Posts Tagged ‘trick’

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)}.


Missing digit – short puzzle

April 15, 2017 Leave a comment

The number 2^{29} has 9 digits, all different; which digit is missing?

Mathematical Mind-Benders, Peter Winkler

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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]);


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…

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Monotonic bijection from naturals to pairs of natural numbers

October 23, 2014 Leave a comment

This is a cute problem I found this evening.

Suppose {\phi : \Bbb{N}^* \rightarrow \Bbb{N}^*\times \Bbb{N}^*} is a bijection such that if {\phi(k) = (i,j),\ \phi(k')=(i',j')} and {k \leq k'}, then {ij \leq i'j'}.

Prove that if {k = \phi(i,j)} then {k \geq ij}.

Proof: The trick is to divide the pairs of positive integers into families with the same product.

\displaystyle \begin{matrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) & \cdots \\ & (2,1) & (3,1) & (2,2) & (5,1) & (2,3) & \cdots \\ & & &( 4,1) & & (3,2) & \cdots \\ & & & & & (6,1) & \cdots \end{matrix}

Note that the {M}-th column contains as many elements as the number of divisors of {M}. Now we just just use a simple observation. Let {\phi(k)=(i,j)} be on the {M}-th column (i.e. {ij = M}). If {n \geq 1} then {\phi(k+n)=(i',j')} cannot be on one of the first {M-1} columns. Indeed, the monotonicity property implies {M = ij \leq i'j'}. The fact that {\phi} is a bijection assures us that {\phi(1),...,\phi(k)} cover the first {M-1} columns. Moreover, one element from the {M}-th column is surely covered, namely {(i,j) = \phi(k)}. This means that

\displaystyle k \geq d(1)+...+d(M-1)+1 \geq M = ij,

where we have denoted by {d(n)} the number of positive divisors of {n}.

Necessary condition for the uniform convergence of a certain series

October 8, 2014 Leave a comment

Let {a_n} be a decreasing sequence of positive numbers such that the series

\displaystyle \sum_{n = 1}^\infty a_n \sin(nx)

is uniformly convergent. Then {(a_n)} must satisfy {(na_n) \rightarrow 0}.

Note that this result implies that the series {\sum_{n=1}^\infty a_n \sin(nx)} is not uniformly convergent on {\Bbb{R}}. It is surprisngly similar to the following result:

Suppose that {(a_n)} is a decreasing sequence of positive real numbers such that the series {\sum_{n=1}^\infty a_n} is convergence. Then {(na_n) \rightarrow 0}. It is no surprise that the proofs of these two results are similar.

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Nice inequality similar to IMO 2011

November 23, 2012 Leave a comment

Prove that

(a+b+c)^2 \geq \sum \sqrt{a^4+8a^2bc}

Note the similarity to the IMO 2011 inequality, and surprisingly the same method works.

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Categories: Inequalities, Olympiad Tags: ,
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