An inequality involving complex numbers

Consider and complex numbers . Show that

if and only if .

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

is a convex function (linear combination of distances in the plane). The inequality is equivalent to , which means that is the global minimum of the function.

For a convex, differentiable function global minimality is equivalent to verifying first order optimality conditions. Denoting , the partial derivatives of the function with respect to are

If then 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 . A quick computation using properties of the modulus gives:

Thus . Of course, the classical inequality implies

If the inequality in the statement of the problem holds, the above relation becomes an equality and for all . Therefore points belong to the mediatrix of the segment . Therefore the centroid also belongs to this mediatrix and to , which implies , as requested.

Conversely, if consider the inequality

to conclude.

Romanian Regional Olympiad 2024 – 10th grade

Problem 1. Let , , . Find the smallest real number such that

Problem 2. Consider a triangle inscribed in the circle of center and radius . For any denote by where are the orthocenters of the triangles , respectively.

a) Prove that if the triangle is equilateral then for every .

b) Show that if there exist three distinct points such that , then the triangle is equilateral. 

Problem 3. Let be three non-zero complex numbers with the same modulus for which and are real numbers. Show that for every positive integer the number is real. 

Problem 4. Let . Determine all functions which verify

for every and such that has a unique solution. 

Hints:

Problem 1: study the monotonicity of the function . Then observe that the inequality is equivalent to . 

Problem 2: Recall the identity whenever is the orthocenter of with circumcenter . This can be proved using complex numbers and recalling that , where is the center of gravity. Therefore

and the analogue equalities. Summing we get

a) When is equilateral and inscribed in a circle of radius we have . Moreover, the identity In particular, one can prove the following:

applied to the triangle equilateral triangle with centroid shows that

Thus

b) Assume there exist three distinct points such that . This implies that

The relation above concerning the center of gravity of shows that . Since the points are distinct, it follows that coincides with , the circumcenter of , therefore is equilateral.

Problem 3: Denote the common value of the modulus: . Then

Since it follows that . Then of course . Finally, we know that are roots of

Since the coefficients of this polynomial are real, an inductive argument shows that if are real then is real, finishing the proof. 

Problem 4. Take and get . Thus, is the identity mapping on its image!! Take and observe that . Therefore for any . Since the equation has a unique solution, it follows that and for . Take and get . Therefore

for any . Since takes all positive values in it follows that

for every , . Coupled with this implies that

for all . If it follows that therefore . From we obtain for all reals . It should be noted that this function obviously verifies the functional equation!

Algebraic proof of the Finsler-Hadwiger inequality

Weitzenbock’s inequality states that if are the side lengths of a triangle with area then

A strengthening of this result due to Finsler and Hadwiger states

A variety of proofs rely on various trigonometric or geometric arguments. Below you can find a purely algebraic argument based on the classical characterization: are side lengths of a triangle if and only if there exist such that , , . If are strictly positive then the triangle will be non-degenerate.

Replacing with the above formulas replaces an inequality in a triangle with a general inequality where only positivity of the variables is involved. With this substitution, using classical notation for cyclic sums gives

and

On the other hand the area given by Heron’s formula is

Thus, Weitzenbock’s inequality is equivalent to

and the Finsler-Hadwiger inequality is equivalent to

This inequality follows at once, since squaring both sides gives

which is a well known consequence of

Equality holds, of course, if and only if . If the triangle is non-degenerate then it must be equilateral. Thus, Weitzenbock and Finsler-Hadwiger inequalities follow at once from classical inequalities, once the side lengths of a triangle are replaced with unconstrained variables.

A proof of the Hadwiger Finsler inequality

The Hadwiger-Finsler inequality states that if are the side lengths of a triangle with area then

This was discussed previously on the blog. This post shows a translation of the original paper by Hadwiger and Finsler and this post shows a surprising geometrical proof.

Various proofs of the inequality are known. However, since an equality always beats an inequality, let us prove the identity

It is immediate to see that Jensen’s inequality applied to the tangent function, which is convex on is enough to deduce the Hadwiger-Finsler inequality from the above identity. To prove the identity, simply compute

Replacing the usual formula gives

Summing these identities for the three angles gives precisely the desired result. The same proof can be found, for example, here.

Area of a spherical rectangle

A spherical rectangle is a spherical geodesic quadrilateral whose vertices form an Euclidean rectangle. In other words, the opposite edges are equal and all angles are equal. Suppose the side lengths of pairs of opposite sides are known. Show that the area of the rectangle is given by

IMO 2023 Problem 2

Problem 2. Let be an acute-angled triangle with . Let be the circumcircle of . Let be the midpoint of the arc of containing . The perpendicular from to meets at and meets again at . The line through parallel to meets line at . Denote the circumcircle of triangle by . Let meet again at .

Prove that the line tangent to at meets line on the internal angle bisector of .

Solution: Let us first do some angle chasing. Since we have and since is cyclic we have . Therefore, if we have . Therefore the arcs and are equal.

Denote by the midpoint of the short arc of . Then is the angle bisector of . Moreover, and are symmetric with respect to and is a diameter in .

Let us denote . It is straightforward to see that , and since we have .

Moreover, . Considering , since we find that is the midpoint of . Then, since in we find that is the midpoint of . But , since is a diameter. It follows that .

On the other hand, , showing that is cyclic and is tangent to the circle circumscribed to . 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 and are concurrent, where is the symmetric of with respect to .

Problems of the International Mathematical Olympiad 2023

Problem 1. Determine all composite integers that satisfy the following property: if are all the positive divisors of with , then divides for every . 

Problem 2. Let be an acute-angled triangle with . Let be the circumcircle of . Let be the midpoint of the arc of containing . The perpendicular from to meets at and meets again at . The line through parallel to meets line at . Denote the circumcircle of triangle by . Let meet again at . Prove that the line tangent to at meets line on the internal angle bisector of . 

Problem 3. For each integer , determine all infinite sequences of positive integers for which there exists a polynomial of the form , where are non-negative integers, such that

for every integer . 

Problem 4. Let be pairwise different positive real numbers such that

is an integer for every Prove that  

Problem 5. Let be a positive integer. A Japanese triangle consists of circles arranged in an equilateral triangular shape such that for each , , , , the row contains exactly circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 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 , along with a ninja path in that triangle containing two red circles.

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

Problem 6. Let be an equilateral triangle. Let be interior points of such that , , , and

Let and meet at let and meet at and let and meet at Prove that if triangle is scalene, then the three circumcircles of triangles and 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

Polygon with an odd number of sides

Let be a convex polygon. Suppose there exists a point inside the polygon such that each one of the segments , intersects exactly one side of the polygon in its interior. Prove that is odd.

Alternative reformulation: Let be a convex polygon with an even number of sides and be an interior point. Then there exist two of the segments which intersect the same side.

Romanian Team Selection test 2007

The Meissner Tetrahedra

Constant width shapes have the same width in every direction. I already underlined here that the disk is not the only such shape in dimension two. Moreover, the Reuleaux triangle is extremal for many geometric quantities. See this paper for a recent development in this direction.

In dimension three, however, the body minimizing the volume is still unknown. Nevertheless, a conjecture is available saying that the Meissner bodies are optimal. Numerical evidence seems to suggest this is the case, however, no proof exists for now. See the paper “Meissner’s Mysterious Bodies” by Bernd Kawohl & Christof Weber on the subject.

The construction of the Meissner bodies is quite interesting and although it is well known, the references where a complete description of the procedure is given are quite scarce. Moreover, pictures showing the explicit procedure are also rare (see for example the book “Bodies of Constant width” by Martini, Montejano, Oliveros). The complete description is shown in the book “Convex Figures” by Yaglom and Boltyanskii.

I will show a few pictures and try to give a few informal details on the construction. In dimension two, the Reuleaux triangle is the intersection of three disks of radius one situated at the vertices of an equilateral triangle of edge length one. This is enough to obtain a shape of constant width one.

Given a regular tetrahedron of edge length one, the intersection of balls of radius one situated at the vertices of the tetrahedron is not a constant width body. Indeed, it is enough to compute the distance between two midpoints of opposing arcs to find two points at distance strictly greater than one. In order to obtain a constant width body each pair of opposite edges should be “smoothed” in the sense that it should be replaced with another type of surface. The difference between smoothed and non-smoothed edges is shown below.

The smoothed parts for a side AB are intersection of spheres of radius one with centers C belonging on the arc opposite to AB. These parts form a surface of revolution of a circle of radius one around the side AB. The resulting bodies are of two types: the smoothed edges meet at a vertex or form a triangle. Both resulting bodies have the same volume and the figures below will make this aspect clearer. I will come back with more details regarding the computation of the volume and the surface area for the Meissner bodies in a following post.

Applications of Helly’s theorem

1. Prove that if the plane can be covered with half planes then there exist three of these which also cover the plane.
2. 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?
3. Consider 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!

Helly’s Theorem

Helly’s theorem. Let convex figures be given in the plane and suppose each three of them have a common point. Prove that all figures have a common point.

Can the convexity hypothesis be removed? 

Computing Restricted Voronoi Cells with Geogram

Given a shape and a family of points in , the Voronoi diagram associated to this set of points consists in a partition of such that each cell contains points closest to the current point than to any other point. Efficient algorithms exist for computing Voronoi diagrams, however in common implementations, the Voronoi cells are not clipped to a bounded region. Indeed, cells corresponding to a “boundary” point among the points considered will be infinite, in this case. Clipping to a bounded region is not difficult, but might require some careful coding.

The software Geogram (https://github.com/BrunoLevy/geogram) has a routine for building the clipped Voronoi diagrams. It gets as inputs the Voronoi points and a triangulation of the bounding box . The reason behind this choice is probably motivated by the existence of efficient clipping algorithm for intersections between polygons and a triangle. Below I show how Geogram can be called from Matlab in a basic situation where is a square. I tested this on a linux system. Keep in mind that Geogram needs to be installed on the machine prior to launching this code.

The code is tested and works very well for thousands of Voronoi cells. The computation in Geogram is really fast. Most of the time is the post-processing in Matlab and the input-output stage.

Inscribed squares: Symmetric case

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 is symmetric with respect to the origin. Since the curve is simple, the origin does not lie on the curve. Consider the curve , rotated with degrees around the origin. Then curves must intersect. Indeed, by continuity, there exists a point which is closest to the origin on the curve and a point furthest away from the origin on the same curve. The rotated curve has the same minimal and maximal distances to the origin. Therefore, point lies in the interior of (or on the boundary) and point lies outside of (or on the boundary). In either case, going from to on we must intersect the curve .

If is a point of intersection, it lies both on and on its rotation with degrees. Moreover, the symmetric points also lie on 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!

Build three particular equal segments in a triangle

I recently stumbled upon the following problem:

Consider a triangle . Construct points on , respectively such that .

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.

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

Consider the triangle with and . Consider such that . Prove that .

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. 

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

Consider a triangle with angles , , . Let be the bisector of the angle . Consider such that . Prove that the triangle is isosceles. 

Graham’s Biggest Little Hexagon

Think of the following problem: What is the largest area of a Hexagon with diameter equal to 1?

As is the case with many questions similar to the one above, called polygonal isoperimetric problems, the first guess is the regular hexagon. For example, the largest area of a Hexagon with fixed perimeter is obtained for the regular hexagon. However, for the initial question, the regular hexagon is not the best one. Graham proved in his paper “The largest small hexagon” that there exists a better competitor and he showed precisely which hexagon is optimal. More details on the history of the problem and more references can be found in Graham’s paper, the Wikipedia page or the Mathworld page.

I recently wanted to use this hexagon in some computations and I was surprised I could not find explicitly the coordinates of such a hexagon. The paper “Isodiametric Problems for Polygons” by Mossinghoff was as close as possible to what I was looking for, although the construction is not explicit. Therefore, below I present a strategy to find what is the optimal hexagon and I will give a precise (although approximate) variant for the coordinates of Graham’s hexagon.

IMO 2022 – Problem 4 – Geometry

Problem 4. Let be a convex pentagon such that . Assume that there is a point inside with , and . Let line intersect lines and at points and , respectively. Assume that the points occur on their line in that order. Let line intersect lines and at points and , respectively. Assume that the points occur on their line in that order. Prove that the points lie on a circle. 

Hadwiger-Finsler inequality – translation of the original paper

The Hadwiger-Finsler inequality states that if are the side lengths of a triangle and is its area then we have

.

I recently decided to take a look at the original paper: Finsler, Paul; Hadwiger, Hugo (1937). “Einige Relationen im Dreieck”. Commentarii Mathematici Helvetici. 10 (1): 316–326. doi:10.1007/BF01214300

This motivated me to try and translate it. Below you can find the document translated into English. Since do not speak German, do not hesitate to suggest corrections and improvements!

Some relations in the triangle by H. Hadwiger and Paul FinslerDownload

Three circles meet again

Consider a triangle and the reflections of with respect to the opposite sides. If is the orthocenter of then prove that the circles have another common point.

I’m sure this problem is classical, but I found it at the following link. There, a solution using inversion is given. Here’s a different one.

Note that one can identify with the midpoints of the triangle and becomes the circumcenter of . Thus we get the following equivalent problem:

Let be a triangle with circumcenter and denote by the midpoints of the corresponding opposite sides. Then the circles have another common point.

A quick drawing shows that three concurrent circles have another common point if and only if their centers are colinear. Let us try to show this. To identify a diameter of , construct the tangent in to the circumcenter of . This tangent meets at . It is not hard to show that is cyclic and is the diameter of . Construct in the same way.

Now we arrive at a different well known problem: the tangents to the circumcircle at vertices meet opposite sides at colinear points. This can be proved using Menelaus’s theorem, working out the ratios in which the tangents cut the opposite sides. Another proof can be given using Pascal’s theorem. Therefore are colinear. But since are diameters of the circles that interest us, it follows at once that the centers of these circles are colinear, therefore these circles meet again.