| # | id | problem_markdown | solutions_markdown | images | country | competition | topics_flat | language | problem_type | final_answer |
|---|
1
|
00hy
|
There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:
First, he chooses an endpoint of each segment as a "sink". Then he places the present at the endpoint of the segment he is at. The present moves as follows:
- If it is on a line segment, it moves towards the sink.
- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.
If the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2 n-1$ friends.
|
[
"Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will prove that the friends eligible to receive presents are the ones on even-numbered intersection points.\n\nFirst part: at most $n$ friends can receive a present.\n\nThe solution relies on a well-known result: the $n$ lines determine regions inside the circle; then it is possible to paint the regions with two colors such that no regions with a common (line) boundary have the same color. The proof is an induction on $n$ : the fact immediately holds for $n=0$, and the induction step consists on taking away one line $\\ell$, painting the regions obtained with $n-1$ lines, drawing $\\ell$ again and flipping all colors on exactly one half plane determined by $\\ell$.\n\nNow consider the line starting on point 1. Color the regions in red and blue such that neighboring regions have different colors, and such that the two regions that have point 1 as a vertex are red on the right and blue on the left, from Tony's point of view. Finally, assign to each red region the clockwise direction and to each blue region the anticlockwise direction. Because of the coloring, every boundary will have two directions assigned, but the directions are the same since every boundary divides regions of different colors. Then the present will follow the directions assigned to the regions: it certainly does for both regions in the beginning, and when the present reaches an intersection it will keep bordering one of the two regions it was dividing. To finish this part of the problem, consider the regions that share a boundary with the circle. The directions alternate between outcoming and incoming, starting from 1 (outcoming), so all even-numbered vertices are directed as incoming and are the only ones able to receive presents.\n\nSecond part: all even-numbered vertices can receive a present.\n\nFirst notice that, since every two chords intersect, every chord separates the endpoints of each of the other $n-1$ chords. Therefore, there are $n-1$ vertices on each side of every chord, and each chord connects vertices $k$ and $k+n, 1 \\leq k \\leq n$.\n\nWe prove a stronger result by induction in $n$ : let $k$ be an integer, $1 \\leq k \\leq n$. Direct each chord from $i$ to $i+n$ if $1 \\leq i \\leq k$ and from $i+n$ to $i$ otherwise; in other words, the sinks are $k+1, k+2, \\ldots, k+n$. Now suppose that each chord sends a present, starting from the vertex opposite to each sink, and all presents move with the same rules. Then $k-i$ sends a present to $k+i+1, i=0,1, \\ldots, n-1$ (indices taken modulo $2 n$ ). In particular, for $i=k-1$, Tony, in vertex 1, send a present to vertex $2 k$. Also, the $n$ paths the presents make do not cross (but they may touch.) More formally, for all $i, 1 \\leq i \\leq n$, if one path takes a present from $k-i$ to $k+i+1$, separating the circle into two regions, all paths taking a present from $k-j$ to $k+j+1, j<i$, are completely contained in one region, and all paths taking a present from $k-j$ to $k+j+1, j>i$, are completely contained in the other region. For instance, possible $\\{ \\}^\\{1\\}$ paths for $k=3$ and $n=5$ follow:\n\n\n\nThe result is true for $n=1$. Let $n>1$ and assume the result is true for less chords. Consider the chord that takes $k$ to $k+n$ and remove it. Apply the induction hypothesis to the remaining $n-1$ lines: after relabeling, presents would go from $k-i$ to $k+i+2,1 \\leq i \\leq n-1$ if the chord were not there.\n\nReintroduce the chord that takes $k$ to $k+n$. From the induction hypothesis, the chord intersects the paths of the presents in the following order: the $i$-th path the chord intersects is the the one that takes $k-i$ to $k+i, i=1,2, \\ldots, n-1$.
|
[
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|
Asia Pacific Mathematics Olympiad (APMO)
|
APMO
|
[
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
]
|
English
|
proof only
|
|
2
|
00k4
|
Let $ABCDEF$ be a regular octahedron with lower vertex $E$, upper vertex $F$, middle slice plane $ABCD$, center $M$ and circumsphere $k$. Furthermore let $X$ be an arbitrary point within side $ABF$. The line $EX$ intersects $k$ in $E$ and $Z$ and the plane $ABCD$ in $Y$.
$$
\text{Show that } \langle EMZ \rangle = \langle EYF \rangle.
$$
|
[
"\nWe intersect the entire figure with the plane through the points $X$, $E$ and $F$. Since $M$ is on line $EF$, it is part of that plane. Also points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle $k'$, which also has $M$ as its center. The intersection of $ABCD$ with the plane is the perpendicular bisector of line $EF$, and $Y$ is on that bisector.\n\n\nWe denote $\\angle ZEF = \\alpha$.\nSince $ZM$ and $EM$ are both radii of $k$ (and $k'$), the triangle $ZME$ is isosceles, therefore\n$\\angle EZM = \\angle ZEM = \\alpha$.\nSince $Y$ is on the bisector of $EF$, also triangle $EYF$ is isosceles, therefore $\\angle YFE = \\angle YEF = \\alpha$.\nTherefore the triangles $ZME$ and $EYF$ are similar, and their corresponding angles $\\angle EMZ$ and $\\angle EYF$ are identical. $\\square$",
"We intersect the figure with the plane $XEF$ as we did in solution 1. Since $\\angle YMF = 90^\\circ$, and by Thales also $\\angle FZY = 90^\\circ$, the quadrilateral $YMFZ$ has a circumcircle. Therefore $\\angle YZM = \\angle YFM$, and we again get that the triangles $ZME$ and $EYF$ are similar. $\\square$"
]
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[
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|
Austria
|
AustriaMO2013
|
[
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
English
|
proof only
|
|
3
|
00xp
|
Problem:
A square is divided into 16 equal squares, obtaining the set of 25 different vertices. What is the least number of vertices one must remove from this set, so that no 4 points of the remaining set are the vertices of any square with sides parallel to the sides of the initial square?
|
[
"Solution:\n\nThe example in Figure 3a demonstrates that it suffices to remove 8 vertices to \"destroy\" all squares. Assume now that we have managed to do that by removing only 6 vertices. Denote the horizontal and vertical lines by $A, B, \\ldots, E$ and $1,2, \\ldots, 5$ respectively. Obviously, one of the removed vertices must be a vertex of the big square - let this be vertex $A 1$. Then, in order to \"destroy\" all the squares shown in Figure $3 \\mathrm{~b}-\\mathrm{e}$ we have to remove vertices $B 2, C 3, D 4, D 2$ and $B 4$. Thus we have removed 6 vertices without having any choice but a square shown in Figure $3 \\mathrm{f}$ is still left intact.\n\n\n$\\mathrm{a}$\n\n$\\mathrm{b}$\n\nc\n\nd\n\ne\n\nf\nFigure 3"
]
|
[
{
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|
Baltic Way
|
Baltic Way 1993
|
[
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
]
|
|
proof and answer
|
8
|
4
|
01l1
|
Exactly one integer number is written in each cell of an $8 \times 8$ square table. Per move it is allowed to choose any $n \times n$ square, $1 < n < 8$, and either to increase by 1 or to decrease by 1 all numbers in the cells of the chosen square.
Is it possible to get the table with zeros in all its cells from the arbitrary initial table?
(V. Kaskevich)
|
[
"We separate the table into five parts: the central $4 \\times 4$ square, two $2 \\times 8$ rectangles, and two $4 \\times 2$ rectangles (see Fig. 1).\nWe can obtain the 0's in all cells of the $2 \\times 8$ rectangles. It suffices to show how we can change (increase by 1 or decrease by 1) the value in any cell of the rectangle so that all other cells of this rectangle keep their value. The corresponding procedures using $2 \\times 2$ and $3 \\times 3$ squares are shown in Fig. 2 and Fig. 3.\n\n\n\n\nFig. 3\n\nIn similar way we can change (increase by 1 or decrease by 1) the value in any cell of the $2 \\times 4$ rectangles. It suffices to show how we can change the value in any cell of these rectangles so that all other cells of these rectangles and all cells of the $2 \\times 8$ rectangles keep their value. The corresponding procedures are shown in Fig. 4 (see, in addition, Fig 3).\n\n\n\nIt remains to change the numbers in the cells of the central $4 \\times 4$ square. It suffices to show how we can change the value of any cell of this square so that all other cells of the table keep their value. The corresponding procedures are shown in Fig. 5.\n\n\nFig. 5"
]
|
[
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|
Belarus
|
60th Belarusian Mathematical Olympiad
|
[
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
]
|
English
|
proof and answer
|
Yes
|
5
|
0217
|
Problem:
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $N$ denote the second point of intersection of line $AI$ and $\omega$. The line through $I$ perpendicular to $AI$ intersects line $BC$, segment $[AB]$, and segment $[AC]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $AEF$ meets $\omega$ again at $P$, and lines $PN$ and $BC$ intersect at $Q$. Prove that lines $IQ$ and $DN$ intersect on $\omega$.
|
[
"Solution:\nBy construction, $APEF$ and $APBC$ are cyclic, and so\n$$\n\\begin{aligned}\n\\angle BDE &= \\angle CDF = \\angle AFD - \\angle FCD = \\angle AFE - \\angle ACB = (180^\\circ - \\angle EPA) - (180^\\circ - \\angle BPA) \\\\\n&= \\angle BPA - \\angle EPA = \\angle BPE.\n\\end{aligned}\n$$\nHence $DBEP$ is cyclic, too. It follows that $\\angle IDP = \\angle EDP = \\angle EBP = \\angle ABP = \\angle ANP = \\angle INP$ since $APBN$ is cyclic, and so $PDNI$ is also cyclic. In particular, $\\angle DPN = \\angle DIN = 90^\\circ$. Let $R$ denote the second intersection of $DP$ and $\\omega$, so $NQ \\perp DR$. Then $\\angle NPR = 90^\\circ$, so $RN$ is a diameter of $\\omega$. It is well-known that $N$ is the midpoint of the arc $\\widehat{BC}$ not containing $A$, whence $RN \\perp BC$. Thus $DQ$ and $NQ$ are altitudes of triangle $RDN$, and so $Q$ is its orthocentre. This implies that $RQ \\perp DN$, whence, since $RN$ is a diameter of $\\omega$, the intersection $X$ of $RQ$ and $DN$ lies on $\\omega$.\n\n\n\nIt is also well-known that $N$ is the centre of the circumcircle $\\Omega$ of triangle $BCI$. Since $DI \\perp IN$ by construction, $DI$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $BC$ of $\\omega$ and $\\Omega$, it follows that $|DI|^2 = |DB||DC| = |DX||DN|$. Hence triangles $DNI$ and $DIX$ are similar; in particular, $\\angle DXI = \\angle DIN = 90^\\circ$. All of this shows that $R, I, Q, X$ lie on a line perpendicular to $DN$ that intersects $DN$ at $X \\in \\omega$. This completes the proof.",
"Solution:\nLet $K$ be the midpoint of segment $[BC]$. It is well-known that $N$ is the midpoint of the small arc $\\widehat{BC}$ of $\\omega$, so $BC \\perp KN$. In particular, $\\angle DKN = 90^\\circ$. But $\\angle DIN = 90^\\circ$ by construction, so $DIKN$ is cyclic, with circumcircle $\\Gamma$. Moreover, $\\angle PEF = 180^\\circ - \\angle PAF = 180^\\circ - \\angle PAB = \\angle PBC$ and $\\angle PFE = \\angle PAE = \\angle PAB = \\angle PCB$ since $AFEP$ and $ACBP$ are cyclic, so triangles $PEF$ and $PBC$ are similar. Now, by construction, $I$ is the midpoint of segment $[EF]$, so, $K$ being the midpoint of $[BC]$, triangles $PIF$ and $PKC$ are similar, too. It follows that $\\angle PID = 180^\\circ - \\angle PIF = 180^\\circ - \\angle PKC = \\angle PKD$, whence $P$ lies on $\\Gamma$.\n\n\n\nLet $\\Omega$ be the circumcircle of triangle $BCI$. By construction, $Q$ lies on the radical axes $PN$ of $\\omega, \\Gamma$ and $BC$ of $\\omega, \\Omega$, so is the radical centre of $\\omega, \\Gamma, \\Omega$. In particular, $IQ$ is the radical axis of $\\Gamma, \\Omega$, so is perpendicular to the line joining the centres of $\\Gamma, \\Omega$. Now it is well-known that $N$ is the centre of $\\Omega$, and, since $\\angle DIN = 90^\\circ$, the centre of $\\Gamma$ is the midpoint of segment $[DN]$. This shows that $IQ \\perp DN$.\nFinally, let $DN$ meet $\\omega$ again at $X$. Since $DI \\perp IN$ by construction and $N$ is the centre of $\\Omega$, $DI$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $BC$ of $\\omega, \\Omega$, it follows that $|DI|^2 = |DB||DC| = |DX||DN|$. Hence triangles $DNI$ and $DIX$ are similar; in particular, $\\angle DXI = \\angle DIN = 90^\\circ$, i.e. $IX \\perp DN$. Since $IQ \\perp DN$, it follows that $X$ is the intersection of $IQ$ and $DN$. Since $X$ lies on $\\omega$ by construction, this completes the proof.",
"Solution:\nSince $APEF$ and $APBC$ are cyclic,\n$$\n\\begin{aligned}\n\\angle CPF &= \\angle BPA - \\angle BPC - \\angle FPA = (180^\\circ - \\angle BCA) - \\angle BAC - \\angle FEA \\\\\n&= (180^\\circ - \\angle BCA - \\angle BAC) - \\angle BED = \\angle CBA - \\angle BED = \\angle CBE - \\angle BED = \\angle BDE = \\angle CDF,\n\\end{aligned}\n$$\nso $DPFC$ is cyclic, too. Thence $\\angle CPD = \\angle CFD = 180^\\circ - \\angle IFA = 90^\\circ + \\angle IAF = 90^\\circ + \\angle CAN = 90
|
[
{
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|
Benelux Mathematical Olympiad
|
15th Benelux Mathematical Olympiad
|
[
"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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
|
proof only
|
|
6
|
03e9
|
A set of points in the plane is called good if the distance between any two points in it is at most $1$. Let $f(n, d)$ be the largest positive integer such that in any good set of $3n$ points, there is a circle of diameter $d$, which contains at least $f(n, d)$ points. Prove that there exists a positive real $\epsilon$, such that for all $d \in (1 - \epsilon, 1)$, the value of $f(n, d)$ does not depend on $d$ and find that value as a function of $n$.
(Kristiyan Vasilev, Konstantin Garov)
|
[
"Since $m(n, d)$ is an increasing function of $d$ and it cannot be larger than $3n$, clearly it hits a constant when $d$ approaches $1$. Fix some $d$, $d < 1$, it may be very close to $1$. Take an equilateral triangle with side length $1$ and put inside it $3n$ points \u2014 $n$ points near each of its vertices so that they be at distance less than $(1 - d)/2$ from the corresponding vertex \u2014 see figure 1.\n\n\n\nThis set of points apparently is a good set. A disk with radius $d$ cannot cover a point that is near to one vertex and a point near another vertex. That is, the maximum number of points that could be covered with such a disk is $n$. This means that $m(n, d) \\le n$ no matter how close to $1$ the value of $d$ is.\n\nWe will prove that $m(n, d) = n$ when $d$ is close to $1$. To prove this we need to study the good set of points. How close together are its points? Can we claim that all of them lie in disk with some not too large radius? Of course, they lie in a disk with radius $1$ since if you take one of the points, the rest are at distance at most $1$ from this point. But can we claim that all of them lie in a disk with a smaller radius? Clearly, they may not lie in a disk with radius $1/2$ \u2014 the equilateral triangle with side length $1$ is an example that disproves this. Crucial to this problem is the following observation.\n**Lemma.** Any good set can be put inside a disk with radius $\\frac{1}{\\sqrt{3}}$. That is, any good set can be put inside a circle circumscribed about an equilateral triangle with a side length $1$.\n\nThe official solution proves it by using Helly's theorem which is instructive. Another method is used here.\n**Proof.** Let $X$ be a good set. We first show that $X$ lies in a circle circumscribed about an acute triangle with side lengths less or equal to $1$. Take the convex hull of $X$ and put it inside a circle with some radius. We begin to tighten this circle until it touches $1$ and then $2$ points of the convex hull. We can tighten the circle further unless these two points lie on its diameter. In this case $A$ lies in a circle with a diameter at most $1$. So, assume we decrease the circle $k$ until it touches $3$ points of $X$, say $A, B, C$. If these points are vertices of an acute triangle we are done, so suppose $\\angle ACB \\ge 90^\\circ$ \u2014 see figure 2. We begin to further decrease the diameter of $k$ keeping the points $A$ and $B$ on the circle. This can be done until the first of the following two events happens:\n1. The circle hits a point $C' \\in X$ (and $C'$ and $C$ lie on different sides of $AB$).\n2. $AB$ becomes a diameter of $k$.\n\n\n\nIf the second event happens first, we are done since $X$ lies in a disk with diameter $1$. In case of the first event, the triangle $ABC'$ is acute. Next, since $\\triangle ABC'$ is acute one of its angles, say $\\angle AC'B$, is greater or equal to $60^\\circ$. It follows the radius of $k' = \\frac{AB}{2 \\sin \\angle C} \\le \\frac{1}{\\sqrt{3}}$.\nSo, we proved that we can cover $X$ with a circle $k$ with radius $1/\\sqrt{3}$. $\\square$\n\nTake a point $A \\in k$ and construct the points $B, C, D \\in k$ such that $AB = BC = CD = d$ \u2014 see figure 3. We construct three disks with diameters $AB, BC, CD$ correspondingly. These disks cover all the points inside $k$ except the red piece in figure 3. Note now that by taking $d$ close to $1$, the point $D$ will be as close to $A$ as we want. This means that we can make the red piece on figure 3 as small as we want. We take $d$ so close to $1$ so that the measure of the arc $AD$ (on figure 3) is less than $\\frac{2\\pi}{3n+1}$. If the red piece does not contain a point of $X$ we are done. Otherwise we begin to rotate the point $A$ and the entire configuration around $k$. It is not possible that every time the red piece hits a point of $X$ since otherwise the number of points in $X$ would be more than $3n$. Therefore we can
|
[
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|
Bulgaria
|
Autumn tournament
|
[
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
]
|
English
|
proof and answer
|
n
|
7
|
04aq
|
Let $M$ and $N$ be positive integers. Consider an $N \times N$ square array consisting of $N^2$ lamps that can be in two states - on or off. At the beginning all lamps are turned off.
A *move* consists of choosing a row or a column of the array and changing the state of $M$ consecutive lamps in the chosen row or column, i.e. turning on the lamps that are turned off and vice versa.
Determine the necessary and sufficient condition for which it can be achieved that after a finite number of moves all lamps are turned on. (Tonći Kokan)
|
[
"The sought condition is that $M$ divides $N$.\n\nIt is easy to see that if $M$ divides $N$ we can choose a sequence of moves after which all lamps will be turned on. Really, if $N = Mk$ for some positive integer $k$ then we choose each row $k$ times. In the $(ik+j)$-th move, for $i = 0, 1, \\dots, M-1$, $j = 1, \\dots, k$ we choose $M$ consecutive lamps from $((j-1)k+1)$-th to $jk$-th place in $(i+1)$-th row.\n\nTo prove necessity, colour the lamps in $M$ colors (named $0, 1, 2, \\dots, M-1$) in a way presented on the Figure 4.1, i.e. color the lamp in the $i$-th row and $j$-th column in color $i + j - 2$ (mod $M$).\n\n\nFigure 4.1: Example of the coloring for $N = 10, M = 6$.\n\nIn every move we change the state of exactly one lamp of each color. In the beginning all lamps are turned off so after each step we have the same number of lamps that are turned on in all colors. If it is possible to achieve that after some move all lamps are turned on, then the number of lamps of each color must be the same.\n\nAssume on the contrary that $M$ does not divide $N$ and let $N = Mk + r$, where $1 \\le r \\le M-1$. Divide the $N \\times N$ array into four subarrays of dimensions $Mk \\times Mk, Mk \\times r, r \\times Mk$ and $r \\times r$ as in the Figure 4.2.\nSince each of the subarrays of dimensions $Mk \\times Mk, Mk \\times r$ and $r \\times Mk$ is a disjoint union of sequences of $M$ consecutive lamps in a row or a column, we see that the number of lamps of each color in their union is the same (it equals $Mk^2 + 2kr$).\n\n---\n\n\nFigure 4.2: Dividing the array for $N = 10$, $M = 6$.\n\nConsider the remaining $r \\times r$ subarray. On Figure 4.3 we see that the number of lamps of color $r-1$ equals $r$, but the number of lamps of color $r$ equals $r-1$. Indeed, the lamps of color $r-1$ appear in each row of the subarray exactly once and the lamps of color $r$ appear in each but first row. Also, there is no row with two or more lamps of color $r$ because $r$ is strictly less than $M$ so each row has all the lamps of different color.\n\n\nFigure 4.3: Example of the $r \\times r$ subarray for $N = 10$, $M = 6$ (and $r = 4$).\n\nHence in the whole array we have a different number of lamps of color $r-1$ and color $r$ and we can never achieve that all of the lamps are turned on."
]
|
[
{
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|
Croatia
|
CroatianCompetitions2011
|
[
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
]
|
|
proof and answer
|
M divides N
|
8
|
04u6
|
Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD = CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC$, $ADE$ and $AO_1O_2$ have a common point different from $A$.
|
[
"Let $Z$ be the midpoint of the longer arc $BC$ of the circumcircle $\\omega$ of the triangle $ABC$. The triangles $ZDB$ and $ZEC$ are congruent, because they agree in the sides $BD = CE$ and $ZB = ZC$, as well as in the corresponding angles between them, for both lie over the chord $AZ$ of $\\omega$. It follows that $\\angle ZDA = \\angle ZEA$, which in turn discloses that the quadrilateral $ADEZ$ is cyclic. So it remains to be shown that the quadrilateral $AO_1O_2Z$ is cyclic.\n\n\n\nThe center $O$ of $\\omega$ satisfies $OO_1 \\perp AB$ and $OO_2 \\perp AC$. The projections of $O$ and $O_1$ onto $AC$ are the midpoints of $AC$ and $AE$ respectively. Thus the projection of the segment $OO_1$ onto $AC$ has length $\\frac{1}{2}CE$. For the same reason, the projection of $OO_2$ on $AB$ has length $\\frac{1}{2}BD$, and by hypothesis these two length agree. Moreover, the angle between $OO_1$ and $AC$ is the same as the angle between $OO_2$ and $AB$. It follows that $OO_1 = OO_2$.\nFurther, we have $\\angle AOO_1 = \\angle ACB = \\angle O_2OZ$, the latter being a consequence of $ZO \\perp BC$ and $OO_2 \\perp AC$. So the rays $OA$ and $OZ$ are isogonal in the angle $O_2OO_1$. In the combination with $AO = ZO$ and $OO_1 = OO_2$ this proves that\n\nthe quadrilateral $AO_1O_2Z$ is an isosceles trapezium and thus in particular cyclic.\nThereby the problem is solved.",
"Let the circumcircles of triangles $ABE$ and $ADC$ intersect each other again at $F \\neq A$. Then the triangles $BFD$ and $EFC$ are congruent, for they agree in their sides $BD = CE$ as well as in their corresponding adjacent angles, i.e., $\\angle FBD = \\angle FEC$ and $\\angle BDF = \\angle ECF$. It follows that the altitudes of these triangles passing through $F$ have the same lengths, wherefore $AF$ is the bisector of the angle $BAC$.\n\n\n\nNow construct the point $S$ such that $SO_1FO_2$ is a parallelogram. We will show that $S$ is the desired point.\nTo prove that $S$ lies on the circumcircle of triangle $AO_1O_2$, we note that the triangles $AO_1O_2$ and $FO_1O_2$ are congruent due to $AO_1 = FO_1$ and $AO_2 = FO_2$. It follows that $AO_1O_2S$ is an isosceles trapezium and hence in particular a cyclic quadrilateral, as claimed. Later, it will help us to have observed that the facts used in this paragraph imply $AS \\parallel O_1O_2 \\perp AF$.\nNext, we prove that $S$ lies on the circumcircle of the triangle $ABC$ and that it is actually the midpoint of its longer arc $BC$; this will also show $S \\neq A$, as needed. Our first intermediate step is to observe that the triangles $O_1SB$ and $O_2CS$ are congruent. Indeed they agree in a pair of sides, $BO_1 = FO_1 = SO_2$ and $SO_1 = FO_2 = CO_2$. Moreover the corresponding angles between these sides are equal, because their complements to $360^\\circ$ are equal as a consequence of\n$$\n\\angle BO_1F = 2\\angle BAF = 2\\angle FAC = \\angle FO_2C\n$$\nand $\\angle FO_1S = \\angle SO_2F$. This concludes the verification of $\\triangle O_1SB \\cong \\triangle O_2CS$, and it follows that $BS = CS$. Further, since $AF$ is the bisector of $\\angle BAC$ and $AF \\perp AS$, the line $AS$ is the exterior bisector of $\\angle BAC$. Altogether we obtain that $S$ is the point described above. The fact that $S$ lies on the circumcircle of the triangle $ADE$ can be shown as in the first solution.",
"Denote by $O$ and $P$ the circumcentres of triangles $ABC$ and $ADE$, respectively. The lines $OO_1$ and $O_2P$ (being the perpendicular bisectors of $AB$ and $AD$, respectively) are both perpendicular to $AB$ and their distance is $\\frac{1}{2}BD$. Similarly, the lines $OO_2$ and $O_1P$ are both perpendicular to $AC$ and their distance is $\\frac{1}{2}CE$. Since $\\frac{1}{2}BD = \\frac{1}{2}CE$, the quadrilateral $O_1OO_2P$ is a parallelogram with equal altitudes, hence a rhombus. It follows that $OP$ is the perpendicular bisector of $O_1O_2$, so all the three circumcentres
|
[
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|
Czech Republic
|
Czech-Polish-Slovak Match
|
[
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
]
|
|
proof only
|
|
9
|
04xa
|
Given a circle $k$ and its chord $AB$ which is not the diameter. Let $C$ be any point inside the longer arc of $k$. We denote by $K$ and $L$ the reflections of $A$ and $B$ with respect to the axes $BC$ and $AC$. Prove that the distance of the midpoints of the line segments $KL$ and $AB$ is independent of the location of the point $C$.
|
[
"Denote by $S$ the midpoint of $AB$, by $M$ the midpoint of $KL$, and by $P$ and $Q$ the feet of the altitudes from $A$ and $B$ in the triangle $ABC$. Obviously, $P$ and $Q$ are the midpoints of $AK$ and $BL$ respectively (fig. 3). Therefore $QS$ is the mid-segment in the triangle $LAB$ and $MP$ is the mid-segment in the triangle $LAK$. We have\n$$\nQS = \\frac{1}{2} LA, \\quad MP = \\frac{1}{2} LA \\quad \\text{and} \\quad QS \\parallel LA \\parallel MP.\n$$\nthus $SPMQ$ is a parallelogram (this is true even in the case of \"degenerate\" triangles $LAB$ or $LAK$).\n\nFig. 3\nThe points $P$ and $Q$ lie on the Thales circle over $AB$, so $SP = SQ = \\frac{1}{2} AB$. This implies the parallelogram $SPMQ$ is a rhombus and the length of its side is independent of the location of $C$. To prove that also the length of its diagonal $SM$ is independent of $C$, it suffices to show that the size of the angle spanned by its sides $SP$ and $SQ$ is constant when moving $C$ along $k$ (then all the rhombuses $SPMQ$ and also their diagonals $SM$ are congruent).\n\nFig. 4a\n\nFig. 4b\nIf the angle $\\alpha$ in the triangle $ABC$ is acute, the point $Q$ lies inside of $AC$ (the angle $\\gamma$ is always acute by the statement of the problem) and the angle $PSQ$ is central to the angle $PAQ$, which is inscribed over the chord $PQ$ of the Thales circle over $AB$ (fig. 4a). Hence\n$$\n\\angle PSQ = 2 \\angle PAQ = 2(90^\\circ - \\gamma) = 180^\\circ - 2\\gamma.\n$$\n\nThe same formula we get when $\\alpha$ is non-acute, since in this case $\\beta$ is acute and we can use the inscribed angle $PBQ$ instead of $PAQ$ (fig. 4b):\n$$\n\\angle PSQ = 2 \\angle PBQ = 2(90^\\circ - \\gamma) = 180^\\circ - 2\\gamma.\n$$\nAs the size of $\\gamma$ does not change when moving $C$ along $k$ (it is an inscribed angle over the fixed chord $AB$), the size of the angle $PSQ$ also does not change.",
"Denote by $\\alpha, \\beta, \\gamma$ the angles of the triangle $ABC$. We only consider the case $\\alpha < 90^\\circ$; the second case ($\\beta < 90^\\circ$) is similar. Let $S, M, U, V$ be the midpoints of the segments $AB, KL, AL, BK$ respectively. Let $H$ be the common point of the lines $AK$ and $BL$ (fig. 5a, b). Note that the quadrilateral $USVM$ is a parallelogram ($SU \\parallel BL \\parallel MV, SV \\parallel AK \\parallel MU$) and\n$$\nSV = AB \\sin \\beta, \\quad MV = SU = AB \\sin \\alpha, \\quad \\angle SVM = \\angle AIL = \\angle ACB\n$$\nBy the law of sines applied to the triangle $ABC$,\n$$\n\\frac{AC}{SV} = \\frac{1}{AB} \\cdot \\frac{AC}{\\sin \\beta} = \\frac{1}{AB} \\cdot \\frac{BC}{\\sin \\alpha} = \\frac{BC}{MV}\n$$\nso the triangles $SVM$ and $ACB$ are proportional (two sides proportional and the same angle between them). Then again by the law of sines.\n$$\nSM = AB \\cdot \\frac{SV}{AC} = \\frac{AB^2 \\sin \\beta}{AC} = \\frac{AB^2 \\sin \\gamma}{AB} = AB \\sin \\gamma\n$$\nwhich obviously does not depend of the location of the point $C$.\n\nFig. 5a\n\nFig. 5b"
]
|
[
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|
Czech-Polish-Slovak Mathematical Match
|
Cesko-Slovacko-Poljsko
|
[
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
]
|
English
|
proof only
|
|
10
|
051m
|
Consider hexagons whose internal angles are all equal.
(i) Prove that for any such hexagon the sum of the lengths of any two neighbouring sides is equal to the sum of the lengths of their opposite sides.
(ii) Does there exist such a hexagon with side lengths 1, 2, 3, 4, 5 and 6 in some order?
|
[
"(i) Let the hexagon be $ABCDEF$. It suffices to show that $|AB| + |BC| = |DE| + |EF|$.\nLet $K$ be the intersection point of rays $FA$ and $CB$ and let $L$ be the intersection point of rays $FE$ and $CD$ (Fig. 8). The size of every internal angle of the hexagon is $120^\\circ$, whence triangles $KAB$ and $LDE$ are equilateral. The quadrilateral $FKCL$ is a parallelogram since its opposite sides are parallel. This implies $|KC| = |LF|$ or $|KB| + |BC| = |LE| + |EF|$, which together with $|KB| = |AB|$ and $|LE| = |DE|$ implies the desired equality $|AB| + |BC| = |DE| + |EF|$.\n\n(ii) Take a parallelogram with side lengths 7 and 5 and internal angles $60^\\circ$ and $120^\\circ$, and cut off equilateral triangles with side lengths 1 and 2 at its acute angles. This gives rise to a hexagon with all internal angles having size $120^\\circ$ and side lengths 1, 4, 5, 2, 3, 6 (Fig. 9).\n\n\nFigure 8\n\nFigure 9",
"(i) Note that the external angles of the hexagon have size $60^\\circ$. Any two opposite sides of the hexagon are parallel, as they are separated by exactly three external angles.\nConsider a line $s$ perpendicular to opposite sides $CD$ and $FA$ of the hexagon $ABCDEF$ (Fig. 10). As all other sides form the same angle $30^\\circ$ with line $s$, the lengths of these sides are proportional to the lengths of the projections of the sides to line $s$. The sum of the lengths of the projections of sides $AB$ and $BC$ is equal to the distance between the parallel lines $CD$ and $FA$ and the same holds also for the opposite sides $DE$ and $EF$. Therefore the sum of the lengths of the projections of sides $AB$ and $BC$ is equal to that of sides $DE$ and $EF$, whence the sums of the lengths of the sides are equal as well.\n\n\nFigure 10\n\n(ii) Figure 11 shows a hexagon in a triangular grid with distance between neighbouring nodes being 1. The side lengths of the hexagon are 1, 4, 5, 2, 3 and 6.\n\n\nFigure 11",
"Consider two types of transformations on hexagons that maintain the property that all internal angles are of the same size.\n(1) Prolonging two opposite sides by the same quantity $x$ (Fig. 12); this causes the side lengths to change according to the template\n$$\n(a, b, c, d, e, f) \\longleftrightarrow (a+x, b, c, d+x, e, f).\n$$\n(2) Prolonging the two neighbouring sides of one particular side by the same quantity $x$ (Fig. 13); this causes the side lengths to change according to the template\n$$\n(a, b, c, d, e, f) \\longleftrightarrow (a+x, b-x, c+x, d, e, f)\n$$\nA straightforward check shows that both transformations maintain the desired property, no matter of in which direction the transformations are applied.\n\n(i) As the internal angles of all regular hexagons are equal, it suffices to show that an arbitrary hexagon with all internal angles equal can be turned into a regular hexagon by a finite sequence of the transformations above.\nIndeed, let the side lengths of a given hexagon with all internal angles equal be $(a, b, c, d, e, f)$. Assume w.l.o.g. that $d \\ge a$ and $f \\ge c$. Choose a quantity $s$ such that $s \\ge \\max(a, b, c)$; by applying the transformation (1) thrice, we can obtain a hexagon with three consecutive sides having the same length:\n$$\n(a, b, c, d, e, f) \\xrightarrow{(1)} (s, b, c, d', e, f) \\xrightarrow{(1)} (s, s, c, d', e', f) \\xrightarrow{(1)} (s, s, s, d', e', f').\n$$\nBy the assumption made above we have $d' \\ge s$ and $f' \\ge s$. W.l.o.g., assume also $d' \\le f'$. The transformation\n$$\n(s, s, s, d', e', f') \\xrightarrow{(2)} (s, s, s, s, e'', f'')\n$$\nleads to a hexagon with four consecutive sides of equal length. But this must be regular since all of its internal angles are equal (Fig. 14).\n\n(ii) Such a hexagon can be obtained from a regular hexagon with side length 1 by the following transformations:\n$$\n(1, 1, 1, 1, 1, 1) \\xrightarrow{(1)} (1, 4, 1, 1, 4, 1) \
|
[
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|
Estonia
|
Final Round of National Olympiad
|
[
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
]
|
English
|
proof and answer
|
Yes; for example a hexagon with all interior angles equal and side lengths in order 1, 4, 5, 2, 3, 6 exists.
|