Olympiad | Beni Bogoşel's blog

An inequality involving complex numbers

March 19, 2024 beni22sof Leave a comment

Consider ${n\geq 1}$ and ${n}$ complex numbers ${z_1,...,z_n \in \Bbb C}$ . Show that

$\displaystyle \sum_{k =1}^n |z_k||z-z_k|\geq \sum_{k=1}^n |z_k|^2, \text{ for every } z \in \Bbb{C},$

if and only if ${z_1+...+z_n = 0}$ .

Proposed by Dan Stefan Marinescu in the Romanian Mathematical Gazette

Solution: For someone familiar with optimality conditions in multi-variable calculus this is straightforward. Notice that

$\displaystyle f(z) = \sum_{k =1}^n |z_k||z-z_k|$

is a convex function (linear combination of distances in the plane). The inequality is equivalent to ${f(z) \geq f(0)}$ , which means that ${0}$ is the global minimum of the function.

For a convex, differentiable function global minimality is equivalent to verifying first order optimality conditions. Denoting ${z = x+iy}$ , ${z_k = x_k+iy_k}$ the partial derivatives of the function ${f}$ with respect to ${x,y}$ are

$\displaystyle \frac{\partial f}{\partial x}(x,y) = \sum_{k=1}^n \sqrt{x_k^2+y_k^2}\frac{x-x_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}},$

$\displaystyle \frac{\partial f}{\partial y}(x,y) = \sum_{k=1}^n \sqrt{x_k^2+y_k^2}\frac{y-y_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}}.$

If ${\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0)}$ then ${\sum x_k=\sum y_k = 0}$ and the conclusion follows. The converse also holds, obviously.

Since this problem was proposed for 10th grade, let’s use some simpler arguments to arrive to a proof. Denote by ${g = \frac{1}{n}(z_1+...+z_n)}$ . A quick computation using properties of the modulus gives:

$\displaystyle \sum_{k=1}^n |z-z_k|^2 = n|z-g|^2+\sum_{i=1}^n |z_k|^2$

Thus ${\sum_{k=1}^n |g-z_k|^2 = \sum_{i=1}^n |z_k|^2}$ . Of course, the classical inequality ${a^2+b^2 \geq 2ab}$ implies

$\displaystyle 2\sum_{k=1}^n |z_k|^2 = \sum_{k=1}^n |g-z_k|^2+\sum_{i=1}^n |z_k|^2\geq 2 \sum_{k=1}^n |z_k||g-z_k|.$

If the inequality in the statement of the problem holds, the above relation becomes an equality and ${|z_k|=|g-z_k|}$ for all ${k=1,...,n}$ . Therefore points ${z_k}$ belong to the mediatrix of the segment ${0g}$ . Therefore the centroid ${g}$ also belongs to this mediatrix and to ${0g}$ , which implies ${g=0}$ , as requested.

Conversely, if ${z_1+...+z_k = 0}$ consider the inequality

$\displaystyle |a||b| \geq \frac{1}{2}(\overline a b + a\overline b)$ to conclude.

Categories: Complex analysis, Geometry, Inequalities, Olympiad Tags: complex, Geometry, inequality, olympiad

Romanian Regional Olympiad 2024 – 10th grade

March 12, 2024 beni22sof Leave a comment

Problem 1. Let ${a,b \in \Bbb{R}}$ , ${a>1}$ , ${b>0}$ . Find the smallest real number ${\alpha}$ such that

$\displaystyle (a+b)^x \geq a^x+b, \forall x \geq \alpha.$

Problem 2. Consider ${ABC}$ a triangle inscribed in the circle ${\mathcal C}$ of center ${O}$ and radius ${1}$ . For any ${M \in \mathcal C\setminus \{A,B,C\}}$ denote by ${S(M) = OH_1^2+OH_2^2+OH_3^2}$ where ${H_1,H_2,H_3}$ are the orthocenters of the triangles ${MAB,MBC,MCA}$ , respectively.

a) Prove that if the triangle ${ABC}$ is equilateral then ${s(M)=6}$ for every ${M \in \mathcal C \setminus \{A,B,C\}}$ .

b) Show that if there exist three distinct points ${M_1,M_2,M_3 \in \mathcal C\setminus \{A,B,C\}}$ such that ${s(M_1)=s(M_2)=s(M_3)}$ , then the triangle ${ABC}$ is equilateral.

Problem 3. Let ${a,b,c}$ be three non-zero complex numbers with the same modulus for which ${A=a+b+c}$ and ${B=abc}$ are real numbers. Show that for every positive integer ${n}$ the number ${C_n = a^n+b^n+c^n}$ is real.

Problem 4. Let ${n \in \Bbb N^*}$ . Determine all functions ${f:\Bbb{R} \rightarrow \Bbb{R}}$ which verify

$\displaystyle f(x+y^{2n})=f(f(x))+y^{2n-1}f(y),$

for every ${x,y \in \Bbb{R}}$ and such that ${f(x)=0}$ has a unique solution.

Hints:

Problem 1: study the monotonicity of the function ${g(x)= (a+b)^x-a^x}$ . Then observe that the inequality is equivalent to ${g(x) \geq g(1)}$ .

Problem 2: Recall the identity ${OH^2 = 9R^2-AB^2-BC^2-CA^2}$ whenever ${H}$ is the orthocenter of ${ABC}$ with circumcenter ${O}$ . This can be proved using complex numbers and recalling that ${OH = 3OG}$ , where ${G}$ is the center of gravity. Therefore

$\displaystyle OH_1^2 = 9-MA^2-MB^2-AB^2$ and the analogue equalities. Summing we get

$\displaystyle s(M) = 27-2\sum MA^2-\sum AB^2.$ a) When ${ABC}$ is equilateral and inscribed in a circle of radius ${1}$ we have ${AB=BC=CA=\sqrt{3}}$ . Moreover, the identity In particular, one can prove the following:

$\displaystyle AM^2+BM^2+CM^2 = AG^2+BG^2+CG^2+3MG^2$ applied to the triangle equilateral triangle ${ABC}$ with centroid ${O}$ shows that

$\displaystyle MA^2+MB^2+MC^2 = AO^2+BO^2+CO^2+3MO^2=6.$ Thus

$\displaystyle s(M) = 27-12-9 = 6.$ b) Assume there exist three distinct points such that ${s(M_1)=s(M_2)=s(M_3)}$ . This implies that

$\displaystyle \sum M_1A^2 = \sum M_2A^2 = \sum M_3A^2.$ The relation above concerning the center of gravity of ${ABC}$ shows that ${M_1G=M_2G=M_3G}$ . Since the points are distinct, it follows that ${G}$ coincides with ${O}$ , the circumcenter of ${ABC}$ , therefore ${ABC}$ is equilateral.

Problem 3: Denote ${r>0}$ the common value of the modulus: ${|a|=|b|=|c|=r}$ . Then

$\displaystyle \overline{a+b+c} = r^2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = r^2 \frac{ab+bc+ca}{abc}.$ Since ${r, a+b+c, abc \in \Bbb{R}}$ it follows that ${ab+bc+ca\in \Bbb{R}}$ . Then of course ${a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) \in \Bbb{R}}$ . Finally, we know that ${a,b,c}$ are roots of

$\displaystyle (z-a)(z-b)(z-c)=0 \Longleftrightarrow z^3-(a+b+c)z^2+(ab+bc+ca)z-abc=0.$ Since the coefficients of this polynomial are real, an inductive argument shows that if ${C_n, C_{n+1}, C_{n+2}}$ are real then ${C_{n+3}}$ is real, finishing the proof.

Problem 4. Take ${y=0}$ and get ${f(x) = f(f(x))}$ . Thus, ${f}$ is the identity mapping on its image!! Take ${y\mapsto -y}$ and observe that ${y^{2n-1}f(y) = -y^{2n-1}f(-y)}$ . Therefore ${f(-y)=-f(y)}$ for any ${y \neq 0}$ . Since the equation ${f(x)=0}$ has a unique solution, it follows that ${f(0)=0}$ and ${f(x) \neq 0}$ for ${x \neq 0}$ . Take ${x=0}$ and get ${f(y^{2n}) = y^{2n-1}f(y)}$ . Therefore

$\displaystyle f(x+y^{2n})=f(x)+f(y^{2n})$ for any ${x,y}$ . Since ${y^{2n}}$ takes all positive values in ${\Bbb{R}}$ it follows that

$\displaystyle f(x+y) = f(x)+f(y)$ for every ${x\in \Bbb{R}}$ , ${y \geq 0}$ . Coupled with ${f(-y)=-f(y)}$ this implies that

$\displaystyle f(x+y)= f(x)+f(y).$ for all ${x,y}$ . If ${f(x_1)=f(x_2)}$ it follows that ${f(x_1-x_2)=0}$ therefore ${x_1=x_2}$ . From ${f(f(x))=f(x)}$ we obtain ${f(x)=x}$ for all reals ${x}$ . It should be noted that this function obviously verifies the functional equation!

Categories: Algebra, Geometry, Olympiad Tags: Algebra, complex, Geometry, olympiad

Putnam 2023 A1-A3

February 15, 2024 beni22sof Leave a comment

A1. For a positive integer ${n}$ , let ${f_n(x) = \cos(x) \cos(2x) \cos(3x) \cdots \cos(nx)}$ . Find the smallest ${n}$ such that ${|f_n''(0)| > 2023}$ .

Hints: Observe that

$\displaystyle f_n'(x) = -f_n(x)\sum_{k=1}^n k \tan(kx).$

Differentiate again and observe that

$\displaystyle f_n''(x) = -f_n'(x)\sum_{k=1}^n k \tan(kx)+f_n(x) \sum_{k=1}^n k^2 (1+\tan^2(kx)).$

It is straightforward now to evaluate ${f_n''(0)}$ and to answer the question.

A2. Let ${n}$ be an even positive integer. Let ${p}$ be a monic, real polynomial of degree ${2n}$ ; that is to say, ${p(x) = x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0}$ for some real coefficients ${a_0, \dots, a_{2n-1}}$ . Suppose that ${p(1/k) = k^2}$ for all integers ${k}$ such that ${1 \leq |k| \leq n}$ . Find all other real numbers ${x}$ for which ${p(1/x) = x^2}$ .

Hints: Denote ${g(y) = p(y)y^2}$ . Then ${y \mapsto g(y)}$ is a polynomial of degree ${2n+2}$ . Moreover, ${g(1/k) = 1}$ for ${1\leq |k|\leq n}$ , showing that ${g-1}$ has the ${2n}$ roots ${\pm \frac{1}{k}}$ .

It follows that ${g(y) = 1+\prod_{k=1}^n (y-\frac{1}{k})(y+\frac{1}{k}) \times (ay^2+by+c)}$ . Identifying coefficients one finds ${a=1}$ , ${b=0}$ and the value of ${c}$ is obtained by observing that the constant term in ${g}$ should be zero. Then the answer to the question is obtained by investigating the roots of ${ay^2+by+c}$ . I guess a distinction should be made between the cases ${n}$ even or odd, since it seems that the sign of ${c}$ depends on that.

A3. Determine the smallest positive real number ${r}$ such that there exist differentiable functions ${f\colon \mathbb{R} \rightarrow \mathbb{R}}$ and ${g\colon \mathbb{R} \rightarrow \mathbb{R}}$ satisfying

(a) ${f(0) > 0}$ ,
(b) ${g(0) = 0}$ ,
(c) ${|f'(x)| \leq |g(x)|}$ for all ${x}$ ,
(d) ${|g'(x)| \leq |f(x)|}$ for all ${x}$ , and
(e) ${f(r) = 0}$ .

Hints: Of course, an example of functions ${f,g}$ are ${\cos}$ and ${\sin}$ . This suggests that the answer could be ${r = \pi/2}$ . In any case, this example shows that the smallest ${r}$ verifies ${r \leq \pi/2}$ .

Assuming that ${r}$ is the smallest zero, on ${[0,r]}$ we have ${f(t)\geq 0}$ . Now let’s try to obtain some differential inequality using the hypothesis:

$\displaystyle f'(x)+\int_0^x f(t)dt \geq -|g(x)|+\int_0^x |g'(t)|dt \geq 0,$

where the last inequality follows from ${g(x) = \int_0^x g'(t)}$ .

Therefore ${F(x) = \int_0^x f(t)dt-f(0)\sin x}$ verifies ${F(0)=0, F'(0)=0}$ and

$\displaystyle F''(x)+F(x) \geq 0.$

Therefore we have ${F''(x)+F(x) = q(x)\geq 0}$ on ${[0,r]}$ with ${F'(0)=F(0)=0}$ . The general solution is

$\displaystyle F(x) = A\cos x+B\sin x+\int_0^x q(t)\sin (x-t)dt.$

The initial conditions show that ${A=B=0}$ . Therefore

$\displaystyle F(x) = \int_0^x q(t)\sin(x-t)dt, x\in [0,r]$

and since ${r\leq \pi/2}$ we have ${F(x) \geq 0}$ on ${[0,r]}$ . Moreover ${F'(x) = \int_0^x q(t)\cos(x-t)dt}$ which is also non-negative on ${[0,r]}$ . This implies

$\displaystyle f(x)-f(0)\cos x\geq 0, x \in [0,r].$

Thus the smallest ${r}$ such that ${f(r)=0}$ is indeed ${r = \pi/2}$ .

Categories: Algebra, Analysis, Inequalities, Olympiad, Problem Solving, Undergraduate Tags: Analysis, inequality, putnam

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 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

Polygon with an odd number of sides

July 6, 2023 beni22sof Leave a comment

Let ${A_1...A_n}$ be a convex polygon. Suppose there exists a point ${P}$ inside the polygon such that each one of the segments ${PA_i}$ , ${i=1,...,n}$ intersects exactly one side of the polygon in its interior. Prove that ${n}$ is odd.

Alternative reformulation: Let ${A_1...A_{2n}}$ be a convex polygon with an even number of sides and ${P}$ be an interior point. Then there exist two of the segments ${PA_1,...,PA_{2n}}$ which intersect the same side.

Romanian Team Selection test 2007

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

A series involving a multiplicative function

May 24, 2023 beni22sof 8 comments

Let ${f:\Bbb{N}^* \rightarrow \Bbb{N}^*}$ be a function with the property that ${f(m\cdot n) = f(n)f(m)}$ for any ${n,m\in \Bbb N^*}$ . Moreover, suppose that ${f(p_n)=n+1}$ for all ${n \geq 1}$ , where ${p_n}$ is the ${n}$ -th prime number.

Prove that

$\displaystyle \sum_{n=1}^\infty \frac{1}{f(n)^2} = 2.$

I posted this problem quite a while back. Here is a solution I found recently.

Categories: Analysis, Olympiad, Problem Solving, Real Analysis Tags: Analysis, convergence, multiplicative, series

Maximize the circumradius of a unit diameter simplex

May 22, 2023 beni22sof Leave a comment

Given a simplex in dimension ${d}$ with diameter at most equal to one, what is its largest possible circumradius? The question is well posed, since a ball of diameter ${2}$ will cover all the points (a ball of radius ${1}$ centered at one of the given points).

Consider ${d+1}$ points in dimension ${d}$ and suppose the origin is the center of the circumscribed ball. Then given two vertices ${x_i,x_j}$ , the circumradius verifies ${|x_i-x_j|^2 = |x_i|^2+|x_j|^2-2x_i\cdot x_j = R^2(2-2\cos \widehat{(x_i,x_j)})}$ . Therefore, one should maximize the minimal cosine of angles of two such vectors.

An immediate upper bound, which is tight can be obtained from the identity ${|x_1+...+x_{d+1}|^2 = (d+1)R^2+2R^2 \sum_{i<j} \cos \widehat{x_i,x_j}}$ . It can be seen that ${\min \cos \widehat{x_i,x_j} \leq -\frac{d+1}{2} / {d+1\choose 2}=-\frac{1}{d}}$ . Thus the maximal circumradius verifies

$\displaystyle R^2 \leq \frac{1}{2+\frac{2}{d}}= \frac{d}{2d+2}.$ This coincides with the case where all angles between different vectors are equal and their cosine is ${-1/d}$ , i.e. the simplex is regular.

Categories: Calculus of Variations, Computational Geometry, Olympiad, Optimization, Problem Solving, shape optimization Tags: circumcircle, diameter, jung, optimization

Some properties of constant width shapes in the plane

March 17, 2023 beni22sof 1 comment

Shapes of constant width in the plane have the property that they are convex and any pair of parallel tangent or supporting lines are at a fixed constant width apart. The circle is the most obvious example of shapes having constant width. However, many more such shapes exist.

Feynman recalls in his book What Do You Care What Other People Think? that the Challenger disaster might be caused by the fact that the section of the booster rockets may not have been perfectly circular. Check out this link for more details. The procedure to check for roundness was to measure the diameter of the cylinder at different angles around the tank. As you may imagine, if the tank’s section has non-circular constant width, this technique does not detect anything wrong.

The most famous examples of constant width shapes are the Reuleaux triangle and more generally, Reuleaux polygons in general. These shapes are not just mathematical curiosities, but have various applications. Reuleaux triangles are used in the rotary Wankel engine design and square drilling machines, while the twenty pence British coin is a Reuleaux heptagon.

Applications of Helly’s theorem

March 12, 2023 beni22sof Leave a comment

Prove that if the plane can be covered with $n \geq 3$ half planes then there exist three of these which also cover the plane.
On a circle consider a finite set of arcs which do not cover the circle, such that any two of them have non-void intersection. Show that all arcs have a common points. If the arcs cover the circle does the conclusion still hold?
Consider $n \geq 3$ half circles which cover the whole circle. Show that we can pick three of them which still cover the circle.

I’ll not provide the solutions for now. The title should be a strong indication towards finding a solution!

Categories: Combinatorics, Geometry, Olympiad Tags: convex, Geometry, helly, Problem Solving

Helly’s Theorem

March 10, 2023 beni22sof Leave a comment

Helly’s theorem. Let ${n\geq 4}$ convex figures be given in the plane and suppose each three of them have a common point. Prove that all ${n}$ figures have a common point.

Can the convexity hypothesis be removed?

Categories: Combinatorics, Geometry, Olympiad, Problem Solving Tags: convex, Geometry, helly

Inscribed squares: Symmetric case

January 19, 2023 beni22sof Leave a comment

I recently heard about the “inscribed square” problem which states that on any closed, continuous curve without self intersections there should exist four points which are the vertices of a (non-degenerate) square. This problem, first raised by Otto Toeplitz in 1911, is still unsolved today. Many particular cases are solved (convex curves, smooth curves, etc.). Nevertheless, things quickly get complicated as one gets close to the original problems where only continuity is assumed.

I will show below a quick argument for a very particular case. Suppose the curve $C$ is symmetric with respect to the origin. Since the curve is simple, the origin does not lie on the curve. Consider the curve $C'$ , rotated with $90$ degrees around the origin. Then curves $C,\ C'$ must intersect. Indeed, by continuity, there exists a point $A$ which is closest to the origin on the curve $C$ and a point $B$ furthest away from the origin on the same curve. The rotated curve has the same minimal and maximal distances to the origin. Therefore, point $A$ lies in the interior of $C'$ (or on the boundary) and point $B$ lies outside of $C'$ (or on the boundary). In either case, going from $A$ to $B$ on $C$ we must intersect the curve $C'$ .

If $X$ is a point of intersection, it lies both on $C$ and on its rotation with $90$ degrees. Moreover, the symmetric points also lie on $C$ by symmetry. Therefore, we have four points at equal distance from the origin, for which the rays from the origin form equal angles. These points are the vertices of a square!

Categories: Geometry, IMO, Olympiad, Optimization, puzzles Tags: continuity, Geometry, square

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

Geometry problem: a property of the 40-40-100 triangle

December 5, 2022 beni22sof 1 comment

Consider the triangle ${ABC}$ with ${\angle A = 100^\circ}$ and ${\angle B = \angle C = 40^\circ}$ . Consider ${D \in AC}$ such that ${\angle CBD = 10^\circ}$ . Prove that ${AD = BC}$ .

This is a surprising result and easy to remember. I first saw this back in the days I was solving olympiad problems. The 40-40-100 triangle has this nice characteristic property. There is a very nice geometric proof and a one liner trigonometry argument.

Categories: Geometry, IMO, Olympiad Tags: angle chasing, Geometry, olympiad

Geometry problem: a property of the 30-70-80 triangle

December 4, 2022 beni22sof Leave a comment

Consider a triangle ${XYZ}$ with angles ${\angle X = 70^\circ}$ , ${\angle Y = 30^\circ}$ , ${\angle Z = 80^\circ}$ . Let ${ZT}$ be the bisector of the angle ${\angle XZY}$ . Consider ${U \in XZ}$ such that ${UT || YZ}$ . Prove that the triangle ${YUZ}$ is isosceles.

Categories: Geometry, IMO, Olympiad Tags: angle chasing, Geometry, trigonometry

Minimal volume of some particular convex sets in the cube

November 21, 2022 beni22sof Leave a comment

Consider a cube and a convex body $K$ contained in it such that the projections of this body on every face of the cube completely covers that face. Prove that the volume of the convex body is at least equal with a third of the volume of the cube.

Categories: Calculus of Variations, Combinatorics, Geometry, IMO, Inequalities, Olympiad, Optimization Tags: convex, cube, minimal, olympiad, optimization, volume

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.