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Archive for the ‘Geometry’ Category

## IMO 2018 Problems – Day 2

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

## IMO 2018 Problems – Day 1

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.

Problem 1. A quadrilateral ${ABCD}$ is inscribed in a circle ${k}$, where ${AB>CD}$ and ${AB}$ is not parallel to ${CD}$. Point ${M}$ is the intersection of the diagonals ${AC}$ and ${BD}$ and the perpendicular from ${M}$ to ${AB}$ intersects the segment ${AB}$ at the point ${E}$. If ${EM}$ bisects the angle ${CED}$, prove that ${AB}$ is a diameter of the circle ${k}$.

Problem 2. Let ${q}$ be a positive rational number. Two ants are initially at the same point ${X}$ in the plane. In the ${n}$-th minute ${(n=1,2,...)}$ each of them chooses whether to walk due north, east, south or west and then walks the distance of ${q^n}$ meters. After a whole number of minutes, they are at the same point in the plane (non necessarily ${X}$), but have not taken exactly the same route within that time. Determine all the possible values of ${q}$.

Problem 3. Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The came ends if a player cannot move, in which case the other player wins.

Determine all pairs ${(a,b)}$ of positive integers such that if initially the two piles have ${a}$ and ${b}$ coins, respectively, then Bob has a winning strategy.

Problem 4. Find all primes ${p}$ and ${q}$ such that ${3p^{q-1}+1}$ divides ${11^q+17^p}$.

## Spherical triangles of area Pi

Recently I stumbled upon this page and found out a very nice result:

If a spherical triangle has area $\pi$ then four copies of it can tile the sphere.

Here we are talking about triangles on the unit sphere whose edges are geodesics. The above result is a simple consequence of the following facts:

1. If a spherical triangle has angles $\alpha,\beta,\gamma$ then its area is $\alpha+\beta+\gamma-\pi$. Therefore if a triangle has area $\pi$, then $\alpha+\beta+\gamma = 2\pi$.
2. If $ABC$ is a triangle of area $\pi$ and $D$ is obtained by symmetrizing $A$ with respect to the midpoint of $BC$ on the sphere, then $DCB$ and $ABC$ are congruent triangles.
3. Using angles and the fact that on the sphere similar triangles are congruent, we obtain that the triangles $ABC,BCD,CDA,DAC$ are all congruent.

Here are a few examples of such partitions:

Categories: Combinatorics, Geometry

## Mesh a hollow sphere

This is an interesting experiment I’ve done today and I’d like to share it. There are nice software out there which allow you to mesh regular surfaces, like the sphere, torus or other examples.

I needed something different: I wanted to mesh a sphere with a hole and an inner surface. This kind of surfaces may be useful for those who want to design objects for 3D printing. When using 3D printers, the cost is usually proportional to the volume of material used, since this also consumes impression time. If you manage to make a hollow object, the cost goes down immediately (as the resistance of the object…). However, to make a hole, you need to consider an inner parallel surface. I’ll not handle this here, since it’s more complicated. I’ll just mesh a hollow sphere.

I did this in Matlab using the nice library Distmesh, which is free and really easy to use. The whole difficulty was making a function whose zero level set is the hollow sphere. Recall that it is possible to find the union of two shapes using the minimum of two level set function and the intersection using the maximum. This is all that you need to understand what I did below. You can see the details in the Matlab code below.

function [p,t] = Hole_Spher_Mesh(mh)

R = 0.5;
r = 0.4;
rh = 0.1;

fh=@(p) sqrt(min(p(:,1).^2+p(:,2).^2,ones(size(p(:,1)))));
fh=@(p) sqrt((p(:,1).^2+p(:,2).^2));
[p,t]=distmeshsurface(@(x)fd(x,R,r,rh),@huniform,mh,[-1.1,-1.1,-1.1;1.1,1.1,1.1]);
points = p';
npt = prod(size(points))/3

clf
patch('Faces',t,'Vertices',p,'FaceColor',[0,0.7,0.9],'EdgeColor','none','FaceAlpha',0.5);

function res = fd(p,R,r,rh)

res1 = sqrt(sum(p.^2,2))-R;
res2 = (rh-sqrt(p(:,1).^2+p(:,2).^2));
res3 = p(:,3);
res2 = max(-res2,res3);
res4 = sqrt(sum(p.^2,2))-r;
res2 = min(res2,res4);
res = max(res1,-res2);


And here is the resulting surface for the parameter mh = 0.01 (if you put a higher precision computations will take longer).

## Romanian Masters in Mathematics contest – 2018

Problem 1. Let ${ABCD}$ be a cyclic quadrilateral an let ${P}$ be a point on the side ${AB.}$ The diagonals ${AC}$ meets the segments ${DP}$ at ${Q.}$ The line through ${P}$ parallel to ${CD}$ mmets the extension of the side ${CB}$ beyond ${B}$ at ${K.}$ The line through ${Q}$ parallel to ${BD}$ meets the extension of the side ${CB}$ beyond ${B}$ at ${L.}$ Prove that the circumcircles of the triangles ${BKP}$ and ${CLQ}$ are tangent .

Problem 2. Determine whether there exist non-constant polynomials ${P(x)}$ and ${Q(x)}$ with real coefficients satisfying

$\displaystyle P(x)^{10}+P(x)^9 = Q(x)^{21}+Q(x)^{20}.$

Problem 3. Ann and Bob play a game on the edges of an infinite square grid, playing in turns. Ann plays the first move. A move consists of orienting any edge that has not yet been given an orientation. Bob wins if at any point a cycle has been created. Does Bob have a winning strategy?

Problem 4. Let ${a,b,c,d}$ be positive integers such that ${ad \neq bc}$ and ${gcd(a,b,c,d)=1}$. Let ${S}$ be the set of values attained by ${\gcd(an+b,cn+d)}$ as ${n}$ runs through the positive integers. Show that ${S}$ is the set of all positive divisors of some positive integer.

Problem 5. Let ${n}$ be positive integer and fix ${2n}$ distinct points on a circle. Determine the number of ways to connect the points with ${n}$ arrows (oriented line segments) such that all of the following conditions hold:

• each of the ${2n}$ points is a startpoint or endpoint of an arrow;
• no two arrows intersect;
• there are no two arrows ${\overrightarrow{AB}}$ and ${\overrightarrow{CD}}$ such that ${A}$, ${B}$, ${C}$ and ${D}$ appear in clockwise order around the circle (not necessarily consecutively).

Problem 6. Fix a circle ${\Gamma}$, a line ${\ell}$ to tangent ${\Gamma}$, and another circle ${\Omega}$ disjoint from ${\ell}$ such that ${\Gamma}$ and ${\Omega}$ lie on opposite sides of ${\ell}$. The tangents to ${\Gamma}$ from a variable point ${X}$ on ${\Omega}$ meet ${\ell}$ at ${Y}$ and ${Z}$. Prove that, as ${X}$ varies over ${\Omega}$, the circumcircle of ${XYZ}$ is tangent to two fixed circles.

Source: Art of Problem Solving forums

Some quick ideas: For Problem 1 just consider the intersection of the circle ${(BKP)}$ with the circle ${(ABCD)}$. You’ll notice immediately that this point belongs to the circle ${(CLQ)}$. Furthermore, there is a common tangent to the two circles at this point.

For Problem 2 we have ${10\deg P = 21 \deg Q}$. Eliminate the highest order term from both sides and look at the next one to get a contradiction.

Problem 4 becomes easy after noticing that if ${q}$ divides ${an+b}$ and ${cn+d}$ then ${q}$ divides ${ad-bc}$.

In Problem 5 try to prove that the choice of start points determines that of the endpoints. Then you have a simple combinatorial proof.

Problem 6 is interesting and official solutions use inversions. Those are quite nice, but it may be worthwhile to understand what happens in the non-inverted configuration.

I will come back to some of these problems in some future posts.

## Regular tetrahedron – computing various quantities in terms of the side-length

Sometimes one needs to find certain quantities related to the regular tetrahedron in ${\Bbb{R}^3}$, like volume, radius of the circumscribed sphere, radius of inscribed sphere, distance between opposite sides, etc. in terms of the side-length which we’ll note in the following with ${a}$. In the past I needed to find the angle under which every side is seen when looking from the center of the regular tetrahedron.

Here’s a trick which can help you find rather easily everything you need related to the regular tetrahedron: just embed it into a cube. We can see rather easily that when drawing certain diagonals of the faces of a cube, like in the figure below, we can recover a regular tetrahedron. Now it becomes rather easy to solve all questions above. We note that the ratio between the side of the cube (denoted by ${\ell}$) and the side of the embedded tetrahedron (denoted by ${a}$) is ${\sqrt{2}}$: ${a = \ell \sqrt{2}}$.

Here are a few ideas:

1. Finding the volume of the tetrahedron in terms of its sides.

The volume of the cube is ${\ell^3}$. The volume of the tetrahedron can be obtained by cutting four corner pyramids with volumes ${\ell^3/6}$ (recall that volume of a pyramid is (area of base) ${\times}$ (height) ${/3}$). Therefore the volume of the regular tetrahedron is ${2\ell^3/6 = l^3/3}$. Replacing ${\ell = a/\sqrt{2} }$ we get that the volume of the tetrahedron is ${a^3/(6\sqrt{2})}$.

2. Finding the circumradius ${R}$.

It is not difficult to see that the sphere passing through the vertices of the tetrahedron also passes through the vertices of the cube. Therefore its radius is a long diagonal of the cube divided by ${2}$. This gives ${\ell \sqrt{3}/2}$. Replacing ${\ell=a/\sqrt{2}}$ we get that the circumradius is ${R = a\sqrt{6}/4}$.

3. Finding the inradius ${r}$.

Once we have the volume and the circumradius, finding the inradius is not that difficult, since the symmetry of the figure shows that ${r+R = h}$, where ${h}$ is the distance from a vertex to the opposite face, also called the height. We can find the height from the formula of the volume (recalled above), and then find ${r = h-R}$. Also note that since the center of the tetrahedron is also the centroid, we must have ${R=3r}$, so we have another quick finish solution.

However, let’s use the cube to do this. We can choose a system of coordinates putting the origin at the center of the cube (and tetrahedron). Put the cube so that its vertices have coordinates ${\pm \ell/2,\pm \ell/2,\pm \ell/2}$ and suppose that the tetrahedron corresponds to the vertices ${A(\ell/2,\ell/2,\ell/2)}$, ${B(-\ell/2,-\ell/2,\ell/2)}$, ${C(\ell/2,-\ell/2,-\ell/2)}$, ${D(-\ell/2,\ell/2,-\ell/2)}$. All we need to do is to compute the distance from the origin to the plane ${(BCD)}$. This is immediate if we know the equation of this plane. Fortunately, it is really easy to see that the coordinates of ${B,C,D}$ verify ${x+y+z+\ell/2 = 0}$ (if not, then note that the normal to ${(B,C,D)}$ is the vector ${(1,1,1)}$ and figure out the remaining translation constant). We know that if a plane is defined by the equation ${ax+by+cz+d=0}$ then the distance from ${(x_0,y_0,z_0)}$ to this plane is

$\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}.$

Apply this to our problem and the distance from the origin to ${(BCD)}$ becomes ${\ell/(2\sqrt{3})}$. Replace ${\ell = a/\sqrt{2}}$ and we get ${r = a/(2\sqrt{6})=a\sqrt{6}/12}$.

4. Find distance between opposite sides.

This is particularly easy with the cube. The distance between the opposite sides is exactly the distance between two parallel faces of the cube and that is ${\ell = a/\sqrt{2}}$.

5. Find angle made by two rays connecting the center with vertices.

Use the coordinate system introduced in 3. and just compute the angle between vectors ${\vec u = (\ell,\ell,\ell)}$ and ${\vec v = (-\ell,-\ell,\ell)}$, for example. If ${\alpha}$ is the angle between ${\vec u}$ and ${\vec v}$ we get that

$\displaystyle \cos \alpha = \frac{\langle \vec u,\vec v\rangle}{\|\vec u\| \|\vec v\|}=\frac{-\ell^2}{3\ell^2} = -\frac{1}{3}.$

Therefore ${\alpha = \arccos (-1/3)}$.

Categories: Geometry, Problem Solving