$. Alors la première équation donne $c=0$ et la dernière équation donne $b=0$. Réciproquement, le triplet $a=b=c=0$ est bien solution, on suppose à présent que les trois réels sont non nuls.\n\nSupposons maintenant que deux des réel... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (a,b,c)=(0,0,0) or (1/2,1/2,1/2) | |
0hjd | Problem:
In triangle $A B C$, let $D$ be the midpoint of side $B C$. Let $E$ and $F$ be the feet of the perpendiculars to $A D$ from $B$ and $C$, respectively. Prove that $B E = C F$. | [
"Solution:\n\nWe have $\\angle D F C = \\pi / 2 = \\angle D E B$. Also, $\\angle C D F = \\angle B D E$ since they are vertical angles. (It seems possible that $E$ and $F$ could lie on the same side of $D$, so that $\\angle C D F$ and $\\angle B D E$ would be supplementary rather than equal; however, if they were s... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
03ag | Някои от градовете в една държава са свързани с директни пътища. Нека $t$ е най-малкото естествено число, за което съществува град, от който до всеки друг град може да се стигне, минавайки по най-много $t$ пътя. Да се докаже, че съществуват градове $A_1$, $A_2$, ..., $A_{2t-1}$, за които за всеки $i \ne j$, $i = 1, 2, ... | [
"Разглеждаме граф $G$ с върхове градовете в държавата и ребра пътищата между тях. От всички подграфи на $G$ да изберем граф $H$, който има свойството на $G$ и е минимален по отношение на броя на върховете.\nДа изберем произволен връх $v_t$ на $H$, който не разделя графа на две несвързани компоненти (не е трудно да ... | Bulgaria | Team selection test for 50. IMO | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
0f1y | Problem:
Can you label each vertex of a cube with a different three digit binary number so that the numbers at any two adjacent vertices differ in at least two digits? | [] | Soviet Union | ASU | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
07p7 | Prove for all integers $N > 1$ that $(N^2)^{2014} - (N^5)^{106}$ is divisible by $N^3 - 1$. | [
"Observe that $2 \\cdot 2014 - 5 \\cdot 106 = 3498 = 3 \\cdot 1166$, hence\n$$\n(N^2)^{2014} - (N^5)^{106} = (N^5)^{106} ((N^3)^{1166} - 1) \\\\ = (N^5)^{106} (N^3 - 1) ((N^3)^{1165} + (N^3)^{1164} + \\dots + 1)\n$$\nis divisible by $N^3 - 1$."
] | Ireland | Irska 2014 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0hcx | Let $ABC$ be a triangle. Let $C_1$, $A_1$ and $B_1$ be the points of $AB$, $BC$ and $AC$, respectively. Let $K$ be the projection of $B_1$ on the line $A_1C_1$. Let the points $M$ and $N$ lie on the rays $B_1A$ and $B_1C$ respectively, so that $\angle B_1A_1C_1 = 2\angle KNB_1$ and $\angle B_1C_1A_1 = 2\angle KMB_1$. P... | [
"Let $M_1$ be the point of the ray $A_1C_1$ such that $M_1C_1 = B_1C_1$, $N_1$ be the point of the ray $C_1A_1$ such that $N_1A_1 = B_1A_1$ (Fig. 39). Then\n$\\angle B_1M_1K = \\frac{1}{2} \\angle A_1C_1B_1 = \\angle KMB_1$,\nwhence $KM_1MB_1$ is inscribed and $M$ is the projection of $M_1$ on the line $AC$.\nAnalo... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0380 | Problem:
the incircle $k$ of $\triangle ABC$ is tangent to the sides $AB$, $BC$ and $CA$ at points $C_1$, $A_1$ and $B_1$, respectively. The points $C_2$, $A_2$ and $B_2$ are diametrically opposite to $C_1$, $A_1$ and $B_1$ in $k$.
a) Prove that the lines $AA_2$, $BB_2$ and $CC_2$ are concurrent.
b) If the line $AA_... | [
"Solution:\n\na) Let $A_4 = AA_2 \\cap BC$ and let the tangent line to $k$ at $A_2$ meet $AB$ and $AC$ at points $X$ and $Y$, respectively. Since $A_2A_1$ is a diameter, we have $XY \\parallel BC$, i.e. $\\triangle AXY \\sim \\triangle ABC$.\n\nSince $k$ is an excircle of $\\triangle AXY$, it follows from above tha... | Bulgaria | Team selection test for 23. BMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > ... | null | proof and answer | 1:1 | |
0f1m | Problem:
a, b, c are positive reals. Show that $a^{3} + b^{3} + c^{3} + 3abc \geq ab(a + b) + bc(b + c) + ca(c + a)$. | [
"Solution:\n\nThe inequality is homogeneous, so we can take $a = 1$ and put $b = 1 + x$, $c = 1 + y$, where $x, y \\geq 0$. Then after some reduction the inequality is equivalent to $x^{3} + y^{3} + x^{2} + y^{2} - x^{2} - y - xy^{2} - xy \\geq 0$, or (after factorising $x^{3} + y^{3}$) to $(x + y + 1)(x - y)^{2} +... | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0451 | It is known that $\triangle ABC$ satisfies $AB = 1$, $AC = 2$ and $\cos B + \sin C = 1$. Find the length of side $BC$. | [
"Denote $a = BC$, $b = AC$, $c = AB$, so $b = 2$, $c = 1$. By the law of sines, we have $\\frac{\\sin B}{\\sin C} = \\frac{b}{c} = 2$, namely, $\\sin B = 2 \\sin C$. And since $\\cos B = 1 - \\sin C$, there is\n$$\n(2 \\sin C)^2 + (1 - \\sin C)^2 = \\sin^2 B + \\cos^2 B = 1,\n$$\nand simplifying it gives $5\\sin^2 ... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | (3 + 2√21)/5 | |
01qu | $n$ points are marked on a plane. Each pair of these points is connected with a segment. Each segment is painted one of four different colors.
Find the largest possible value of $n$ such that one can paint the segments so that for any four points there are four segments (connecting these four points) of four different ... | [
"Answer: the largest possible value of $n$ is $9$."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
09zq | We start with a square with side length $1$. During the first minute, small squares with side length $\frac{1}{3}$ grow on the middle of the vertical sides. During the next minute, on the middle of each vertical line segment in the new figure, a new small square grows, whose sides have length $\frac{1}{3}$ of these lin... | [
"Let us analyze the process step by step.\n\nLet $C_n$ be the circumference after $n$ minutes.\n\nAt the start ($n = 0$):\nThe figure is a square with side $1$, so the circumference is $4 \\times 1 = 4$.\n\nMinute 1:\nOn each vertical side, a square of side $\\frac{1}{3}$ is added in the middle. Each such square ad... | Netherlands | Second Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 8 - 4*(1/3)^60 | |
03fv | Given is a triangle $ABC$ and the points $M$, $P$ lie on the segments $AB$, $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM = \angle ABC$, find the measure of $\angle ABC$. | [
"Let $D$ be the reflection of $P$ across $C$, so $AB = BD$ and $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore $... | Bulgaria | 5 Bulgarian National Olympiad - Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 90° | |
0jcn | Problem:
Hexagon $A B C D E F$ has a circumscribed circle and an inscribed circle. If $A B=9$, $B C=6$, $C D=2$, and $E F=4$. Find $\{D E, F A\}$. | [
"Solution:\n\nAnswer: $\\left\\{\\frac{9+\\sqrt{33}}{2}, \\frac{9-\\sqrt{33}}{2}\\right\\}$\n\nBy Brianchon's Theorem, $A D$, $B E$, $C F$ concur at some point $P$. Also, it follows from the fact that tangents from a point to a circle have equal lengths that $A B + C D + E F = B C + D E + F A$.\n\nLet $D E = x$, so... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | {(9+sqrt(33))/2, (9-sqrt(33))/2} | |
0lan | Prove that for every positive integer $n$ the equation
$$
x^2 + 15y^2 = 4^n
$$
has at least $n$ non-negative integer solutions $(x, y)$. | [
"Consider equation: $x^2 + 15y^2 = 4^n$ (*).\nWe have the following remarks:\n\n*Remark 1:* If $(x, y)$ is a non-negative integer solution of (*) for $n = k$, $k \\ge 1$, then $(2x, 2y)$ is a non-negative integer solution of (*) for $n = k + 1$.\n\n*Remark 2:* For each $n \\ge 2$, equation (*) always has 1 non-nega... | Vietnam | Vietnamese Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0344 | Problem:
Find all odd prime numbers $p$ which divide the number $1^{p-1}+2^{p-1}+\cdots+2004^{p-1}$. | [
"Solution:\nNote that $k^{p-1} \\equiv 0 \\pmod{p}$ if $p$ divides $k$ and $k^{p-1} \\equiv 1 \\pmod{p}$ otherwise (by Fermat's little theorem). Then\n$$\n0 \\equiv 1^{p-1}+2^{p-1}+\\cdots+2004^{p-1} \\equiv 0 \\cdot\\left[\\frac{2004}{p}\\right]+1 \\cdot\\left(2004-\\left[\\frac{2004}{p}\\right]\\right) \\pmod{p}\... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 17, 2003 | |
03qs | Let $M$ be a set consisting of $n$ points in the plane, and satisfying:
(1) there exist 7 points in $M$ which constitute the vertices of a convex heptagon;
(2) if for any 5 points in $M$ which constitute the vertices of a convex pentagon, then there is a point in $M$ which lies in the interior of the pentagon.
Find th... | [
"First, we prove that $n \\ge 11$. Suppose a convex heptagon has its vertices in $M$ given by $A_1A_2A_3A_4A_5A_6A_7$. Using Condition (1), we get that there exists one point $P_1$ belonging to $M$ in the interior of convex pentagon $A_1A_2A_3A_4A_5$. Connecting $P_1A_1$ and $P_1A_5$, we obtain that there exists on... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 11 | |
09j4 | Let $ABC$ be a scalene triangle. Let $A_1$, $B_1$, $C_1$ be the points on the sides $BC$, $CA$, $AB$, respectively such that $\angle B_1AC_1 = \angle B_1A_1C_1$, $\angle C_1BA_1 = \angle C_1B_1A_1$ and $\angle A_1CB_1 = \angle A_1C_1B_1$. Let $A_2$, $B_2$, $C_2$ be the second intersections of the circumcircle of $A_1B_... | [
"**Claim.** The lines $AA_2$, $BB_2$, $CC_2$ are concurrent if and only if the lines $AA_1$, $BB_1$, $CC_1$ are concurrent.\n*Proof.* Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$ and $\\angle ACB = \\gamma$. Angle chasing, $\\angle C_1B_2A = \\angle C_1A_1B_1 = \\alpha$. Since $\\angle C_1AB_2 = \\alpha$ we ... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Tri... | null | proof only | null | |
0jyq | Problem:
A box contains twelve balls, each of a different color. Every minute, Randall randomly draws a ball from the box, notes its color, and then returns it to the box. Consider the following two conditions:
(1) Some ball has been drawn at least twelve times (not necessarily consecutively).
(2) Every ball has been d... | [
"Solution:\nBelow is a python implementation to compute the probability, using the same method as the solution to the easier version (with three balls).\n\n```\nfrom fractions import Fraction\nN = 12\nprobs = [{} for i in range((N-1)*(N-1)+2)]\nprob1 = Fraction()\nprob2 = Fraction()\ninit = tuple(0 for i in range(N... | United States | HMMT November | [
"Discrete Mathematics > Algorithms"
] | null | final answer only | 663659309086473387879121984765654681548533307869748367531919050571107782711246694886954585701687513519369602069583/29675177620217171380656410198651124206162093498768869463821672067789922444492392280614561539198623553884143178743808 | |
0hjg | Problem:
Given a quadrilateral $ABCD$, show that the midpoints of its four edges form the vertices of a parallelogram. | [
"Solution:\n\nLet $M, N, P, Q$ be the midpoints of $AB, BC, CD$, and $DA$, respectively. Then $MN$ is the midline of $\\triangle ABC$ opposite $AC$, so it is parallel to $AC$ and of length $\\frac{1}{2} AC$. Similarly, $PQ$ is the midline of $\\triangle ACD$, so $PQ \\parallel AC$ and $PQ =... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
068q | Twelve friends play a tennis tournament, where each player plays exactly one game with each of the eleven other players. The winner wins one point and the loser gets zero points, while there is no tie. If the number of points of the players are $B_1, B_2, ..., B_{12}$, find the maximum value of $\Sigma_3 = B_1^3 + B_2^... | [
"The 12 friends will play $\\binom{12}{2} = \\frac{12 \\cdot 11}{2} = 66$ games, so the total number of points from all games is $66$. One possible outcome of the tournament is to put all players in an order and each of them to win all players that are after of him in the order. Then the first one has $11$ points, ... | Greece | Selection Examination | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 4356 | |
0jn9 | Problem:
Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$. $R$ can be divided into two unit squares, as shown.
Pro selects a point $P$ uniformly at random in the interior of $R$. Find the probability that the line through $P$ with slope... | [
"Solution:\n\nAnswer: $\\frac{3}{4}$\n\n\n\nPrecisely the middle two (of four) regions satisfy the problem conditions, and it's easy to compute the (fraction of) areas as $\\frac{3}{4}$."
] | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3/4 | |
0dwa | Problem:
Naj bosta $E$ in $F$ razpolovišči stranic $AD$ in $DC$ pravokotnika $ABCD$. Označimo z $G$ presečišče daljic $AF$ in $EC$. Dokaži, da je $\Varangle CGF = \Varangle FBE$. | [
"Solution:\n\nKer je $\\Varangle FBA = \\Varangle BAF = \\Varangle DFA$ in $\\Varangle EBA = \\Varangle DCE$, je\n$$\n\\begin{aligned}\n\\Varangle CGF & = \\pi - \\Varangle FCG - \\Varangle GFC = \\\\\n& = \\pi - \\Varangle DCE - (\\pi - \\Varangle DFG) = \\\\\n& = -\\Varangle DCE + \\Varangle DFA = \\\\\n& = \\Var... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ctn | Let $n$ be a positive integer. We have $2n+1$ (not necessarily distinct) nonzero integers with a nonzero sum. One needs to put them in places of the stars into an expression $*x^{2n} + *x^{2n-1} + \cdots + *x + *$ so that the resulting polynomial would have no integer root. Is it always possible?
Пусть $n$ — натуральн... | [
"Yes.\nPut the number with the largest absolute value at $x^{2n}$; then the only possible integer root is $-1$. This is easily avoidable by permuting the remaining coefficients.\n\n\nSolution:\nДа, обязательно.\nПусть $p_0, p_1, \\dots, p_{2n}$ — числа на карточках, причём $p_{2n}$ — наибольшее по модулю из них. По... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials"
] | English; Russian | proof and answer | Yes | |
0az5 | Problem:
Consider the function $f: \mathbb{N} \rightarrow \mathbb{Z}$ satisfying, for all $n \in \mathbb{N}$,
a. $|f(n)|=n$
b. $0 \leq \sum_{k=1}^{n} f(k)<2 n$.
Evaluate $\sum_{n=1}^{2018} f(n)$. | [
"Solution:\n\nLet $S_{n}=\\sum_{k=1}^{n} f(k)$. We want the value of $S_{2018}$.\n\nClaim: $f(n)= \\begin{cases}n & \\text{ if } S_{n-1}<n \\\\ -n & \\text{ if } S_{n-1} \\geq n\\end{cases}$\n\nProof: The inequality condition is $0 \\leq S_{n-1}+f(n)<2 n$.\n\n- If $n>S_{n-1}$, then $0 \\leq S_{n-1}+f(n)<n+f(n)$ so ... | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2649 | |
0hv6 | Problem:
Trevor and Edward play a game in which they take turns adding or removing beans from a pile. On each turn, a player must either add or remove the largest perfect square number of beans that is in the heap. The player who empties the pile wins. For example, if Trevor goes first with a pile of 5 beans, he can ei... | [
"Solution:\nThe correct answers are 0 (worth imaginary points), 5 (worth 0 points), 20 (4 points), 29, 45 (5 points), 80 (6 points), 101, 116, 135, 145, 165, 173 (7 points), 236, 257 (8 points), 397, 404, 445, 477, 540, 565, 580, 629, 666 (9 points), $836, 845, 885, 909, 944$, 949, 954, 975 (10 points). This game i... | United States | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 5 | |
0ldy | Consider a positive integer $m$ and a rectangle board of size $m \times 2018$ which consists of $m$ rows and $2018$ columns. We write $0$ or $1$ into some cells of the board (only one number in a cell) and the rest are left empty. The board is *complete* if for an arbitrary binary sequence $S$ of length $2018$, we can ... | [
"a.\nFirst, consider an empty rectangle board of size $2^k \\times 2018$ and $2^k$ binary sequences of length $k$. We write those sequences to the left of the board so that each row consists of a sequence. Thus, the rest to the right of the board are $2018 - k$ empty columns. It is obvious that each of the first $k... | Vietnam | VN IMO Booklet | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0i7o | Problem:
Several positive integers are given, not necessarily all different. Their sum is $2003$. Suppose that $n_{1}$ of the given numbers are equal to $1$, $n_{2}$ of them are equal to $2$, $\ldots$, $n_{2003}$ of them are equal to $2003$. Find the largest possible value of
$$
n_{2}+2 n_{3}+3 n_{4}+\cdots+2002 n_{20... | [
"Solution:\n\nThe sum of all the numbers is $n_{1}+2 n_{2}+\\cdots+2003 n_{2003}$, while the number of numbers is $n_{1}+n_{2}+\\cdots+n_{2003}$. Hence, the desired quantity equals\n$$\n\\begin{gathered}\n\\left(n_{1}+2 n_{2}+\\cdots+2003 n_{2003}\\right)-\\left(n_{1}+n_{2}+\\cdots+n_{2003}\\right) \\\\\n=(\\text{s... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2002 | |
01k1 | Find all $x$ satisfying the equality
$$
[x^2] - [-x^2] - 8[x] + 2 = 0.
$$
(Here $[x]$ stands for the greatest integer not exceeding $x$.) | [
"Answer: $\\sqrt{3}$, $\\sqrt{7}$, $\\sqrt{11}$.\nSince $8[x]$ is an integer number, we can rewrite the initial equation as\n$[x^2 - 4[x] + 1] - [-x^2 - 4[x] + 1] = 0$.\nSet $y = x^2 - 4[x] + 1$, then $[y] = [-y]$.\nIt is easy to see that for $y > 0$ we have $0 \\le [y] = [-y] \\le -1$, which is impossible.\nSimila... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | sqrt(3), sqrt(7), sqrt(11) | |
07v9 | For $n \ge 2$, an *n-spinner* is a “fidget spinner” with $n$ identical arms and an angle of $360/n$ degrees between pairs of adjacent arms. For instance, a $3$-spinner and a $4$-spinner are illustrated below:
We wish to place each of the numbers $1, \ldots, nm$ on the arms of an $n$-spinner, ... | [
"First, place the spinner so that the arm containing the number $1$ is at the top. There, it can be put in any of the $m$ positions on that arm. This orients the $n$-spinner and breaks the symmetry. The other $nm - 1$ numbers can now be placed in any other positions, so the answer is $m(nm - 1)!$.\n\nAlternatively,... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | m(nm - 1)! = (nm)!/n | |
0a82 | Problem:
Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2x) = 2f(x)$, $f(4x+1) = 4f(x) + 3$, and $f(4x-1) = 2f(2x-1) - 1$. Show that $f$ is an injection, i.e. if $f(x) = f(y)$, then $x = y$. | [
"Solution:\nIf $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \\equiv 1 \\bmod 4$, then $f(x) \\equiv 3 \\bmod 4$, and if $x \\equiv 3 \\bmod 4$, then $f(x) \\equiv 1 \\bmod 4$. Clearly $f(0) = 0$, $f(1) = 3$, $f(2) = 6$, and $f(3) = 5$. So at least $f$ restricted to the se... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 11 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09te | Problem:
Zij $n \geq 2$ een geheel getal. Bepaal de kleinste positieve gehele $m$ zodat geldt: gegeven $n$ punten in het vlak, geen drie op een lijn, zijn er $m$ lijnen te vinden, zodat geen enkele lijn door één van de gegeven punten gaat en zodat voor elk tweetal gegeven punten $X \neq Y$ geldt dat er een lijn is waa... | [
"Solution:\n\nWe bewijzen dat de kleinste $m$ gelijk is aan $\\frac{n}{2}$ als $n$ even is en $\\frac{n+1}{2}$ als $n$ oneven. Kies de $n$ punten allemaal op dezelfde cirkel en noem ze $P_{1}, P_{2}, \\ldots, P_{n}$ in de volgorde waarin ze op de cirkel liggen. De $n$ lijnstukken $P_{1} P_{2}, P_{2} P_{3}, \\ldots,... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m = n/2 if n is even; m = (n+1)/2 if n is odd | |
03gb | Problem:
Let $ABC$ be a triangle with sides of lengths $a$, $b$ and $c$. Let the bisector of the angle $C$ cut $AB$ in $D$. Prove that the length of $CD$ is
$$
\frac{2ab \cos \frac{C}{2}}{a+b}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2ab cos(C/2)/(a+b) | |
00ad | Point $D$ is chosen on side $BC$ of the acute triangle $ABC$ so that $AD = AC$. Let $P$ and $Q$ be respectively the feet of the perpendiculars from $C$ and $D$ to $\overline{AB}$. It is known that
$$AP^2 + 3BP^2 = AQ^2 + 3BQ^2.$$ Find $\triangle ABC$. | [
"$$\nAP^2 + 3BP^2 = AQ^2 + 3BQ^2\n$$\n$$\nAQ^2 - AP^2 = 3(BP^2 - BQ^2)\n$$\nExpress $AQ^2$ and $AP^2$ by Pythagoras theorem for the right-angled triangles $ADQ$ and $ACP$: $AQ^2 = AD^2 - DQ^2$, $AP^2 = AC^2 - CP^2$. Since $AC = AD$, it follows that $AQ^2 - AP^2 = CP^2 - DQ^2$. Likewise the right-angled triangles $B... | Argentina | Argentine National Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Angle ABC equals 60 degrees. | |
035h | Problem:
The points $P$ and $Q$ lie in the interior of $\triangle ABC$, $\angle ACP = \angle BCQ$ and $\angle CAP = \angle BAQ$. The feet of the perpendiculars from $P$ to the lines $BC$, $CA$ and $AB$ are denoted by $D$, $E$ and $F$, respectively. Prove that if $\angle DEF = 90^{\circ}$, then $Q$ is the orthocenter o... | [
"Solution:\n\nWe have $\\angle BCQ = \\angle ACP = \\angle EDP$. Since $PD \\perp BC$, it follows that $ED \\perp CQ$. Analogously we have $AQ \\perp EF$. Since $\\angle DEF = 90^{\\circ}$, we conclude that $\\angle AQC = 90^{\\circ}$ as well. Then $\\triangle QCD \\sim \\triangle ACP$, because $\\angle QCD = \\ang... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
037d | Problem:
Let $m \geq 5$ and $n$ be positive integers and $M$ be a regular $(2n+1)$-gon. Find the number of convex $m$-gons with vertices among the vertices of $M$ and having at least one acute angle. | [
"Solution:\nIt is easy to observe that there are at most two acute angles in every convex $m$-gon. Moreover, if there are two acute angles then they are located at one and the same side.\n\nFix $l = 0, 1, \\ldots, n-1$ and let $A$ and $B$ be two vertices of $M$ such that there are $l$ vertices on the arc $\\widehat... | Bulgaria | Team selection test for 47. IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | ((2n+1)(mn - 2n - 1)/(m - 1)) * C(n, m-2) | |
08uw | How many ways of lining up $2010$ white stones and $2010$ black stones along a horizontal straight line are there so as to satisfy the following condition?
* In the line-up there are odd number of pairs consisting of one white stone and one black stone with the white one lying on the right of the black one. | [
"$$\n\\frac{\\binom{4020}{2010} - \\binom{2010}{1005}}{2} \\text{ ways}\n$$\n\nWhen you line-up $4020$ stones, $2010$ of them white and $2010$ others black, along a horizontal straight line, you have the following two possibilities for the resulting configuration, regardless of the condition of the problem:\n\n(1) ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | (binom(4020,2010) - binom(2010,1005))/2 | |
08wf | A point $E$ is located on the side $DA$ of a quadrilateral $ABCD$ in such a way that the lines $AB$ and $EC$ are parallel. If $AB = 3$, $BC = 3$, $CD = 5$, $DE = 3$ and $EA = 2$, determine $EC$.
Here we denote for a line segment $XY$ its length also be $XY$.
 | [
"$$\n\\boxed{\\frac{24}{5}}\n$$\nLet $F$ be the point of intersection of $BD$ and $EC$. From $AB // EF$, we get $AB : EF = DA : DE$. Therefore, we have $EF = \\frac{AB \\cdot DE}{DA} = \\frac{9}{5}$.\n\nWe also have $AB = CB = 3$, $AD = CD = 5$, which imply that the triangles $ABD$ and $CBD$ are congruent as the si... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 24/5 | |
0f63 | Problem:
Show that any cross-section of a cube through its center has area not less than the area of a face. | [
"Solution:\n\nLet the cube have side length $a$. The area of a face is $a^2$.\n\nAny cross-section through the center of the cube is a plane passing through the center. The largest possible cross-section is when the plane is perpendicular to the space diagonal, which gives a regular hexagon. The smallest possible c... | Soviet Union | 18th ASU | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0fp3 | Se tienen dos progresiones de números reales, una aritmética $(a_n)_{n \in \mathbb{N}}$ y otra geométrica $(g_n)_{n \in \mathbb{N}}$ no constante. Se cumple que $a_1 = g_1 \neq 0$, $a_2 = g_2$ y $a_{10} = g_3$. Decidir, razonadamente, si para cada entero positivo $p$, existe un entero positivo $m$, tal que $g_p = a_m$. | [
"Sean $d$ y $r \\neq 1$ la diferencia y la razón, respectivamente, de las progresiones aritmética $(a_n)_{n \\in \\mathbb{N}}$ y geométrica $(g_n)_{n \\in \\mathbb{N}}$. En primer lugar tenemos $g_1 r = g_2 = a_2 = a_1 + d = g_1 + d$ de donde $d = g_1(r - 1)$. En segundo lugar $g_1 r^2 = g_3 = a_{10} = a_1 + 9d = g... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Spanish | proof and answer | Yes. For every positive integer p, an index m exists with g_p = a_m, explicitly m = (8^{p-1} + 6)/7. | |
0bnw | Let $ABCD$ be a cyclic quadrangle, let $\gamma$ be its circumcircle, and let $M$ be the midpoint of the arc $AB$ of $\gamma$, not containing the vertices $C$ and $D$. The line through $M$ and the point where the diagonals $AC$ and $BD$ cross one another, crosses $\gamma$ again at $N$. Let $P$ and $Q$ be points on the s... | [
"Since the lines $AC$, $BD$ and $MN$ are concurrent, $(AM/MB)(BC/CN)(ND/DA) = 1$, by Ceva's theorem in trigonometric form along with the sine law in the triangle $ABN$; and since $MA = MB$, it follows that $CN/DN = BC/AD$.\n\nNext, the triangles $ADP$ and $QDA$ are similar, so $AD^2 = DP \\cdot DQ$. Similarly, $BC^... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Advanced Configur... | English | proof only | null | |
0grm | There are $n$ stone piles each consisting of $2018$ stones. The weight of each stone is equal to one of the numbers $1, 2, \ldots, 25$ and the total weights of any two piles are different. It is given that if we choose any two piles and remove the heaviest and lightest stones from each of these two piles then the pile ... | [
"The answer: $n = 12$. We numerate the piles according to their weights in increasing order. Let $S_i$ and $S'_i$ be the weights of pile number $i$ before and after removing of two stones. Then $S_1 < S_2 < \\dots < S_n$ and $S'_1 > S'_2 > \\dots > S'_n$. Let $S_1 - S'_1 = x$. Then $S_2 - S'_2 \\ge x+2, \\dots, S_n... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 12 | |
008b | In the triangle $ABC$, the points $M$ and $N$ lay onto the segments $AB$ and $AC$, respectively, so that $MN$ is parallel to $BC$ and tangent to the incircle of the triangle $ABC$. Let $K$ be the point where the incircle of the triangle $AMN$ is tangent to $MN$. It is known that $MN=4$, $BC=12$, and the segments $MK$ a... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0iw7 | Problem:
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log_{2}(S)$? | [
"Solution:\nThe sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\\frac{1}{2}\\left((1+i)^{2009}+(1-i)^{2009}\\right)=2^{1004}$. Thus $\\log_{2}(S)=1004$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 1004 | |
0jie | Problem:
Pentagon $A B C D E$ is given with the following conditions:
(a) $\angle C B D + \angle D A E = \angle B A D = 45^{\circ}, \angle B C D + \angle D E A = 300^{\circ}$
(b) $\frac{B A}{D A} = \frac{2 \sqrt{2}}{3},\ C D = \frac{7 \sqrt{5}}{3}$, and $D E = \frac{15 \sqrt{2}}{4}$
(c) $A D^{2} \cdot B C = A B \cdot ... | [
"Solution:\n\nAnswer: $\\sqrt{\\sqrt{39}}$ As a preliminary, we may compute that by the law of cosines, the ratio $\\frac{A D}{B D} = \\frac{3}{\\sqrt{5}}$. Now, construct the point $P$ in triangle $A B D$ such that $\\triangle A P B \\sim \\triangle A E D$. Observe that $\\frac{A P}{A D} = \\frac{A E \\cdot A B}{A... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | sqrt(39) | |
05pz | Problem:
Soient $ABCD$ un quadrilatère inscrit dans un cercle. On note $\Delta_{1}$ la droite parallèle à $(BC)$ passant par $A$ et $\Delta_{2}$ la droite parallèle à $(AD)$ passant par $B$. On note $E$ le point d'intersection de $\Delta_{1}$ avec $(CD)$ et $F$ le point d'intersection de $\Delta_{2}$ avec $(CD)$. La d... | [
"\nFIGURE 1 - Preuve de la première implication\n\nOn commence par quelques remarques préliminaires. On a $\\widehat{EDA}=180^{\\circ}-\\widehat{ADC}=\\widehat{ABC}=180^{\\circ}-\\widehat{EAB}$, donc $(AB)$ et $\\Gamma_{1}$ sont tangents. De même, $\\Gamma_{2}$ et $(AB)$ sont tangents.\n\nE... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jni | Problem:
Let $b(x) = x^{2} + x + 1$. The polynomial $x^{2015} + x^{2014} + \cdots + x + 1$ has a unique "base $b(x)$" representation
$$
x^{2015} + x^{2014} + \cdots + x + 1 = \sum_{k=0}^{N} a_{k}(x) b(x)^{k}
$$
where
- $N$ is a nonnegative integer;
- each "digit" $a_{k}(x)$ (for $0 \leq k \leq N$) is either the zero po... | [
"Solution:\nAnswer: $-1006$\n\nComparing degrees easily gives $N = 1007$. By ignoring terms of degree at most $2013$, we see\n$$\na_{N}(x)\\left(x^{2} + x + 1\\right)^{1007} \\in x^{2015} + x^{2014} + O\\left(x^{2013}\\right)\n$$\nWrite $a_{N}(x) = u x + v$, so\n$$\n\\begin{aligned}\na_{N}(x)\\left(x^{2} + x + 1\\r... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -1006 | |
0avp | Problem:
In an $n \times n$ checkerboard, the rows are numbered 1 to $n$ from top to bottom, and the columns are numbered 1 to $n$ from left to right. Chips are to be placed on this board so that each square has a number of chips equal to the absolute value of the difference of the row and column numbers. If the total... | [
"Solution:\n\nThe total number of chips for an $n \\times n$ board is equal to\n$$\n\\begin{aligned}\nn \\times 0 + 2 \\times (n-1) \\times 1 + 2 \\times (n-2) \\times 2 + \\cdots + 2 \\times 1 \\times (n-1) &= \\sum_{i=1}^{n} 2 \\times (n-i) \\times i \\\\\n&= 2\\left(n \\sum_{i=1}^{n} i - \\sum_{i=1}^{n} i^{2}\\r... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 20 | |
090s | Decimos que un polinomio $p(x)$, con coeficientes reales, es *almeriense* si tiene la forma
$$
p(x) = x^3 + ax^2 + bx + a
$$
y sus tres raíces son números reales positivos en progresión aritmética. Halla todos los polinomios almerienses tales que $p(7/4) = 0$. | [
"Llamemos $\\alpha \\le \\beta \\le \\gamma$ a las raíces del polinomio. De la condición de estar en progresión aritmética tenemos que existe un número real no negativo $\\delta$ de manera que $\\alpha = \\beta - \\delta$ y $\\gamma = \\beta + \\delta$. Por su parte, utilizando las fórmulas de Cardano–Viète, result... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | Spanish | proof and answer | p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4} and p(x) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}. | |
07ig | Prove that among any 9 distinct real numbers, there exist 4 distinct numbers $a$, $b$, $c$, $d$ such that
$$
(ac + bd)^2 \geq \frac{9}{10}(a^2 + b^2)(c^2 + d^2).
$$ | [
"According to the Lagrange's identity, we have $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$. So we have to prove that:\n$$\n(ac + bd)^2 \\geq 9(ad - bc)^2 \\iff |ac + bd| \\geq 3|ad - bc|\n$$\nFirst, we check the case where none of the numbers are zero. By dividing both sides by $|ac|$, we have:\n$$\n|1 + \... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0b11 | Problem:
Determine the number of ordered quadruples $(a, b, c, d)$ of odd positive integers that satisfy the equation $a + b + c + d = 30$. | [
"Solution:\n\nLet $a, b, c, d$ be odd positive integers. Any odd positive integer can be written as $2k + 1$ for some integer $k \\geq 0$.\n\nLet $a = 2k_1 + 1$, $b = 2k_2 + 1$, $c = 2k_3 + 1$, $d = 2k_4 + 1$, where $k_1, k_2, k_3, k_4 \\geq 0$.\n\nThen:\n$$\na + b + c + d = (2k_1 + 1) + (2k_2 + 1) + (2k_3 + 1) + (... | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 560 | |
006h | De un cuadrado de papel de lado $1$ hay que recortar dos triángulos equiláteros iguales. Hallar el máximo valor posible del lado de los triángulos. | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Spanish | proof and answer | sqrt(3) - 1 | |
06dh | If $a \ge b \ge c \ge 0$ and $a + b + c = 3$, then prove that $ab^2 + bc^2 + ca^2 \le \frac{27}{8}$ and determine the equality case(s). | [
"Let $f(a, b, c) = ab^2 + bc^2 + ca^2$. Note that it suffices to show\n$$\nf(a, b, c) - f(a, c, b) \\le 0, \\qquad (1)\n$$\n$$\nf(a, b, c) + f(a, c, b) \\le \\frac{27}{4}, \\qquad (2)\n$$\nsince adding these yields the result. To prove (1), observe that\n$$\nf(a, b, c) - f(a, c, b) = (a - b)(b - c)(c - a) \\le 0.\n... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | Maximum is 27/8, with equality when a = b = 3/2 and c = 0. | |
0335 | Problem:
Find all values of $a$ such that the maximum of the function $f(x) = \frac{a x - 1}{x^{4} - x^{2} + 1}$ is equal to $1$. | [
"Solution:\nSince the denominator of the function is positive, the given condition means that $a x - 1 \\leq x^{4} - x^{2} + 1$ for any $x$, and that the equality is attained for some $x$.\n\nLet $a \\geq 0$. Then $a x \\leq 0$ for $x \\leq 0$ and hence $a$ is the minimum of the function $g(x) = \\frac{x^{4} - x^{2... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | a = 2 or a = -2 | |
027z | Problem:
Seja $n$ um inteiro positivo.
a) Um quadrado de lado $n$ é dividido em $n^{2}$ quadradinhos de lados unitários por retas paralelas aos seus lados. Determine o número de retângulos cujos vértices são vértices de quadradinhos e que possuem lados paralelos aos lados do quadrado original.
b) Três quadrados de lad... | [
"Solution:\n\na) Os vértices dos retângulos são unicamente determinados pelas interseções de duas das $n+1$ retas verticais e paralelas com duas das $n+1$ retas horizontais paralelas aos lados do quadrado que delimitam os quadradinhos unitários. Podemos escolher a primeira reta vertical de $n+1$ maneiras e a segund... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | a) ((n(n+1))/2)^2. b) n^2(2n+1)^2 − n^4 − n^3(n+1) − (n(n+1)/2)^2. | |
0fuh | Problem:
1. Sei $ABCD$ ein Rechteck mit $|AD| \leq |AB|$. Sei $M$ der Mittelpunkt der Strecke $AD$ und $N$ der Mittelpunkt der Strecke $BC$. Der Punkt $E$ sei die Projektion von $B$ auf die Gerade $CM$.
a. Zeige, dass $ANEM$ ein gleichschenkliges Trapez ist.
b. Zeige, dass die Fläche des Vierecks $ABNE$ halb so gros... | [
"Solution:\n\nWegen $|AD| \\leq |AB|$ liegt der Punkt $E$ im Innern des Rechtecks $ABCD$.\n\na. Da die Strecke $AN$ durch eine Translation um den Vektor $\\overrightarrow{AM}$ in $MC$ übergeht, sind $AN$ und $ME$ parallel. Zu zeigen ist noch $\\angle MAN = \\angle ENA$. Sei $F$ der Schnittpunkt von $AN$ mit $EB$ un... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0006 | Dado cualquier conjunto de 9 puntos en el plano de los cuales no hay tres colineales, demuestre que para cada punto $P$ del conjunto, el número de triángulos que tienen como vértices a tres de los ocho puntos restantes y a $P$ en su interior, es par. | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof only | null | |
0icl | Problem:
Suppose one is given $n$ real numbers, not all zero, but such that their sum is zero. Prove that one can label these numbers $a_{1}, a_{2}, \ldots, a_{n}$ in such a manner that
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n-1} a_{n}+a_{n} a_{1}<0 .
$$ | [
"Solution:\nLet the given numbers (in an arbitrary order) be $b_{1}, b_{2}, \\ldots, b_{n}$. For every possible permutation $\\pi$ of $\\{1,2, \\ldots, n\\}$, consider the sum\n$$\nb_{\\pi(1)} b_{\\pi(2)}+b_{\\pi(2)} b_{\\pi(3)}+\\cdots+b_{\\pi(n-1)} b_{\\pi(n)}+b_{\\pi(n)} b_{\\pi(1)} .\n$$\nWe wish to show that s... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
016m | Let $M$ be the centroid of a non-equilateral triangle $ABC$. Let $A$ and $A'$ lie on opposite sides of the line $BC$ such that the triangle $BCA'$ is equilateral, and let $A''$ be such an internal point of the segment $AA'$ that $A''A' = 2AA''$. Let the points $B', B'', C', C''$ be defined analogously. Prove that the t... | [
"From $\\overrightarrow{AM} = \\frac{1}{3}\\overrightarrow{AA''}$ and $\\overrightarrow{AA''} = \\frac{1}{3}\\overrightarrow{AA'}$ we get $\\overrightarrow{MA''} = \\frac{1}{3}\\overrightarrow{A'''A'}$. Similarly $\\overrightarrow{MB''} = \\frac{1}{3}\\overrightarrow{B'''B'}$ and $\\overrightarrow{MC''} = \\frac{1}... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
025t | Problem:
Se $(x, y)$ é uma solução do sistema
$$
\left\{\begin{array}{l}
x y=6 \\
x^{2} y+x y^{2}+x+y=63
\end{array}\right.
$$
determine o valor de $x^{2}+y^{2}$. | [
"Solution:\nTemos\n$$\n\\begin{aligned}\n63 & =x^{2} y+x y^{2}+x+y \\\\\n& =x y(x+y)+(x+y) \\\\\n& =6(x+y)+(x+y) \\\\\n& =7(x+y)\n\\end{aligned}\n$$\nPortanto, $x+y=9$. Assim,\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y \\\\\n& =81-12 \\\\\n& =69\n\\end{aligned}\n$$"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 69 | |
0kkm | Problem:
Let $m$ be a positive integer. Show that there exists a positive integer $n$ such that each of the $2m+1$ integers
$$
2^{n}-m, 2^{n}-(m-1), \ldots, 2^{n}+(m-1), 2^{n}+m
$$
is positive and composite. | [
"Solution:\nLet $P$ be the set of prime divisors of the $2m+1$ numbers\n$$\n2^{m+1}-m, 2^{m+1}-m+1, \\ldots, 2^{m+1}+m\n$$\nWe claim that\n$$\nn = m+1 + \\prod_{p \\in P}(p-1)\n$$\nworks. To check this, let $k$ be any integer with $|k| \\leq m$. We can take some prime $q \\mid 2^{m+1}+k$, as $2^{m+1}+k \\geq 2^{m+1... | United States | HMMT Spring 2021 Team Round | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
01kc | Given a right-angled triangle $ABC$ with $\angle C = 90^\circ$. Its perimeter is equal to $30$ cm. $CH$ is the altitude of this triangle, $CK$ and $CL$ are the bisectors of the angles $ACH$ and $BCH$ respectively.
Find the length of the hypotenuse $AB$ if $KL = 4$ cm. | [
"Let $a = BC$, $b = AC$, $c = AB$, $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, $P = a+b+c$. Then $\\angle ACH = \\beta$, $\\angle BCH = \\alpha$, thus $\\angle ACK = \\angle KCH = \\beta/2$, $\\angle HCL = \\angle LCB = \\alpha/2$. Since $\\angle HKC = \\angle KAC + \\angle ACK$ (exterior angle of triangle $AC... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 13 | |
03un | (1) Prove that
$$
\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \ge 1,
$$
for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz = 1$.
(2) Prove that the equality holds for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xy... | [
"(1) Let\n$$\n\\frac{x}{x-1} = a, \\frac{y}{y-1} = b, \\frac{z}{z-1} = c,\n$$\nthen\n$$\nx = \\frac{a}{a-1}, \\quad y = \\frac{b}{b-1}, \\quad z = \\frac{c}{c-1}.\n$$\nSince $xyz = 1$, we have\n$$\nabc = (a-1)(b-1)(c-1),\n$$\nthat is\n$$\na + b + c - 1 = ab + bc + ca.\n$$\nTherefore\n$$\n\\begin{aligned}\na^2 + b^2... | China | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0k46 | Problem:
Let $G$ be an undirected simple graph. Let $f(G)$ be the number of ways to orient all of the edges of $G$ in one of the two possible directions so that the resulting directed graph has no directed cycles. Show that $f(G)$ is a multiple of $3$ if and only if $G$ has a cycle of odd length. | [
"Solution:\n\nLet $f_{G}(q)$ be the number of ways to color $G$ with $q$ colors. This is the chromatic polynomial of $G$, and turns out to be polynomial in $q$. Indeed, choose an edge $e$ and let $G \\backslash e$ be the graph $G$ with $e$ removed and let $G / e$ be graph $G$ with the vertices on either side of $e$... | United States | HMIC 2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0aby | The sides of the triangle are consecutive terms in the arithmetic progression. Prove that the line connecting the centroid and the center of the incircle is parallel to the side of the triangle with middle length. | [
"Let $\\triangle ABC$ be a triangle with sides $\\overline{AB} = c$, $\\overline{a} = \\overline{BC} = c + d$ and $\\overline{b} = \\overline{AC} = c + 2d$. Let $O$ be the center of the incircle and $O_1$ be the centroid, and $D, E$ be the feet of the side $BC$, respectively. Let $\\overline{OD} = r$, and the area ... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09c2 | Хурц өнцөгт $ABC$ гурвалжны $\angle A = 30^\circ$, $H$ нь орто төв ба $M$ нь $BC$ талын дундаж байг. $T$ нь $HM$ шулуун дээр орших $HM = MT$ байх цэг бол $AT = 2BC$ гэж батал. | [
"\n\n$BM = MC$, $HM = MT$ гэдгээс $BH \\parallel TC$ болно. Иймд $\\angle ACT = 90^\\circ$. Мөн $BT \\parallel HC$ гэдгээс $\\angle ABT = 90^\\circ$. Энэ 2 нөхцөлөөс $ABTC$ 4 өнцөгт тойрогт багтах ба тойргийн төв нь $AT$-ийн дундаж 0 цэг болно.\n\n$OB = OC$ ба $\\angle BOC = 2 \\cdot \\angl... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | Mongolian | proof only | null | |
0cen | Fix integers $n \ge 2$ and $1 \le m \le n-1$. Let $a_0$, $a_1$, $\dots$, $a_n$ be non-negative real numbers satisfying $a_0 + a_1 + \dots + a_n = 1$. Prove that, if $\sum_{k=0}^n a_k x^k < x^m$ for some $0 < x < 1$, then $\sum_{k=0}^{m-1} (m-k)a_k < \sum_{k=m+1}^n (k-m)a_k$.
The Problem Selection Committee | [
"As $a_0 + a_1 + \\dots + a_n = 1$, the required inequality is equivalent to $\\sum_{k=0}^n k a_k > m$. To prove this inequality, note that the exponential $t \\mapsto x^t$, $t \\in \\mathbb{R}$, is convex and apply Jensen's inequality to write $x^{\\sum_{k=0}^n k a_k} \\le \\sum_{k=0}^n a_k x^k < x^m$. As $0 < x <... | Romania | Twentieth IMAR Mathematical Competition | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0i1m | Problem:
Alex picks his favorite point $(x, y)$ in the first quadrant on the unit circle $x^{2}+y^{2}=1$, such that a ray from the origin through $(x, y)$ is $\theta$ radians counterclockwise from the positive $x$-axis. He then computes $\cos^{-1}\left(\frac{4x+3y}{5}\right)$ and is surprised to get $\theta$. What is ... | [
"Solution:\n\n$x = \\cos(\\theta),\\ y = \\sin(\\theta)$. By the trig identity you never thought you'd need, $\\frac{4x+3y}{5} = \\cos(\\theta-\\phi)$, where $\\phi$ has sine $3/5$ and cosine $4/5$. Now $\\theta-\\phi=\\theta$ is impossible, since $\\phi \\neq 0$, so we must have $\\theta-\\phi=-\\theta$, hence $\\... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 1/3 | |
0fvc | Problem:
Sei $n$ eine natürliche Zahl. Bestimme die Anzahl Teilmengen $A \subset \{1,2, \ldots, 2 n\}$, sodass für keine zwei Elemente $x, y \in A$ gilt $x+y=2 n+1$. | [
"Solution:\n\nDie einzigen Lösungen der Gleichung $x+y=2 n+1$ in $\\{1,2, \\ldots, 2 n\\}$ sind die Paare $(x, y)=(k, 2 n+1-k)$, wobei $1 \\leq k \\leq 2 n$. Betrachte nun die disjunkte Zerlegung\n$$\n\\{1,2, \\ldots, 2 n\\}=\\{1,2 n\\} \\cup \\{2,2 n-1\\} \\cup \\ldots \\cup \\{n-1, n+2\\} \\cup \\{n, n+1\\}\n$$\n... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 3^n | |
0d5s | Prove that there exist infinitely many non prime positive integers $n$ such that $7^{n-1} - 3^{n-1}$ is divisible by $n$. | [
"We will look for integers of the form $n = 7^{a} - 3^{a}$ with $a$ dividing $n-1$. Clearly, if $a$ exists then $n = 7^{a} - 3^{a}$ divides $7^{n-1} - 3^{n-1}$.\n\nLet $a = 3^{r}$ for $r \\geq 1$. We have $n = 7^{a} - 3^{a} \\equiv (-1)^{3^{r}} - 3^{3^{r}} \\equiv 4 \\pmod{8}$. We deduce that $n$ is a composite num... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English, Arabic | proof only | null | |
0cn9 | Prove or disprove that for every integer $n \ge 2009$ one can choose two distinct pairs of fractions from the sequence $\frac{1}{n}, \frac{2}{n-1}, \frac{3}{n-2}, \dots, \frac{n-1}{2}, \frac{1}{n}$ so that the sums of fractions in pairs are equal. (A. Shapovalov, K. Knop) | [
"Каждая из данных дробей имеет вид $\\frac{n+1-a}{a} = \\frac{n+1}{a} - 1$, где $1 \\le a \\le n$. Стало быть, нам требуется найти такие различные натуральные числа $a, b, c$ и $d$, не большие 2009, для которых $(\\frac{n+1}{a} - 1) + (\\frac{n+1}{b} - 1) = (\\frac{n+1}{c} - 1) + (\\frac{n+1}{d} - 1)$, что равносил... | Russia | Euler olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English; Russian | proof only | null | |
062p | Problem:
Man beweise oder widerlege, dass für alle positiven reellen Zahlen $a$, $b$ und $c$ die Ungleichung
$$
3 \leq \frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}<\frac{33}{4}
$$
gilt. | [
"Solution:\nBeweis von $3 \\leq \\frac{4 a+b}{a+4 b}+\\frac{4 b+c}{b+4 c}+\\frac{4 c+a}{c+4 a}$ :\n\n1. Variante: Multiplizieren mit Hauptnenner und Vereinfachen führt zur äquivalenten Ungleichung $45 a b c \\leq 7\\left(a^{2} b+b^{2} c+c^{2} a\\right)+8\\left(a b^{2}+b c^{2}+c a^{2}\\right)$. Dies folgt aus der Un... | Germany | 1. IMO-Auswahlklausur | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03tx | Let $a_1, a_2, \dots, a_{11}$ be $11$ distinct positive integers with their sum less than $2007$, and write the numbers $1, 2, \dots, 2007$ in order on the blackboard. Now we define a group of $22$ ordered operations:
The $i$-th operation is to take any number on the blackboard, and then add $a_i$ to it, if $1 \le i \l... | [
"The answer is: The \"good\" groups is more than the \"second good\" groups by $\\prod_{i=1}^{11} a_i$.\n\nMore generally, we write numbers $1, 2, \\dots, n$ in order on the blackboard, and define a group of $l$ ordered operations: The $i$-th operation is to take any number on the blackboard, and then add $b_i$ ($b... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | ∏_{i=1}^{11} a_i | |
02kt | Problem:
Uma companhia tem um lucro de $6\%$ nos primeiros $R\$ 1000,00$ reais de venda diária, e $5\%$ em todas as vendas que excedem $R\$ 1000,00$ reais, nesse mesmo dia. Qual é o lucro dessa companhia num dia que as vendas alcançam $R\$ 6000,00$ reais?
(a) $R\$ 250$
(b) $R\$ 300$
(c) $\$ 310$
(d) $R\$ 320$
(e) $R\... | [
"Solution:\n\nNos primeiros $R\\$ 1000$ reais a companhia tem lucro de $R\\$ 60$ reais, e para os $R\\$ 5000$ reais restantes tem lucro de $5000 \\times 5\\% = 250$ reais. Logo o lucro da empresa nesse dia é $R\\$ 310$ reais."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | c | |
05ei | Problem:
Trouver tous les triplets $ (a, b, c) $ de réels strictement positifs tels que
$$
\left\{\begin{array}{l}
a \sqrt{b} - c = a \\
b \sqrt{c} - a = b \\
c \sqrt{a} - b = c
\end{array}\right.
$$ | [
"Solution:\nOn remarque que $a = b = c = 4$ est solution. On va montrer que c'est en fait la seule.\n\nSi deux des trois nombres sont égaux à $4$ (disons $a$ et $b$), on vérifie facilement que le troisième aussi, car $4 \\sqrt{4} - c = 4$, donc $c = 4$.\n\nSi un des trois nombres vaut $4$ (disons $a$), alors $c \\s... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (4, 4, 4) | |
06p3 | Find all integer solutions of the equation
$$
\frac{x^{7}-1}{x-1}=y^{5}-1
$$ | [
"The equation has no integer solutions. To show this, we first prove a lemma.\n\nLemma. If $x$ is an integer and $p$ is a prime divisor of $\\frac{x^{7}-1}{x-1}$ then either $p \\equiv 1 \\pmod{7}$ or $p=7$.\n\nProof. Both $x^{7}-1$ and $x^{p-1}-1$ are divisible by $p$, by hypothesis and by Fermat's little theorem,... | IMO | IMO 2006 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, ... | English | proof and answer | No integer solutions exist. | |
0d31 | $ABC$ is a triangle, $G$ its centroid and $A'$, $B'$, $C'$ the midpoints of its sides $BC$, $CA$, $AB$, respectively. Prove that if the quadrilateral $AC'GB'$ is cyclic then
$$
AB \cdot CC' = AC \cdot BB'.
$$ | [
"First solution. Because $AC'GB'$ is cyclic, we have $\\measuredangle GAB' = \\measuredangle GC'B'$. Because $B'C'$ is parallel to $BC$, we have $\\measuredangle CC'B' = \\measuredangle C'CB$. We deduce that\n$$\n\\sin \\measuredangle A'A C = \\sin \\measuredangle C'C B.\n$$\n\nBut triangle... | Saudi Arabia | Preselection tests for the full-time training | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles ... | English | proof only | null | |
05e2 | Problem:
Let $k$ be a positive integer. Lexi has a dictionary $\mathcal{D}$ consisting of some $k$-letter strings containing only the letters $A$ and $B$. Lexi would like to write either the letter $A$ or the letter $B$ in each cell of a $k \times k$ grid so that each column contains a string from $\mathcal{D}$ when re... | [
"Solution:\nWe claim the minimum value of $m$ is $2^{k-1}$.\n\nFirstly, we provide a set $\\mathcal{S}$ of size $2^{k-1}-1$ for which Lexi cannot fill her grid. Consider the set of all length-$k$ strings containing only $A$'s and $B$'s which end with a $B$, and remove the string consisting of $k$ $B$'s. Clearly the... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2^{k-1} | |
0lca | The sequence $(x_n)$ is defined by
$$
x_0 = 2,\ x_{n+1} = [\sqrt{2x_n(x_n+1)}],\ \forall n \in \mathbb{N}^*
$$
Prove that $x_{2n} = 2^n + [2^n\sqrt{2}]$ and $x_{2n+1} = 2^{n+1} + [2^n\sqrt{2}]$ for all $n \in \mathbb{N}^*$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | x_{2n} = 2^n + floor(2^n * sqrt(2)) and x_{2n+1} = 2^{n+1} + floor(2^n * sqrt(2)) for all n. | |
0gtu | In a school having $n$ students any student has exactly $2023$ friends and if two students are not friends then they have exactly $2022$ common friends. Find all possible values of $n$. | [
"Answer: $n = 2024, 2026, 2028, 2696, 4044$.\n\nLet us reformulate a more general version of the problem in terms of the graph theory: In a regular graph $G$ on $n$ vertices each vertex has a degree $k < n-1$ and if two vertices are not neighbours then they have exactly $k-1$ common neighbours. Find all pairs $(n, ... | Turkey | Team Selection Test for IMO 2023 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | n = 2024, 2026, 2028, 2696, 4044 | |
01vw | The point $C$ with an abscissa $-2$ lie on the hyperbola $y = 1/x$. Two lines with slopes $2$ and $1/2$ passes through $C$ and intersect the hyperbola for the second time at the points $A$ and $B$.
Find the coordinates of the circumcenter of the triangle $ABC$. | [
"Answer: $\\left(-\\frac{11}{8};\\ 2\\right)$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | (-11/8, 2) | |
0b9y | Let $x, y$ be real numbers such that $x+y$, $x^3+y^3$, $x^5+y^5$ and $y^5$ are rational. Prove that $x$ and $y$ are rational. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0jxe | Problem:
A number is called cool if it is the sum of two nonnegative perfect squares. For example the numbers $17$ and $25$ are cool because $17=4^{2}+1^{2}$ and $25=5^{2}+0^{2}$, but the number $15$ is not cool.
Show that if $k$ is cool, then $2k$ is cool. | [
"Solution:\nObserve that if $k = a^{2} + b^{2}$, then $2k = (a-b)^{2} + (a+b)^{2}$."
] | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05cd | Inside a regular hexagon $ABCDEF$, equal rectangles $ABZY$, $CDXZ$, and $EFYX$ are drawn. How much of the area of the hexagon $ABCDEF$ do these rectangles cover?
 | [
"Each interior angle of a regular hexagon has a size of $120^\\circ$. Thus, a regular hexagon can be divided into $6$ equilateral triangles with side lengths equal to the hexagon itself (Fig. 23). Denoting the area of such a triangle as $S$, the area of the hexagon $ABCDEF$ is therefore $6S$.\n\nThe opposite sides ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2/3 | |
0a2h | Let $\triangle ABC$ be an acute-angled triangle such that $|AB| < |AC|$ with circumscribed circle $\Gamma$ with centre $O$. Points $D, E$ and $F$ are constructed as the feet of the altitudes from $A, B$ and $C$, respectively. We let $P$ be the intersection point of the tangents to $\Gamma$ through $B$ and $C$. The line... | [
"Since $O$ is the circumcentre, we get that\n$$\n\\begin{align*}\n\\angle BAO &= \\frac{1}{2}(180^\\circ - \\angle AOB) \\\\\n&= 90^\\circ - \\angle ACB \\\\\n&= 90^\\circ - \\angle EFA \\\\\n&= \\angle FAR,\n\\end{align*}\n$$\nwhere we also used that $BCEF$ is a cyclic quadrilateral due to Thales and that $R$ is t... | Netherlands | IMO Team Selection Test 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
08fy | Problem:
Un intero positivo $n$ si dice doppiamente reversibile di tipo $\ell$ se esistono due basi consecutive $b$ e $b+1$ tali che $n$ sia rappresentato da numeri palindromi di $\ell$ cifre sia in base $b$ che in base $b+1$. Ad esempio, 104 è doppiamente reversibile di tipo 3 perché $104_{10}=404_{5}=252_{6}$.
a. D... | [
"Solution:\n\na. Supponiamo per assurdo che esista un intero positivo doppiamente reversibile di tipo 2. Esso si scrive come $x x$ in una certa base $b$ e come $y y$ nella base $b+1$. Per definizione di rappresentazione di un numero in una base, si deve avere\n$$\nx \\cdot b^{1}+x \\cdot b^{0}=y \\cdot(b+1)^{1}+y \... | Italy | Olimpiadi di Matematica - Febbraio | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null |
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