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

February 2, 2018 Leave a comment

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)}.


Plotting 3D level sets in Paraview

December 30, 2017 Leave a comment

Surfaces can be represented as certain level-sets for some 3D functions. Given a set of points with values attached to them a level set associated to a certain number will separate the points into two sets: with values higher or lower than the number considered. It is nice to have good looking plots when working with the level-set method in shape optimization. Paraview is a nice, open source framework, which has the right tools in order to produce high quality plots.

I’ll briefly describe below how to use Paraview to make some nice pictures of level-sets. First of all, you’ll need your data in some format that Paraview can understand. I use vtk file format for which there is a nice automated interface in the software I use for the optimization (FreeFem++). In the vtk file you need to have a set of points and a scalar value attached to them.

If you want to create level-sets associated to certain values, follow the steps below:

    1. Load your file (containing points with at least a scalar value) and click Apply.
    2. Next, go to Filters/Alphabetical and select “Cell Data to Point Data” (if you forget to do this, you’ll get a rough surface where you see the discretization). Click Apply.
    3. Then apply the Contour filter (by clicking the button or going in Filters/Alphabetical). You’ll have to select the field for which the contours will be drawn and then put in the values of the level-sets you want to see. Click Apply. An example can be seen below.lvlview

If your level-set cuts the boundary of the domain, Paraview will draw a hole there. If you want to have a closed region instead, you need to use the IsoVolume filter instead of the Contour one. The difference is that you need to specify two values and Paraview will draw the surface enclosing the points corresponding to these values. Many other features are directly available: you can color the level set following another scalar value, you can set the lighting, etc. You can also symmetrize your geometry using the Reflect filter. Below you can see a result built from my work.


You can also create animations in a pretty  straightforward way. Just go to View and select the Animation box. Then you’ll see the animation options. Add a Camera object with Orbit field selected. You’ll be presented with multiple options, like the center of rotation, direction of the vertical and initial position. Once everything is set, click the play button to see the animation. Then go to File/Save Animation to save it to a file.


I heard that Paraview could to many things when dealing with visualization aspects, but I hesitated to use it until now since the interface is not straightforward. The use of Filters is not clear in the beginning, but after playing with some examples, everything becomes really easy to use. The next step is to automatize all this using scripts.

Happy New Year!

Putnam 2017 A3 – Solution

December 4, 2017 Leave a comment

Problem A3. Denote {\phi = f-g}. Then {\phi} is continuous and {\int_a^b \phi = 0}. We can see that

\displaystyle I_{n+1}-I_n = \int_a^b (f/g)^n \phi = \int_{\phi\geq 0} (f/g)^n \phi+ \int_{\phi<0} (f/g)^n \phi

Now note that on {\{ \phi>=0\}} we have {f/g \geq 1} so {(f/g)^n \phi \geq \phi}. Furthermore, on {\{\phi<0\}} we have {(f/g)^n <1} so multiplying with {\phi<0} we get {(f/g)^n \phi \geq \phi}. Therefore

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} \phi + \int_{\phi<0} \phi = \int \phi = 0.

To prove that {I_n} goes to {+\infty} we can still work with {I_{n+1}-I_n}. Note that the negative part cannot get too big:

\displaystyle \left|\int_{ \phi <0 } (f/g)^n \phi \right| \leq \int_{\phi<0} |\phi| \leq \int_a^b |f-g|.

As for the positive part, taking {0<\varepsilon< \max_{[a,b]} \phi} we have

\displaystyle \int_{\phi\geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon}(f/g)^n \varepsilon.

Next, note that on {\{ \phi \geq \varepsilon\}}

\displaystyle \frac{f}{g} = \frac{g+\phi}{g} \geq \frac{g+ \varepsilon}{g}.

We would need that the last term be larger than {1+\delta}. This is equivalent to {g\delta <\varepsilon}. Since {g} is continuous on {[a,b]}, it is bounded above, so some delta small enough exists in order for this to work.

Grouping all of the above we get that

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon} \varepsilon (1+\delta)^n.

Since {|\phi>\varepsilon|>0} we get that {I_{n+1}-I_n} goes to {+\infty}.

Putnam 2017 A2 – Solution

December 4, 2017 Leave a comment

Problem A2. We have the following recurrence relation

\displaystyle Q_n = \frac{Q_{n-1}^2-1}{Q_{n-2}},

for {n \geq 2}, given {Q_0=1} and {Q_1=x}. In order to prove that {Q_n} is always a polynomial with integer coefficients we should prove that {Q_{n-2}} divides {Q_{n-1}^2-1} somehow. Recurrence does not seem to work very well. Also, root based arguments might work, but you need to take good care in the computation.

A simpler idea, which is classic in this context, is to try and linearize the recurrence relation. In order to do this, let’s write two consecutive recurrence relations

\displaystyle Q_nQ_{n-2} +1 = Q_{n-1}^2

\displaystyle Q_n^2 = Q_{n+1}Q_{n-1}+1

We add them and we obtain the following relation

\displaystyle \frac{Q_n}{Q_{n-1}} = \frac{Q_{n+1}+Q_{n-1}}{Q_n+Q_{n-2}},

which leads straightforward to a telescoping argument. Finally, we are left with a simple linear recurrence with integer coefficient polynomials, and the result follows immediately.

Putnam 2017 – Problem A1

December 4, 2017 Leave a comment

Problem A1. We have {n^2 \in S \Rightarrow N \in S} and {n \in S \Rightarrow (n+5)^2 \in S}. Therefore {n \in S \Rightarrow n+5 \in S}.

Next, let’s note what elements cannot be in {S}. Note that taking square roots and squaring {n+5} cannot change a non-zero remainder modulo {5} into a zero remainder. Therefore, starting from {2} one could never get a multiple of {5} following the allowed operations. Thus we can safely say that multiples of {5} are not in the minimal set {S}.

Furthermore, {1} could only be obtained as a square root of itself with the allowed operations. Starting from {2}, one could never get below by performing square roots or {n \mapsto (n+5)^2}. Therefore, the minimal set {S} does not contain {1} and multiples of {5}.

Now, we show that it contains all the rest. The general idea is as follows: it is enough to find which is the smallest element in a class of remainders modulo {5} to deduce that all larger elements are there (recall the operation {n \in S \Rightarrow n+5\in S}). Now in order to obtain small elements of {S}, one would need to take successive square roots. So if we prove that for some {a} we have {a^{2^n}\in S} for some {n} then we get that {a \in S}.

Now let’s start from the beginning. We have {2 \in S} so {49 = (2+5)^2 \in S}. Since {49+5k} is in {S} for every {k}, we get that all squares of the form {5k+4} greater than {49} are in {S}. Moreover, {(49+5)^2 = 2916} so all numbers of the form {5k+1} greater than {2916} are in {S}. Since {81^2 =6561} it follows that {6561 \in S \Rightarrow 81 \in S \Rightarrow 9 \in S \Rightarrow 3 \in S}. Moreover, {6^{16}} ends in {6} and is greater than {2916} so {6 \in S}. Next, we have {4^8 = 16^4} which ends in {6} and is greater than {2916} so it is also in {S}. Therefore {4 \in S}.

Finally, we have that {n \in S \Rightarrow n+5 \in S}, {\{2,3,4,6\} \subset S} and {1 \notin S}, {5k \notin S}. This means that the minimal set {S} is {\Bbb{Z}_+^* \setminus\{1\} \setminus \{5k: k \in \Bbb{Z}_+\}}.

Categories: Uncategorized

Putnam 2017 – Problems

December 4, 2017 Leave a comment

Source: Art of Problem Solving forum

Problem A1. Let {S} be the smallest set of positive integers such that

  • a) {2} is in {S,}
  • b) {n} is in {S} whenever {n^2} is in {S,} and
  • c) {(n+5)^2} is in {S} whenever {n} is in {S.}

Which positive integers are not in {S?}

(The set {S} is “smallest” in the sense that {S} is contained in any other such set.)

Problem A2. Let {Q_0(x)=1}, {Q_1(x)=x,} and

\displaystyle Q_n(x)=\frac{(Q_{n-1}(x))^2-1}{Q_{n-2}(x)}

for all {n\ge 2.} Show that, whenever {n} is a positive integer, {Q_n(x)} is equal to a polynomial with integer coefficients.

Problem A3. Let {a} and {b} be real numbers with {a<b,} and let {f} and {g} be continuous functions from {[a,b]} to {(0,\infty)} such that {\int_a^b f(x)\,dx=\int_a^b g(x)\,dx} but {f\ne g.} For every positive integer {n,} define

\displaystyle I_n=\int_a^b\frac{(f(x))^{n+1}}{(g(x))^n}\,dx.

Show that {I_1,I_2,I_3,\dots} is an increasing sequence with {\displaystyle\lim_{n\rightarrow\infty}I_n=\infty.}

Problem A4. A class with {2N} students took a quiz, on which the possible scores were {0,1,\dots,10.} Each of these scores occurred at least once, and the average score was exactly {7.4.} Show that the class can be divided into two groups of {N} students in such a way that the average score for each group was exactly {7.4.}

Problem A5. Each of the integers from {1} to {n} is written on a separate card, and then the cards are combined into a deck and shuffled. Three players, {A,B,} and {C,} take turns in the order {A,B,C,A,\dots} choosing one card at random from the deck. (Each card in the deck is equally likely to be chosen.) After a card is chosen, that card and all higher-numbered cards are removed from the deck, and the remaining cards are reshuffled before the next turn. Play continues until one of the three players wins the game by drawing the card numbered {1.}

Show that for each of the three players, there are arbitrarily large values of {n} for which that player has the highest probability among the three players of winning the game.

Problem A6. The {30} edges of a regular icosahedron are distinguished by labeling them {1,2,\dots,30.} How many different ways are there to paint each edge red, white, or blue such that each of the 20 triangular faces of the icosahedron has two edges of the same color and a third edge of a different color?

Problem B1. Let {L_1} and {L_2} be distinct lines in the plane. Prove that {L_1} and {L_2} intersect if and only if, for every real number {\lambda\ne 0} and every point {P} not on {L_1} or {L_2,} there exist points {A_1} on {L_1} and {A_2} on {L_2} such that {\overrightarrow{PA_2}=\lambda\overrightarrow{PA_1}.}

Problem B2. Suppose that a positive integer {N} can be expressed as the sum of {k} consecutive positive integers

\displaystyle N=a+(a+1)+(a+2)+\cdots+(a+k-1)

for {k=2017} but for no other values of {k>1.} Considering all positive integers {N} with this property, what is the smallest positive integer {a} that occurs in any of these expressions?

Problem B3. Suppose that

\displaystyle f(x) = \sum_{i=0}^\infty c_ix^i

is a power series for which each coefficient {c_i} is {0} or {1}. Show that if {f(2/3) = 3/2}, then {f(1/2)} must be irrational.

Problem B4. Evaluate the sum

\displaystyle \sum_{k=0}^{\infty}\left(3\cdot\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right)

\displaystyle =3\cdot\frac{\ln 2}2-\frac{\ln 3}3-\frac{\ln 4}4-\frac{\ln 5}5+3\cdot\frac{\ln 6}6-\frac{\ln 7}7-\frac{\ln 8}8-\frac{\ln 9}9+3\cdot\frac{\ln 10}{10}-\cdots.

(As usual, {\ln x} denotes the natural logarithm of {x.})

Problem B5. A line in the plane of a triangle {T} is called an equalizer if it divides {T} into two regions having equal area and equal perimeter. Find positive integers {a>b>c,} with {a} as small as possible, such that there exists a triangle with side lengths {a,b,c} that has exactly two distinct equalizers.

Problem B6. Find the number of ordered {64}-tuples {\{x_0,x_1,\dots,x_{63}\}} such that {x_0,x_1,\dots,x_{63}} are distinct elements of {\{1,2,\dots,2017\}} and

\displaystyle x_0+x_1+2x_2+3x_3+\cdots+63x_{63}

is divisible by {2017.}

Categories: Uncategorized

A hint for Project Euler Pb 613

November 18, 2017 Leave a comment

The text for Problem 613 can be found here. The hint is the following picture 🙂


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