Archive for the ‘Geometry’ Category

Balkan Mathematical Olympiad 2017 – Problems

May 10, 2017 Leave a comment

Problem 1. Find all ordered pairs of positive integers { (x, y)} such that:

\displaystyle x^3+y^3=x^2+42xy+y^2.

Problem 2. Consider an acute-angled triangle {ABC} with {AB<AC} and let {\omega} be its circumscribed circle. Let {t_B} and {t_C} be the tangents to the circle {\omega} at points {B} and {C}, respectively, and let {L} be their intersection. The straight line passing through the point {B} and parallel to {AC} intersects {t_C} in point {D}. The straight line passing through the point {C} and parallel to {AB} intersects {t_B} in point {E}. The circumcircle of the triangle {BDC} intersects {AC} in {T}, where {T} is located between {A} and {C}. The circumcircle of the triangle {BEC} intersects the line {AB} (or its extension) in {S}, where {B} is located between {S} and {A}.

Prove that {ST}, {AL}, and {BC} are concurrent.

Problem 3. Let {\mathbb{N}} denote the set of positive integers. Find all functions {f:\mathbb{N}\longrightarrow\mathbb{N}} such that

\displaystyle n+f(m)\mid f(n)+nf(m)

for all {m,n\in \mathbb{N}}

Problem 4. On a circular table sit {\displaystyle {n> 2}} students. First, each student has just one candy. At each step, each student chooses one of the following actions:

  • (A) Gives a candy to the student sitting on his left or to the student sitting on his right.
  • (B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps. Find the number of legitimate distributions.

(Two distributions are different if there is a student who has a different number of candy in each of these distributions.)

Source: AoPS

Project Euler Problem 285

March 25, 2017 Leave a comment

Another quite nice problem from Project Euler is number 285. The result of the problem depends on the computation of a certain probability, which in turn is related to the computation of a certain area. Below is an illustration of the computation for k equal to 10.


To save you some time, here’s a picture of the case k=1 which I ignored and spent quite a lot of time debugging… Plus, it only affects the last three digits or so after the decimal point…


Here’s a Matlab code which can construct the pictures above and can compute the result for low cases. To solve the problem, you should compute explicitly all these areas.

function problem285(k)

N = 100000;

a = rand(1,N);
b = rand(1,N);

ind = find(abs(sqrt((k*a+1).^2+(k*b+1).^2)-k)<0.5);

axis equal

M = k;
pl = 1;

for k=1:M
if mod(k,100)==0
r1 = (k+0.5)/k;
r2 = (k-0.5)/k;

f1 = @(x) (x<=(-1/k+r1)).*(x>=(-1/k-r1)).*(sqrt(r1^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f1 = @(x) f1(x).*(f1(x)>=0);
f2 = @(x) (x<=(-1/k+r2)).*(x>=(-1/k-r2)).*(sqrt(r2^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f2 = @(x) f2(x).*(f2(x)>=0);

if k == pl
thetas = linspace(0,pi/2,200);
hold on
plot([0 1 1 0 0],[0 0 1 1 0],'k','LineWidth',3);
hold off
axis off

A(k) = integral(@(x) f1(x)-f2(x),0,1);


xs = xlim;
ys = ylim;

w = 0.01;
axis([xs(1)-w xs(2)+w ys(1)-w ys(2)+w]);


Fireworks – Project Euler 317

March 19, 2017 Leave a comment

You can read the text of the problem here. The idea is that we have an explosion at a given height in a uniform gravitational field (no friction/wind). Supposing that all particles go outwards from the explosion point with a constant initial speed, what is the shape of the body formed by these particles? A hint is given in the picture below with a nice Matlab simulation.


Now what happens if we add a little bit of wind? Here’s the perturbation obtained when adding a 10m/s uniform wind speed vs the initial configuration. Something like this would be a lot more challenging 🙂


Project Euler Problem 144 – Laser light escaping an ellipse

February 5, 2017 Leave a comment


Project Euler – problem 144 – visualization of the solution in Matlab

About Problem 587 – Project Euler

January 27, 2017 Leave a comment

I started looking again at some problems on the Project Euler site. It’s not often that I manage to solve a recent problem. Below are a few hints about the math needed to provide an answer to problem 587.

I’ll not repeat the statement of the problem here, so please read it by following the above link. The idea is to find the ratio of the areas of some triangular shapes made of two segments and one circle arc. The breakdown of the problem is the following:

  • Note that the size of the square does not matter, since we need the result as a ratio, so we can consider that the square has side equal to 1. First note that the L-shape in the corner of the triangle has area equal to 1/8-\pi/32. We’ll need this in the end, to compute the ratio.
  • Secondly, the shape whose area we need to compute can be decomposed into a right-angled triangle and a complement of a part of the circle. In order to have precise information about these shapes we compute the intersection of the line y=x/n and the circle (x-0.5)^2+(y-0.5)^2=0.25. It is possible to find the value explicitly.
  • Once we have the coordinates of the point of intersection we can compute the area of the triangle. The area of the rounded part which remains is the difference between a rectangle and the circle portion PQON. In order to do find this difference it suffices to find the angle PSQ.
  • Once the mathematical part is done, the programming job is not that tough, since the numbers involved are not that high. I managed to write a Matlab solution in no time.fig587

Some of the easy Putnam 2016 Problems

December 11, 2016 Leave a comment

Here are a few of the problems of the Putnam 2016 contest. I choose to only list problems which I managed to solve. Most of them are pretty straightforward, so maybe the solutions posted here may be very similar to the official ones. You can find a complete list of the problems on other sites, for example here.

A1. Find the smallest integer {j} such that for every polynomial {p} with integer coefficients and every integer {k}, the number

\displaystyle p^{(j)}(k),

that is the {j}-th derivative of {p} evaluated at {k}, is divisible by {2016}.

Hints. Successive derivatives give rise to terms containing products of consecutive numbers. The product of {j} consecutive numbers is divisible by {j!}. Find the smallest number such that {2016 | j!}. Prove that {j-1} does not work by choosing {p = x^{j-1}}. Prove that {j} works by working only on monomials…

Read more…

Balkan Mathematical Olympiad – 2016 Problems

Problem 1. Find all injective functions {f: \mathbb R \rightarrow \mathbb R} such that for every real number {x} and every positive integer {n},

\displaystyle \left|\sum_{i=1}^n i\left(f(x+i+1)-f(f(x+i))\right)\right|<2016

Problem 2. Let {ABCD} be a cyclic quadrilateral with {AB<CD}. The diagonals intersect at the point {F} and lines {AD} and {BC} intersect at the point {E}. Let {K} and {L} be the orthogonal projections of {F} onto lines {AD} and {BC} respectively, and let {M}, {S} and {T} be the midpoints of {EF}, {CF} and {DF} respectively. Prove that the second intersection point of the circumcircles of triangles {MKT} and {MLS} lies on the segment {CD}.

Problem 3. Find all monic polynomials {f} with integer coefficients satisfying the following condition: there exists a positive integer {N} such that {p} divides {2(f(p)!)+1} for every prime {p>N} for which {f(p)} is a positive integer.

Problem 4. The plane is divided into squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of {1201} colours so that no rectangle with perimeter {100} contains two squares of the same colour. Show that no rectangle of size {1\times1201} or {1201\times1} contains two squares of the same colour.

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