IMO | Beni Bogoşel's blog

IMO 2023 Problem 2

July 19, 2023 beni22sof 1 comment

Problem 2. Let ${ABC}$ be an acute-angled triangle with ${AB < AC}$ . Let ${\Omega}$ be the circumcircle of ${ABC}$ . Let ${S}$ be the midpoint of the arc ${CB}$ of ${\Omega}$ containing ${A}$ . The perpendicular from ${A}$ to ${BC}$ meets ${BS}$ at ${D}$ and meets ${\Omega}$ again at ${E \neq A}$ . The line through ${D}$ parallel to ${BC}$ meets line ${BE}$ at ${L}$ . Denote the circumcircle of triangle ${BDL}$ by ${\omega}$ . Let ${\omega}$ meet ${\Omega}$ again at ${P \neq B}$ .

Prove that the line tangent to ${\omega}$ at ${P}$ meets line ${BS}$ on the internal angle bisector of ${\angle BAC}$ .

Solution: Let us first do some angle chasing. Since ${BC||LD}$ we have ${\angle EBC=\angle BLD}$ and since ${BLPD}$ is cyclic we have ${\angle BLD = \angle PBD}$ . Therefore, if ${E' \in PD \cap \Omega}$ we have ${\angle BPE'=\angle EBC=\angle EAC}$ . Therefore the arcs ${BE'}$ and ${CE}$ are equal.

Denote by ${F}$ the midpoint of the short arc ${BC}$ of ${\Omega}$ . Then ${AF}$ is the angle bisector of ${\angle BAC}$ . Moreover, ${E}$ and ${E'}$ are symmetric with respect to ${SF}$ and ${AE'}$ is a diameter in ${\Omega}$ .

Let us denote ${G \in BS\cap AF, H \in AF\cap PE'}$ . It is straightforward to see that ${\angle EAF = \angle FSE'}$ , and since ${AE||SF}$ we have ${AF||SE'}$ .

Moreover, ${\angle AEE'=90^\circ}$ . Considering ${Q \in AE\cap SE'}$ , since ${SE=SE'}$ we find that ${S}$ is the midpoint of ${QE'}$ . Then, since ${AH||QE'}$ in ${\Delta DQE'}$ we find that ${G}$ is the midpoint of ${AH}$ . But ${\angle APH=90^\circ}$ , since ${AE'}$ is a diameter. It follows that ${\angle GPH = \angle AHP}$ .

On the other hand, ${\angle BGF = \frac{1}{2}( \text{arc}(AS)+\text{arc}(EF) = \angle PBG}$ , showing that ${PBHG}$ is cyclic and ${PG}$ is tangent to the circle circumscribed to ${PLBD}$ . As shown in the figure below, the geometry of this problem is quite rich.

There are quite a few inscribed hexagon where Pascal’s theorem could be applied. Moreover, to reach the conclusion of the problem it would be enough to prove that ${AF, PE'}$ and ${BP'}$ are concurrent, where ${P'}$ is the symmetric of ${P}$ with respect to ${SF}$ .

Categories: Geometry, IMO, Olympiad Tags: cyclic, Geometry, IMO

IMO 2023 Problem 4

July 12, 2023 beni22sof Leave a comment

Problem 4. Let ${x_1,x_2,\dots,x_{2023}}$ be pairwise different positive real numbers such that

$\displaystyle a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}$ is an integer for every ${n=1,2,\dots,2023.}$ Prove that ${a_{2023} \geqslant 3034.}$

Categories: Algebra, IMO, Inequalities, Problem Solving Tags: Algebra, IMO, inequality

IMO 2023 Problem 1

July 12, 2023 beni22sof Leave a comment

Problem 1. Determine all composite integers ${n>1}$ that satisfy the following property: if ${d_1, d_2, \ldots, d_k}$ are all the positive divisors of ${n}$ with ${1=d_1<d_2<\cdots<d_k=n}$ , then ${d_i}$ divides ${d_{i+1}+d_{i+2}}$ for every ${1 \leqslant i \leqslant k-2}$ .

Categories: Algebra, IMO, Number theory, Olympiad Tags: IMO, number theory, prime, Problem Solving

Problems of the International Mathematical Olympiad 2023

July 11, 2023 beni22sof Leave a comment

Problem 3. For each integer ${k \geqslant 2}$ , determine all infinite sequences of positive integers ${a_1, a_2, \ldots}$ for which there exists a polynomial ${P}$ of the form ${P(x)=x^k+c_{k-1} x^{k-1}+\cdots+c_1 x+c_0}$ , where ${c_0, c_1, \ldots, c_{k-1}}$ are non-negative integers, such that

$\displaystyle P\left(a_n\right)=a_{n+1} a_{n+2} \cdots a_{n+k}$

for every integer ${n \geqslant 1}$ .

Problem 4. Let ${x_1,x_2,\dots,x_{2023}}$ be pairwise different positive real numbers such that

$\displaystyle a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}$

is an integer for every ${n=1,2,\dots,2023.}$ Prove that ${a_{2023} \geqslant 3034.}$

Problem 5. Let ${n}$ be a positive integer. A Japanese triangle consists of ${1 + 2 + \dots + n}$ circles arranged in an equilateral triangular shape such that for each ${i = 1}$ , ${2}$ , ${\dots}$ , ${n}$ , the ${i^{th}}$ row contains exactly ${i}$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of ${n}$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with ${n = 6}$ , along with a ninja path in that triangle containing two red circles.

In terms of ${n}$ , find the greatest ${k}$ such that in each Japanese triangle there is a ninja path containing at least ${k}$ red circles.

Problem 6. Let ${ABC}$ be an equilateral triangle. Let ${A_1,B_1,C_1}$ be interior points of ${ABC}$ such that ${BA_1=A_1C}$ , ${CB_1=B_1A}$ , ${AC_1=C_1B}$ , and

$\displaystyle \angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$

Let ${BC_1}$ and ${CB_1}$ meet at ${A_2,}$ let ${CA_1}$ and ${AC_1}$ meet at ${B_2,}$ and let ${AB_1}$ and ${BA_1}$ meet at ${C_2.}$ Prove that if triangle ${A_1B_1C_1}$ is scalene, then the three circumcircles of triangles ${AA_1A_2, BB_1B_2}$ and ${CC_1C_2}$ all pass through two common points.

(Note: a scalene triangle is one where no two sides have equal length.)

Source: imo-official.org, AOPS forums

Categories: Algebra, Combinatorics, Geometry, IMO, Inequalities, Number theory, Olympiad, Problem Solving Tags: Geometry, IMO, number theory, Problem Solving

Build three particular equal segments in a triangle

December 18, 2022 beni22sof 3 comments

I recently stumbled upon the following problem:

Consider a triangle $ABC$ . Construct points $P,Q$ on $AB, AC$ , respectively such that $BP=PQ=QC$ .

I was not able to solve this myself, so a quick search on Google using “BP=PQ=QC” yielded the following article where the solution to the problem above is presented.

Categories: Geometry, Olympiad Tags: construction, Geometry, homothety, IMO, scaling

IMO 2022 – Problem 4 – Geometry

September 6, 2022 beni22sof Leave a comment

Problem 4. Let ${ABCDE}$ be a convex pentagon such that ${BC = DE}$ . Assume that there is a point ${T}$ inside ${ABCDE}$ with ${TB = TD}$ , ${TC = TE}$ and ${\angle ABT = \angle TEA}$ . Let line ${AB}$ intersect lines ${CD}$ and ${CT}$ at points ${P}$ and ${Q}$ , respectively. Assume that the points ${P, B, A, Q}$ occur on their line in that order. Let line ${AE}$ intersect lines ${CD}$ and ${DT}$ at points ${R}$ and ${S}$ , respectively. Assume that the points ${R, E, A, S}$ occur on their line in that order. Prove that the points ${P, S, Q, R}$ lie on a circle.

Categories: Geometry, IMO, Olympiad, Problem Solving Tags: cyclic, Geometry, IMO, Problem Solving

IMO 2022 – Problem 2 – Non-standard functional equation

September 6, 2022 beni22sof 2 comments

Problem 2. Let ${\Bbb R^+}$ denote the set of positive real numbers. Find all functions ${f : \Bbb{R}^+ \rightarrow \Bbb{R}^+}$ such that for each ${x \in \Bbb R^+}$ , there is exactly one ${y \in \Bbb R^+}$ satisfying ${xf(y) + yf(x) \leq 2}$ .

Categories: Algebra, IMO, Problem Solving Tags: Algebra, functional equation, IMO

IMC 2022 – Day 1 – Problem 1

August 8, 2022 beni22sof Leave a comment

Problem 1. Let ${f:[0,1]\rightarrow (0,\infty)}$ be an integrable function such that ${f(x)\cdot f(1-x) = 1}$ for all ${x\in [0,1]}$ . Prove that

$\displaystyle \int_0^1 f(x) dx \geq 1.$

Solution: If you want just a hint, here it is: Cauchy-Schwarz. For a full solution read below.

Categories: Analysis, Olympiad, Problem Solving Tags: Analysis, IMC, IMO, olympiad

Function with zero average on vertices of all regular polygons

July 4, 2022 beni22sof Leave a comment

Fix a positive integer ${n \geq 3}$ . Let ${f:\Bbb{R}^2 \rightarrow \Bbb{R}}$ be a function such that for any regular ${n}$ -gon ${A_1...A_n}$ we have

$\displaystyle f(A_1)+...+f(A_n) = 0.$

Prove that ${f}$ is identically equal to zero.

Source: Romanian Team Selection Test 1996, see also the Putnam Contest Problems from 2009.

Solution: The solution comes by looking at some examples:

1. Consider an equilateral triangle ${A_1A_2A_3}$ . It is possible to produce another two equilateral triangles ${A_1B_2B_3}$ and ${A_1C_2C_3}$ such that ${A_2B_2C_2}$ , ${A_3B_3C_3}$ are equilateral. Note that we kept a common vertex and we rotated the initial triangle by ${2\pi/3}$ and ${4\pi/3}$ . Applying the result for all the small triangles and summing we obtain

$\displaystyle 3f(A_1)+... = 0$

where the missing terms are again sums of values of ${f}$ on some equilateral triangles. It follows that ${f(A_1)=0}$ .

2. For a square things are even simpler, since considering rotations of a square around one vertex one ends up with a configuration containing a square, its midpoints and its center. A similar reasoning shows that the value of the function ${f}$ at the center needs to be equal to zero, summing the values of the function ${f}$ on the vertices of all small squares.

In the general case, the idea is the same. Consider an initial polygon ${A_1...A_n}$ and rotate it around ${A_1}$ with angles ${2k\pi/n}$ , ${1\leq k \leq n-1}$ . Then sum all the values of the function on the vertices of these regular polygons. Observe that the vertex ${A_1}$ is repeated ${n}$ times while all other vertices are part of some regular polygon. In the end we get

$\displaystyle n f(A_1)+0 = 0$

where the zeroes correspond to sums over vertices of regular polygons.

The same type of reasoning should hold when the sum over vertices of regular polygons is replaced by an integral on a circle. The proof would follow the same lines. Fix a point ${A}$ , then integrate ${f}$ on all rotations of the circle ${C}$ through ${A}$ . On one side this integral should be equal to zero. On the other it contains the value of ${f}$ in ${A}$ and values on concentric circles in ${A}$ . This should imply that ${f(A)}$ is zero for any point ${A}$ .

Categories: Geometry, IMO, Olympiad, puzzles Tags: IMO, integral, olympiad, regular polygons

Balkan Mathematical Olympiad 2022

May 9, 2022 beni22sof Leave a comment

Problem 1. Let ${ABC}$ be an acute triangle such that ${CA \neq CB}$ with circumcircle ${\omega}$ and circumcentre ${O}$ . Let ${t_A}$ and ${t_B}$ be the tangents to ${\omega}$ at ${A}$ and ${B}$ respectively, which meet at ${X}$ . Let ${Y}$ be the foot of the perpendicular from ${O}$ onto the line segment ${CX}$ . The line through ${C}$ parallel to line ${AB}$ meets ${t_A}$ at ${Z}$ . Prove that the line ${YZ}$ passes through the midpoint of the line segment ${AC}$ .

Problem 2. Let ${a, b}$ and ${n}$ be positive integers with ${a>b}$ such that all of the following hold:

i. ${a^{2021}}$ divides ${n}$ ,

ii. ${b^{2021}}$ divides ${n}$ ,

iii. 2022 divides ${a-b}$ .

Prove that there is a subset ${T}$ of the set of positive divisors of the number ${n}$ such that the sum of the elements of ${T}$ is divisible by 2022 but not divisible by ${2022^2}$ .

Problem 3. Find all functions ${f: (0, \infty) \rightarrow (0, \infty)}$ such that $f(y(f(x))^3 + x) = x^3f(y) + f(x)$ for all ${x, y>0}$ .

Problem 4. Consider an ${n \times n}$ grid consisting of ${n^2}$ until cells, where ${n \geq 3}$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places ${k}$ frogs on the cells so that each of the ${n^2}$ cells can be reached by a frog in a finite number (possible zero) of hops. Find the least value of ${k}$ for which this is always possible regardless of the colouring chosen by Dionysus.

Source: AOPS

Categories: Algebra, Combinatorics, Geometry, IMO, Olympiad, Problem Solving Tags: balkan, IMO, Problem Solving

IMO 2020 – Problem 2

September 30, 2020 beni22sof Leave a comment

Problem 2. The real numbers ${a,b,c,d}$ are such that ${a \geq b \geq c \geq d>0}$ and ${a+b+c+d=1}$ . Prove that

$\displaystyle (a+2b+3c+4d)a^ab^bc^cd^d<1.$

Hints: This problem is not difficult if one knows a bit of real analysis. The function ${x \mapsto x^x}$ is convex and can be extended continuously to ${[0,1]}$ noting that ${\lim_{x \rightarrow 0} x^x=1}$ . Therefore, the function ${(a,b,c,d) \mapsto (a+2b+3c+4d)a^ab^bc^cd^d}$ is can be extended continuously to ${[0,1]^4}$ and it attains its minimum and maximum values. Moreover, the constraint ${a+b+c+d=1}$ defines a compact set inside ${[0,1]^4}$ so there exists a minimum and a maximum for this function on the given domain. If at the maximum some of the variables should be ${0}$ , the restriction ${a,b,c,d}$ would make the inequality in the hypothesis to be sharp, but strict.

Having no other idea than to go ahead with the analysis, a natural idea is to see what happens when perturbing two of the variables while maintaining the constraint. First note that the case where ${a=b=c=d=1/4}$ gives the value ${10/16}$ which is obviously less than ${1}$ . Taking ${b,c,d \rightarrow 0}$ gives values as close to ${1}$ as we want, so this the maximum is surely not attained when all variables are equal. So we may well assume that at the maximum there exist two variables, say ${a}$ and ${b}$ such that ${a>b}$ .

Then, change the variables ${a}$ and ${b}$ to ${a-t,b+t}$ , with ${t>0}$ , which preserves the constraint. Consider the function

$\displaystyle f: t \mapsto (a+2b+3c+4d+t)(a-t)^{(a-t)}(b+t)^{(b+t)}c^cd^d.$

The derivative of this function is

$\displaystyle (a-t)^{(a-t)}(b+t)^{(b+t)}c^cd^d[1+(a+2b+3c+4d+t)[\ln(b+t)-\ln(a-t)]].$

It is not difficult to see that the function above is increasing with respect to ${t}$ . Therefore, the function ${f}$ is convex and attains its maxima at the extrema of the function are attained when ${a-t}$ and ${b+t}$ are as far away as possible. Therefore, ${(a,b,c,d)}$ is not a maximum for the above function when ${a<1}$ . A similar argument works when perturbing other pairs of points. Therefore the maximal value is attained when ${a=1}$ and ${b=c=d=0}$ and this maximal value is ${1}$ . Since the hypothesis clearly states ${b,c,d>0}$ , it follows that the upper bound is indeed ${1}$ but is not attained.

Categories: Analysis, IMO, Inequalities, Olympiad, Problem Solving Tags: Analysis, IMO, Inequalities

IMO 2020 – Problem 1

September 24, 2020 beni22sof Leave a comment

Problem 1. Consider the convex quadrilateral ${ABCD}$ . The point ${P}$ is in the interior of ${ABCD}$ . The following ratio equalities hold:

$\displaystyle \angle PAD : \angle PBA : \angle DPA = 1:2:3 = \angle CBP : \angle BAP : \angle BPC.$

Prove that the following three lines meet in a point: the internal bisectors of angles ${\angle ADP}$ and ${\angle PCB}$ and the perpendicular bisector of segment ${AB}$ .

Solution: The key to solving this problem is to understand that what is important is not the quadrilateral ${ABCD}$ , but the triangle ${PAB}$ . Also, note the interesting structure of the triangles ${DPA}$ and ${CPB}$ which have one angle which is three times the other. Also, note that the angle ${\angle DAP}$ is half the angle ${\angle ABP}$ . Therefore, it is possible to build a cyclic quadrilateral here, by considering the point ${X}$ on ${AD}$ such that ${\angle XPA = \angle DAP}$ . (make a figure to understand the notations).

Why is this construction useful for our problem? If you look now in the triangle ${XPD}$ you see that the angles ${\angle X}$ and ${\angle P}$ are equal. Therefore, the angle bisector of ${\angle ADP}$ is the perpendicular bisector of ${XP}$ , which goes through the circumcenter of the cyclic quadrilateral ${AXPB}$ . Note that this is exactly the circumcenter of ${ABP}$ .

Now, repeating the same argument in the triangle ${BPC}$ , we can see that the two angle bisectors in the problem all go through the circumcenter of the triangle ${APB}$ , which obviously lies on the perpendicular bisector of ${AB}$ .

As a conclusion, what helped solve this problem was understanding the structure of the triangles with one angle three times the other, and transforming the angle bisector into something more useful, like a perpendicular bisector which goes through a circumcenter.

Categories: Geometry, IMO, Olympiad, Problem Solving Tags: bisector, circumscribed, Geometry, IMO, triangle

IMO 2020 Problems

September 23, 2020 beni22sof Leave a comment

Problem 1. Consider the convex quadrilateral ${ABCD}$ . The point ${P}$ is in the interior of ${ABCD}$ . The following ratio equalities hold:

$\displaystyle \angle PAD : \angle PBA : \angle DPA = 1:2:3 = \angle CBP : \angle BAP : \angle BPC.$

Prove that the following three lines meet in a point: the internal bisectors of angles ${\angle ADP}$ and ${\angle PCB}$ and the perpendicular bisector of segment ${AB}$ .

Problem 2. The real numbers ${a,b,c,d}$ are such that ${a \geq b \geq c \geq d>0}$ and ${a+b+c+d=1}$ . Prove that

$\displaystyle (a+2b+3c+4d)a^ab^bc^cd^d<1.$

Problem 3. There are ${4n}$ pebbles of weights ${1,2,3,...,4n}$ . Each pebble is coloured in one of ${n}$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied:

The total weights of both piles are the same.
Each pile contains two pebbles of each colour.

Problem 4. There is an integer ${n>1}$ . There are ${n^2}$ stations on a slope of a mountain, all at different altitudes. Each of two cable car companies, ${A}$ and ${B}$ , operaters ${k}$ cable cars; each cable car provides a transfer from one of the stations to a higher one (with no intermediate stops). The ${k}$ cable cars of ${A}$ have ${k}$ different starting points and ${k}$ different finishing points, and a cable car which starts higher also finishes higher. The same conditions hold for ${B}$ . We say that two stations are linked by a company if one can start from the lower station and reach the higher one by using one or more cars of that company (no other movements between stations are allowed). Determine the smallest positive ${k}$ for which one can guarantee that there are two stations that are linked by both companies.

Problem 5. A deck of ${n>1}$ cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.

For which ${n}$ does it follow that the numbers on all cards are all equal?

Problem 6. Prove that there exists a positive constant ${c}$ such that the following statement is true:

Consider an integer ${n>1}$ , and a set ${\mathcal S}$ of ${n}$ points in the plane such that the distance between any two different points in ${\mathcal S}$ is at least ${1}$ . It follows that there is a line ${\ell}$ separating ${\mathcal S}$ such that the distance for any point of ${\mathcal S}$ to ${\ell}$ is at least ${c n^{1/3}}$ .

(A line ${\ell}$ separates a set of points ${\mathcal S}$ if some segment joining two points in ${\mathcal S}$ crosses ${\ell}$ .)

Note. Weaker results with ${cn^{1/3}}$ replaced with ${cn^\alpha}$ may be awarded points depending on the value of the constant ${\alpha>1/3}$ .

Source: imo-official.org

Categories: Algebra, Combinatorics, Geometry, IMO, Inequalities, Olympiad, Problem Solving Tags: Combinatorics, Geometry, IMO

IMO 2019 Problem 1

July 18, 2019 beni22sof 7 comments

Problem 1. Let ${\Bbb{Z}}$ be the set of integers. Determine all functions ${f:\Bbb Z \rightarrow \Bbb Z}$ such that, for all integers ${a}$ and ${b}$ ,

$\displaystyle f(2a)+2f(b) = f(f(a+b)).$

Solution: As usual with this kind of functional equations the first thing that comes into mind is to pick simple cases of ${a}$ and ${b}$ .

Categories: Algebra, IMO, Olympiad Tags: functional equation, IMO, olympiad

IMO 2019 – Problems

July 18, 2019 beni22sof Leave a comment

Problem 1. Let ${\Bbb{Z}}$ be the set of integers. Determine all functions ${f:\Bbb Z \rightarrow \Bbb Z}$ such that, for all integers ${a}$ and ${b}$ ,

$\displaystyle f(2a)+2f(b) = f(f(a+b)).$

Problem 2. In triangle ${ABC}$ , point ${A_1}$ lies on the side ${BC}$ and point ${B_1}$ lies on side ${AC}$ . Let ${P}$ and ${Q}$ be points on segments ${AA_1}$ and ${BB_1}$ , respectively, such that ${PQ}$ is parallel to ${AB}$ . Let ${P_1}$ be a point on line ${PB_1}$ such that ${B_1}$ lies strictly between ${P}$ and ${P_1}$ and ${\angle PP_1C=\angle BAC}$ . Similarly, let ${Q_1}$ be a point on line ${QA_1}$ such that ${A_1}$ lies strictly between ${Q}$ and ${Q_1}$ and ${\angle CQ_1Q = \angle CBA}$ .

Prove that points ${P,Q,P_1}$ and ${Q_1}$ are concyclic.

Problem 3. A social network has ${2019}$ users, some pairs of whom are friends. Whenever user ${A}$ is friends with user ${B}$ , user ${B}$ is also friends with user ${A}$ . Events of the following kind may happen repeatedly, one at a time:

Three users ${A,B,C}$ such that ${A}$ is friends with both ${B}$ and ${C}$ but ${B}$ and ${C}$ are not friends change their friendship statuses such that ${B}$ and ${C}$ are now friends, but ${A}$ is no longer friends with ${B}$ and no longer friends with ${C}$ . All other friendship statuses are unchanged.

Initially, ${1010}$ users have ${1009}$ friends each and ${1009}$ users haf ${1010}$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.

Problem 4. Find all pairs ${(k,n)}$ of positive integers such that

$\displaystyle k! = (2^n-1)(2^n-2)(2^n-4)...(2^n-2^{n-1}).$

Problem 5. The Bank of Bath issues coins with an ${H}$ on a side and a ${T}$ on the other. Harry has ${n}$ of these coins arranged in a line from left to right. He repeatedly performs the following operation: if there are exactly ${k>0}$ coins showing ${H}$ , then he turns over the ${k}$ th coin from the left; otherwise, all coins show ${T}$ and he stops. For example, if ${n=3}$ the process starting with the configuration ${THT}$ would be

$\displaystyle THT \rightarrow HHT \rightarrow HTT \rightarrow TTT,$

which stops after three operations.

(a) Show that, for each initial configuration, Harry stops after a finite number of operations.
(b) For each initial configuration ${C}$ , let ${L(C)}$ be the number of operation before Harry stops. For example ${L(THT)=3}$ and ${L(TTT)=0}$ . Determine the average value of ${L(C)}$ over all ${2^n}$ possible initial configurations ${C}$ .

Problem 6. Let ${I}$ be the incenter of the acute triangle ${ABC}$ with ${AB \neq AC}$ . The incircle ${\omega}$ of ${ABC}$ is tangent to sides ${BC,CA}$ and ${AB}$ at ${D,E,F}$ , respectively. The line through ${D}$ perpendicular to ${EF}$ meets ${\omega}$ again at ${R}$ . The line ${AR}$ meets ${\omega}$ again at ${P}$ . THe circumcircles of triangles ${PCE}$ and ${PBF}$ meet again at ${Q}$ .

Prove that lines ${DI}$ and ${PQ}$ meet on a line through ${A}$ , perpendicular to ${AI}$ .

Source: http://www.imo-official.org

Categories: Algebra, Combinatorics, Geometry, IMO, Number theory, Olympiad, Probability, Problem Solving Tags: Algebra, Combinatorics, Geometry, IMO

IMO 2018 Problems – Day 2

July 10, 2018 beni22sof Leave a comment

Problem 4. A site is any point ${(x, y)}$ in the plane such that ${x}$ and ${y}$ are both positive integers less than or equal to 20.

Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to ${\sqrt{5}}$ . On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone.

Find the greatest ${K}$ such that Amy can ensure that she places at least ${K}$ red stones, no matter how Ben places his blue stones.

Problem 5. Let ${a_1,a_2,\ldots}$ be an infinite sequence of positive integers. Suppose that there is an integer ${N > 1}$ such that, for each ${n \geq N}$ , the number

$\displaystyle \frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$

is an integer. Prove that there is a positive integer ${M}$ such that ${a_m = a_{m+1}}$ for all ${m \geq M}$ .

Problem 6. A convex quadrilateral ${ABCD}$ satisfies ${AB\cdot CD = BC\cdot DA}$ . Point ${X}$ lies inside ${ABCD}$ so that ${\angle{XAB} = \angle{XCD}}$ and ${\angle{XBC} = \angle{XDA}}$ . Prove that ${\angle{BXA} + \angle{DXC} = 180}$ .

Source: AoPS

Categories: Algebra, Geometry, Olympiad Tags: Combinatorics, Geometry, IMO

IMO 2018 Problems – Day 1

July 9, 2018 beni22sof Leave a comment

Problem 1. Let ${\Gamma}$ be the circumcircle of acute triangle ${ABC}$ . Points ${D}$ and ${E}$ are on segments ${AB}$ and ${AC}$ respectively such that ${AD = AE}$ . The perpendicular bisectors of ${BD}$ and ${CE}$ intersect minor arcs ${AB}$ and ${AC}$ of ${\Gamma}$ at points ${F}$ and ${G}$ respectively. Prove that lines ${DE}$ and ${FG}$ are either parallel or they are the same line.

Problem 2. Find all integers ${n \geq 3}$ for which there exist real numbers ${a_1, a_2, \dots a_{n + 2}}$ satisfying ${a_{n + 1} = a_1}$ , ${a_{n + 2} = a_2}$ and

$\displaystyle a_ia_{i + 1} + 1 = a_{i + 2}$

For ${i = 1, 2, \dots, n}$ .

Problem 3. An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from ${1}$ to ${10}$

$\displaystyle 4$

$\displaystyle 2\quad 6$

$\displaystyle 5\quad 7 \quad 1$

$\displaystyle 8\quad 3 \quad 10 \quad 9$

Does there exist an anti-Pascal triangle with ${2018}$ rows which contains every integer from ${1}$ to ${1 + 2 + 3 + \dots + 2018}$ ?

Source: AoPS.

Categories: Geometry, IMO, Olympiad, Uncategorized Tags: Geometry, IMO, olympiad, Problem Solving

IMO 2016 – Problem 1

July 26, 2016 beni22sof Leave a comment

IMO 2016, Problem 1. Triangle ${BCF}$ has a right angle at ${B}$ . Let ${A}$ be the point on line ${CF}$ such that ${FA=FB}$ and ${F}$ lies between ${A}$ and ${C}$ . Point ${D}$ is chosen such that ${DA=DC}$ and ${AC}$ is the bisector of ${\angle DAB}$ . Point ${E}$ is chosen such that ${EA=ED}$ and ${AD}$ is the bisector of ${\angle EAC}$ . Let ${M}$ be the midpoint of ${CF}$ . Let ${X}$ be the point such that ${AMXE}$ is a parallelogram (where ${AM || EX}$ and ${AE || MX}$ ). Prove that the lines ${BD, FX}$ and ${ME}$ are concurrent.

Categories: IMO, Olympiad, Problem Solving, Uncategorized Tags: Geometry, IMO

IMO 2015 Problem 1

July 10, 2015 beni22sof Leave a comment

Problem 1. We say that a finite set ${\mathcal{S}}$ of points in the plane is balanced if, for any two different points ${A}$ and ${B}$ in ${\mathcal{S}}$ , there is a point ${C}$ in ${\mathcal{S}}$ such that ${AC=BC}$ . We say that ${\mathcal{S}}$ is center-free if for any three different points ${A}$ , ${B}$ and ${C}$ in ${\mathcal{S}}$ , there is no points ${P}$ in ${\mathcal{S}}$ such that ${PA=PB=PC}$ .

(a) Show that for all integers ${n\ge 3}$ , there exists a balanced set having ${n}$ points.

(b) Determine all integers ${n\ge 3}$ for which there exists a balanced center-free set having ${n}$ points.

Problem 2. Find all triples of positive integers ${(a, b, c)}$ such that ${ab-c, bc-a, ca-b}$ are all powers of 2.

Problem 3. Let ${ABC}$ be an acute triangle with ${AB > AC}$ . Let ${\Gamma }$ be its cirumcircle., ${H}$ its orthocenter, and ${F}$ the foot of the altitude from ${A}$ . Let ${M}$ be the midpoint of ${BC}$ . Let ${Q }$ be the point on ${ \Gamma }$ such that ${\angle HQA }$ and let ${K }$ be the point on ${\Gamma }$ such that ${\angle HKQ }$ . Assume that the points ${A,B,C,K }$ and ${Q }$ are all different and lie on ${\Gamma}$ in this order.

Prove that the circumcircles of triangles ${KQH }$ and ${FKM }$ are tangent to each other.

Source: AoPS

Categories: Olympiad Tags: Geometry, IMO, number theory

IMO 2014 Problem 6

July 9, 2014 beni22sof 1 comment

A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large ${n}$ , in any set of ${n}$ lines in general position it is possible to colour at least ${\sqrt{n}}$ lines blue in such a way that none of its finite regions has a completely blue boundary.

Note: Results with ${\sqrt{n}}$ replaced by ${c\sqrt{n}}$ will be awarded points depending on the value of the constant ${c}$ .

IMO 2014 Problem 6 (Day 2)

Categories: Combinatorics, Geometry, IMO, Olympiad Tags: Combinatorics, Geometry, IMO

Older Entries

Beni Bogoşel's blog

Archive

IMO 2023 Problem 2

IMO 2023 Problem 4

IMO 2023 Problem 1

Problems of the International Mathematical Olympiad 2023

Build three particular equal segments in a triangle

IMO 2022 – Problem 4 – Geometry

IMO 2022 – Problem 2 – Non-standard functional equation

IMC 2022 – Day 1 – Problem 1

Function with zero average on vertices of all regular polygons

Balkan Mathematical Olympiad 2022

IMO 2020 – Problem 2

IMO 2020 – Problem 1

IMO 2020 Problems

IMO 2019 Problem 1

IMO 2019 – Problems

IMO 2018 Problems – Day 2

IMO 2018 Problems – Day 1

IMO 2016 – Problem 1

IMO 2015 Problem 1

IMO 2014 Problem 6

Beni Bogoşel

Blog Stats

Today’s Visitors

Stay connected

Blogroll

Archive

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Beni Bogoşel

Blog Stats

Today’s Visitors

Stay connected

Blogroll