| # | id | problem_markdown | solutions_markdown | images | country | competition | topics_flat | language | problem_type | final_answer |
|---|
1
|
0a7r
|
Problem:
Let $AB$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $OC$ and $AB$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $BC$ and let the lines $CP$ and $AB$ meet at $Q$. We choose $R$ on $AP$ so that $RQ$ and $AB$ are perpendicular to each other.
Show that $|BQ| = |QR|$.
|
[
"Solution:\n\n(See Figure 7.) Draw $PB$. By the Theorem of Thales, $\\angle RPB = \\angle APB = 90^{\\circ}$. So $P$ and $Q$ both lie on the circle with diameter $RB$. Because $\\angle AOC = 90^{\\circ}$, $\\angle RPQ = \\angle CPA = 45^{\\circ}$. Then $\\angle RBQ = 45^{\\circ}$, too, and $RBQ$ is an isosceles right triangle, or $|BQ| = |QR|$.",
"Solution:\n\nSet $O = (0, 0)$, $A = (-1, 0)$, $B = (1, 0)$, $C = (0, 1)$, and $P = (t, u)$, where $t > 0$, $u > 0$, and $t^2 + u^2 = 1$. The equation of line $CP$ is $y - 1 = \\frac{u - 1}{t} x$. So $Q = \\left( \\frac{t}{1 - u}, 0 \\right)$ and $|BQ| = \\frac{t}{1 - u} - 1 = \\frac{t + u - 1}{1 - u}$. On the other hand, the equation of line $AP$ is $y = \\frac{u}{t + 1}(x + 1)$. The $y$ coordinate of $R$ and also $|QR|$ is $\\frac{u}{t + 1}\\left( \\frac{t}{1 - u} + 1 \\right) = \\frac{ut + u - u^2}{(t + 1)(1 - u)} = \\frac{ut + u - 1 + t^2}{(t + 1)(1 - u)} = \\frac{u + t - 1}{1 - u}$. The claim has been proved."
]
|
[
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|
Nordic Mathematical Olympiad
|
Nordic Mathematical Contest, NMC 9
|
[
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
]
|
|
proof only
|
|
2
|
0eu5
|
Consider the sixteen tiles fixed on a wall as shown below. How many ways are there to write either $0$ or $1$ on each tile so that the product of the two numbers written on every neighboring pair of tiles (sharing a common side) is always $0$?
|
[
"When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*. First, we prove a couple of lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then\n$$\na_n = F_{n+2}\n$$\nfor all $n \\ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \\ge 1$.\n\n*Proof.* We prove by induction on $n \\ge 1$. It is clear that the lemma holds for $n=1$ and $n=2$. Assume the lemma for all $k$, $1 \\le k < n$, where $n \\ge 3$. If $0$ is written on the tile at one end, then any number ($0$ or $1$) can be written on the next tile and hence there are $a_{n-1}$ such z-patterns. If $1$ is written there, then $0$ should be written on the next tile and hence there are $a_{n-2}$ such z-patterns. Thus we get\n$$\na_n = a_{n-1} + a_{n-2} = F_{n+1} + F_n = F_{n+2},\n$$\nwhich proves the lemma.\n\n**Lemma 2.** Let $b_n$ be the number of z-patterns for $n$ tiles fixed on a wall in a ring shape. Then\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}\n$$\nfor all $n \\ge 4$.\n\n*Proof.* Let $n \\ge 4$ and choose any tile among the $n$ tiles fixed on a wall in a ring shape. If $0$ is written on the tile, then any number can be written on the neighboring tiles and hence there are $a_{n-1}$ such z-patterns. If $1$ is written on the tile, then $0$ should be written on the neighboring two tiles and hence there are $a_{n-3}$ such z-patterns. Thus we get\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1},\n$$\nwhich proves the lemma.\n\nWe now return to the problem. Consider the $2 \\times 2$ tiles in the center. Observe that the number of $0$'s that can be written on these four tiles equals $4$, $3$ or $2$.\n\nCase 1) Four $0$'s:\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\nWe may apply Lemma 2 to the $12$ tiles surrounding the four tiles in the center because any number can be written on the $12$ tiles. Therefore, the number of z-patterns in this case equals\n$$\nb_{12} = F_{13} + F_{11} = 233 + 89 = 322.\n$$\n\nCase 2) Three $0$'s:\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 0 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\nIn each of the four subcases, only $0$ can be written on two neighboring tiles of the tile marked by $1$. For the remaining $10$ tiles, it is clear that the number of z-patterns equals to $a_9 \\times 2 = F_{11} \\times 2 = 178$. Therefore, the total number of z-patterns in this case equals\n$$\n178 \\times 4 = 712.\n$$\n\nCase 3) Two $0$'s:\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}\n$$\nIn each of the two subcases, only $0$ can be written on four neighboring tiles of the two tiles marked by $1$. For the remaining $6$ tiles, it is clear that the number of z-patterns equals to $a_3^2 \\times 2^2 = F_5^2 \\times 4 = 100$. Therefore, the total number of z-patterns in this case equals\n$$\n100 \\times 2 = 200.\n$$\n\nCombining the three cases, the answer is: $322 + 712 + 200 = 1234$. $\\square$"
]
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[
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|
South Korea
|
20th Korean Mathematical Olympiad Final Round
|
[
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
]
|
|
proof and answer
|
1234
|
3
|
0l9s
|
Let $m$ and $n$ be two integers greater than $3$ and let be given a rectangular $m \times n$ board. At each step, one puts simultaneously $4$ marbles into $4$ cells of the board (each marble into a cell) so that these four cells form one of the following schemata.
Is it true that starting from a rectangular $m \times n$ board without marbles in it after a finite number of steps of appropriate puttings, one can put marbles into all cells of the board so that each cell is filled with a same (positive) number of marbles for all cells when:
i) $m = 2004$ and $n = 2006$?
ii) $m = 2005$ and $n = 2006$?
(At each step, it is not necessary that the four cells which are selected to put marbles into contained no marbles).
|
[
"i) After two steps: one can put into each cell of a small board of size $(4 \\times 2)$ a marble. One can partition the given board of size $(2004 \\times 2006)$ into small boards of size $(4 \\times 2)$. Therefore, after some steps, one can put marbles into all cells of the given board so that the number of marbles in each cell is the same for all cells.\n\nii) We now prove by contradiction that in the second case, the response to the problem is \"no\". Indeed, suppose that the contrary: after some steps, there would be $k$ marbles ($k > 0$) in each cell of the given board of size $(2005 \\times 2006)$. Color black all cells belonging to the odd rows and consider each non colored cell as white. Then, the number of black cells is equal to $1003 \\times 2006$ and the number of white cells is equal to $1002 \\times 2006$. But at each step we put exactly $2$ marbles into black cells and $2$ marbles in white cells. Therefore, after an arbitrary number of steps, the number of marbles in all black cells must be equal to the number of the marbles in all white cells. Consequently, we would have $1003 \\times 2006 \\times k = 1002 \\times 2006 \\times k$ and so get $1 = 0$. This contradiction proves our assertion."
]
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[
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|
Vietnam
|
Vijetnam 2006
|
[
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
]
|
|
proof and answer
|
i) Yes. ii) No.
|
4
|
03bq
|
Given is a circle $k$ and a point $A$ outside of it. The segment $BC$ is a diameter of $k$. Find the locus of the orthocenter of $\triangle ABC$, when $BC$ is changing.
|
[
"Let $O$ be the center of $k$ and $\\omega$ be the circle with diameter $AO$. Let $AA_1$ and $BB_1$ be the altitudes of $\\triangle ABC$ and $H$ is the orthocenter. Then the power of $H$ with respect to $\\omega$ is equal to $AH \\cdot HA_1$ and the power of $H$ with respect to $k$ is equal to $BH \\cdot HB_1$. Since $AH \\cdot HA_1 = BH \\cdot HB_1$ it follows that $H$ lies on the radical axis $l$ of $k$ and $\\omega$. Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type.",
"Let $AB \\cap k = B'$, $AC \\cap k = C'$ and $BC' \\cap B'C = H$. Then $H$ is the orthocenter of $\\triangle ABC$ and $H$ lies on the polar $l$ of point $A$ with respect to $k$ (*Why?*). Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type."
]
|
[
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|
Bulgaria
|
Bulgarian National Mathematical Olympiad
|
[
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
]
|
English
|
proof and answer
|
The locus is the radical axis of k and the circle with diameter AO, equivalently the polar of A with respect to k.
|
5
|
09wj
|
A square sheet of paper lying on the table is divided into $8 \times 8 = 64$ equal squares. These squares are numbered from **a1** to **h8** as on a chess board (see fig. 1). We now start folding, in such a way that square **a1** always stays in the same spot on the table. First we fold along the horizontal midline (fig. 1). This will cause square **a8** to fold on top of square **a1**. Then we fold along the vertical midline (fig. 2). Next, we fold along the new horizontal midline (fig. 3), et cetera. After folding six times, we have a small package of paper in front of us (fig. 7) that we can consider as a stack of 64 square pieces of paper.
fig. 1
fig. 2
fig. 3
...
fig. 7
The squares in this stack are numbered from bottom to top from 1 to 64.
So square **a1** gets number 1.
Which number does square **f6** get?
|
[
"43"
]
|
[
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|
Netherlands
|
Second Round
|
[
"Discrete Mathematics > Algorithms"
]
|
English
|
final answer only
|
43
|
6
|
0dpx
|
Convex quadrilateral *ABCD* is inscribed in circle *ω*. Rays *AB* and *DC* intersect at *K*. *L* is chosen on the diagonal *BD* so that $\angle BAC = \angle DAL$. *M* is chosen on the segment *KL* so that $CM \parallel BD$. Prove that the line $BM$ touches $\omega$. (Kungozhin M.)
|
[
"Let *N* be a point on the line $AK$ so that $MN \\parallel AL$. Since $\\frac{CM}{DL} = \\frac{NM}{AL} = \\frac{KM}{KL}$ and $\\angle CMN = \\angle DLA$, it follows that $K$ is a center of homothety which sends $\\triangle CMN$ into similar $\\triangle DLA$. On the other hand, $\\triangle DLA$ is similar to $\\triangle CBA$ because $\\angle DAL = \\angle BAC$ and $\\angle ADL = \\angle ACB$. Consequently, $\\angle (BN, BC) = \\angle (MN, MC)$, and thus points $N$, $B$, $M$, $C$ are cyclic. Therefore, $\\angle CBM = \\angle CNM = \\angle CAB$ from which it follows that $BM$ touches $\\omega$.",
"For this solution we need the following theorem.\n\n**Pascal's theorem.** Points $A$, $B$, $C$, $D$, $E$, $F$ (not necessarily in this order) lie on some circle. Then the intersections of the lines $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$ lie on a straight line.\n\nBack to the problem. Let $E$ be the second intersection of the line $AL$ and $\\omega$. Define $l_b$ as a tangent line to $\\omega$ at $B$. Let's apply Pascal's theorem on points $B$, $B_1$, $A$, $E$, $C$, $D$ (here $B_1$ coincides with $B$) and pairs of lines ($BB_1$, $EC$), ($B_1A$, $CD$), ($AE$, $DB$). These pairs of lines coincide with pairs of lines ($l_b$, $EC$), ($BA$, $CD$), ($AE$, $DB$). Then from Pascal's theorem, the straight line connecting $K = BA \\cap CD$ and $L = AE \\cap DB$ will also contain $l_b \\cap EC$. Since $M = EC \\cap KL$, it follows that $BM$ touches $\\omega$, as desired."
]
|
[
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|
Silk Road Mathematics Competition
|
XXI SILK ROAD MATHEMATICAL COMPETITION
|
[
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
English
|
proof only
|
|
7
|
0lfz
|
Problem:
Inside the triangle $A B C$ a point $M$ is given. The line $B M$ meets the side $A C$ at $N$. The point $K$ is symmetrical to $M$ with respect to $A C$. The line $B K$ meets $A C$ at $P$. If $\angle A M P=\angle C M N$, prove that $\angle A B P=\angle C B N$.
|
[
"Solution:\n\nLet $M^{*}$ be the isogonal conjugate of $M$ with respect to $\\triangle A B C$.\n\nLEMMA. Let $\\ell$ be the isogonal conjugate of $A M$ with respect to $\\angle B M C$, and let $\\ell^{*}$ be the isogonal conjugate of $A M^{*}$ with respect to $\\angle B M^{*} C$. The lines $\\ell, \\ell^{*}$ meet on $B C$ and are symmetrical with respect to $B C$.\n\nProof. Let $X$ be the meeting point of $B M^{*}$ and $C M$. Let $Y \\in B C$ be a point such that $X A, X Y$ are isogonal conjugate with respect to $\\angle B X C$.\nSince $C A, C Y$ are isogonal conjugate with respect to $\\angle M^{*} C M$, so $A$ and $Y$ are isogonal conjugate with respect to $\\triangle C X M^{*}$, it follows that $M^{*} Y$, $M^{*} A$ are isogonal conjugate with respect to $\\angle B M^{*} C$. Similarly, we can prove $M Y, M A$ are isogonal conjugate with respect to $\\angle B M C$.\n\nBy easy angle chasing we get $\\angle M Y B=\\angle C Y M^{*}$, thus $M Y \\equiv \\ell$ and $M^{*} Y \\equiv \\ell^{*}$ are symmetrical with respect to $B C$.\n\nFrom the Lemma we now get that $P K$ is the isogonal conjugate of $B M^{*}$ with respect to $\\angle A M^{*} C$, thus from $B \\in P K$ we get $P K$ to be the angle bisector of $\\angle A M^{*} C$ and $M^{*} \\in B K$, therefore $\\angle A B P=\\angle C B N$.",
"Solution:\n\nNotice that the thesis is equivalent (by Steiner's theorem) with $\\frac{B C}{B A}=\\frac{M C}{M A}$; stated in other words, that $B$ lies on the $M$-Apollonius circle of $\\triangle M C A$.\n\nLet $F$ be the foot of the angle bisector of $\\angle C M A$. By dint of the above observation, it is enough to prove that $B \\in \\odot(M F K)$, which is trivial by simple angle chasing; suppose wlog that $M A>M C$, it then follows $\\angle M F K=\\pi+\\angle M A C-\\angle M C A=\\pi-\\angle K B M$."
]
|
[
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|
Zhautykov Olympiad
|
XI International Zhautykov Olympiad in Sciences
|
[
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
|
proof only
|
|
8
|
0edn
|
Take an equilateral triangle $ABC$ with the side of length $a$. Draw the squares $BADE$ and $CBFG$ on the sides $AB$ and $BC$, respectively (see figure). How long is the line segment $DG$?
(A) $(\sqrt{2} + 1)a$
(B) $(2\sqrt{2} + 1)a$
(C) $(\sqrt{3} + 1)a$
(D) $\sqrt{3}a$
(E) 1
|
[
"Denote the intersections of the segment $DG$ with the lines $AB$ and $BC$ by $K$ and $L$ respectively. Due to symmetry the segment $DG$ is parallel to the segment $AC$, so $KBL$ is an equilateral triangle. The triangles $DKA$ and $LGC$ are one half of an equilateral triangle with altitudes of length $a$ and sides $|DK| = |GL| = \\frac{2a}{\\sqrt{3}}$. The side of the equilateral triangle $KBL$ therefore measures\n$$\n|KL| = a - |AK| = a - \\frac{|DK|}{2} = a - \\frac{a}{\\sqrt{3}}.\n$$\nThe length of the segment $DG$ is\n$$\n|DG| = 2|DK| + |KL| = \\frac{4a}{\\sqrt{3}} + \\left(a - \\frac{a}{\\sqrt{3}}\\right) = a + \\frac{3a}{\\sqrt{3}} = (1 + \\sqrt{3})a."
]
|
[
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|
Slovenia
|
Slovenija 2016
|
[
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
|
MCQ
|
(\sqrt{3} + 1)a
|
9
|
05j5
|
Problem:
Sur la diagonale $[BD]$ d'un carré $ABCD$ on a choisi un point $E$. Soient $O_{1}$ et $O_{2}$ les centres des cercles circonscrits aux triangles $ABE$ et $ADE$ respectivement. Montrer que $AO_{1}EO_{2}$ est un carré.
|
[
"Solution:\n\n$\\widehat{AO_{1}E} = 2\\widehat{ABE} = 90^{\\circ}$ et $O_{1}A = O_{1}E$ donc $AO_{1}E$ est un triangle isoc\u00e8le rectangle en $O_{1}$. De m\u00eame, $AO_{2}E$ est un triangle isoc\u00e8le rectangle en $O_{2}$.\n\nComme $O_{1}$ et $O_{2}$ ne sont pas dans le m\u00eame demi-plan d\u00e9limit\u00e9 par $(AE)$, on en d\u00e9duit que $AO_{1}EO_{2}$ est un carr\u00e9."
]
|
[
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|
France
|
OFM
|
[
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
|
proof only
|
|
10
|
09vi
|
Problem:
Zij $ABC$ een scherphoekige driehoek met $O$ het middelpunt van de omgeschreven cirkel. Punt $Q$ ligt op de omgeschreven cirkel van $\triangle BOC$ zodat $OQ$ een middellijn is. Punt $M$ ligt op $CQ$ en punt $N$ ligt inwendig op lijnstuk $BC$ zodat $ANC M$ een parallellogram is. Bewijs dat de omgeschreven cirkel van $\triangle BOC$ en de lijnen $AQ$ en $NM$ door één punt gaan.
|
[
"Solution:\n\nDefinieer $T$ als het snijpunt van de omgeschreven cirkel van $\\triangle BOC$ en de lijn $AQ$. We gaan bewijzen dat $T$ op $NM$ ligt. Schrijf $\\alpha=\\angle BAC$. Dan geldt $\\angle BOC=2\\alpha$ vanwege de middelpuntsomtrekshoekstelling. Aangezien $|OB|=|OC|$ en $\\angle OCQ=90^{\\circ}=\\angle OBQ$ wegens Thales, geldt $\\triangle OBQ \\cong \\triangle OCQ$ (ZZR). Dus geldt $\\angle COQ=\\angle BOQ=\\alpha$.\n\nVanwege parallellogram $ANCM$ geldt $\\angle AMC=180^{\\circ}-\\angle MCN=\\angle QCN=\\angle QCB=\\angle QOB=\\alpha$, dus ook $\\angle CNA=\\angle AMC=\\alpha$. Nu zien we dat $\\angle QTB=\\angle QOB=\\alpha=\\angle CNA$, waaruit volgt dat $\\angle ATB=180^{\\circ}-\\angle QTB=180^{\\circ}-\\angle CNA=\\angle ANB$. Dit betekent dat $ATNB$ een koordenvierhoek is.\n\nOok is $ATCM$ een koordenvierhoek omdat $\\angle AMC=\\alpha=\\angle COQ=\\angle CTQ=180^{\\circ}-\\angle ATC$.\n\nKoordenvierhoek $ATCM$ geeft $\\angle ATM=\\angle ACM=180^{\\circ}-\\angle QCB-\\angle BCA$ vanwege de gestrekte hoek bij $C$. We weten $\\angle QCB=\\angle QOB=\\alpha=\\angle CAB$, dus met een hoekensom in driehoek $ABC$ krijgen we $\\angle ATM=180^{\\circ}-\\angle CAB-\\angle BCA=\\angle ABC$.\n\nNu gebruiken we koordenvierhoek $ATNB$ om te vinden dat $\\angle ABC=\\angle ABN=180^{\\circ}-\\angle ATN$, zodat al met al geldt $\\angle ATM=180^{\\circ}-\\angle ATN$. Dit betekent dat $M$, $T$ en $N$ op een lijn liggen, zoals we wilden bewijzen."
]
|
[
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|
Netherlands
|
IMO-selectietoets II
|
[
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
]
|
|
proof only
|
|