Consider the pyramid $\left \{ (x, y, z): x,y,z\geq 0, x+y+z\leq 1 \right \}$.  This is a pyramid that has $3$ pairwise-orthogonal edges, all of length $1$ (just like a corner of a room):   From now on I'll call it a "unit pyramid" (UP). One can easily check that the volume of a UP is exactly $\frac{1}{6}$. So that raises the question:     Can we take six UPs and fit them all perfectly inside a unit cube?   When I thought about it, my initial answer was an obvious yes. So I tried to imagine such arranging, but I couldn't. I was quite surprised since I usually have a very good spatial vision. Then I tried drawing a solution and failed.  Since I'm not tech-savvy enough to use a 3D graphics software, I decided to take this problem into the real world.  So here's a unit pyramid:    It's composed of 3 edges of length 1 (which I set to be $5$ cm) and 3 edges of length $\sqrt2$ ($\approx 7$ cm).  Now, there's only one reasonable way to attach another one to it, if our goal is a unit cube:     (since now we have our square base). Now I added all the edges necessary to form a cube:   And as your 3D mind can see, now we have four unit pyramids:   It's pretty hard to see the region that has stayed uncovered by our 4 UPs, so here it is, emphasized:   And what do you know - that's a ! A quick reality check: a regular tetrahedron with edges of length $\sqrt 2$ has a volume of $\sqrt 2 (\sqrt 2)^3/12=4/12=1/3$, which suits the fact that we entered 4 UPs inside.   But now, I'm pretty certain that you can't fit another 2 of them inside. To be sure, I constructed them seperately (last picture, I promise!):  On the left there's a UP, and on the right - our regular tetrahedron (would you believe that one is twice the volume of the other?) Now you'll just have to trust me that it won't fit.  Well, that convinced me that the mission of fitting 6 unit pyramids inside a unit cube is impossible, but of course I wouldn't call it a mathematical proof. Does anyone have a more mathematically-convincing argument? I'd be curious to hear your ideas and thoughts about this.  Thank you for your time reading the question!
120631
In grade school we learn that a square can be equidecomposed into two congruent right isosceles triangles. Does the following three dimensional generalization hold?  Consider a trirectangular tetrahedron with vertices at $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and volume $\frac{1}{6}$. (It has three orthogonal right triangular faces and one equilateral triangular face opposite the (tri-right-angled) common vertex.) The inscribing cube of the tetrahedron has unit side length and unit volume. Is there a dissection of this cube into 6 tetrahedra of the above form?   Provided that such a dissection exists and this is the way it is to be done, it's easy to see how 4 tetrahedra can be glued together to give the boundary of the cube, but I don't quite see how the other two should fit inside.
50775
When the circumference of a circle (c) and the perimeter of a square (8h) are equal, we call it a squared circle. Radius (r) is equal to one. (Fig. 1)    Next we simplify to the quarter squared circle which gives us our first triangle A. (Fig. 2)    Because a triangle has three sides it can be normalised in three ways, with exceptions. A normalised triangle is simply a triangle with one or more side-lengths equal to one. All sets of normalised triangles have the same angles. See the example below, normalising the 345 triangle. (Fig. 3)    Notice above that the height of triangle 1 is the reciprocal of the hypotenuse of triangle 2, the base of triangle 1 is the reciprocal of the hypotenuse of triangle 3 and the base of triangle 2 is the reciprocal of the height of triangle 3. This is the case for all sets of normalised triangles.  We can use this rule to normalise triangle A. (Fig. 4)    When the perimeter of a square (8h) and the circumference of a circle (c) are equal, the ratio of the radius (r) to an eighth of the perimeter (h) will always be 4/Pi. (Fig. 5)    We can use this rule to create three nested versions of the quarter squared circle. (Fig. 6)    This gives us a second set of triangles D, E and F. (Fig. 7)    Notice that triangles D and A and triangles E and B are identical but triangles F and C are different. Next we interlock two quarter squared circles as shown below. (Fig. 8)    We then add a quarter circle with a radius (1+x). The arrows represent the tangents. (Fig. 9)    This gives us a third set of triangles J, K and L. (Fig. 10)    The set of triangles (A, B and C) and the set of triangles (J, K and L) should be identical, because both sets are normalised. If we compare the two sets we can extrapolate the following:    As shown below, it makes sense that y equals Pi/4, because we would expect the two marked points to be at the same height. The arrows represent the tangents. (Fig. 11)    If y equals Pi/4, it means that the triangle set (D, E and F) is also normalised. By comparing the three sets we can extrapolate the following:    If we do the calculations using the current value of Pi, the equations above fail. This points to the possibility that Pi is incorrect. So the question becomes, is there a value for Pi that satisfies all of the above equations?  Yes, here it is expressed as a fraction:    If we use the above value for Pi, we find that the three sets of triangles listed above, are in fact the same normalised set. (Fig. 12)    Because the circle and the square are both perfectly symmetrical, we should expect Pi to also have perfect symmetry. This symmetry can be expressed by the following equations:    So it seems, the angle (theta1) that squares the circle, perfect symmetry, occurs when its sine is equal to the reciprocal of its tangent.  Please be kind, this has taken over a year of my life so far. Is this a valid proof?
42664
Find the volume of the tetrahedron with the vertices $P(1,1,1)$, $Q(1, 2, 3)$, $R(3, 1, 2)$, and $S(2, 3, 1)$.
76009
So I have spent some time thinking about Fermat's Last Theorem and about how to come up with a proof for certain cases of n. To begin with I took n = 3.  $$A^3 + B^3 = C^3 (A,B,C \in \mathbb {Z})( A,B,C ≠ 0)$$ We can assume that $A&lt;B&lt;C$. Because of this we can rewrite this as: $$A^3 + xA^3 = C^3$$ We now have three possibilities: $x$ is an integer, $x$ is an irrational, or $x$ is a fraction.  If $x$ is an integer then $x$+1 and $x$ must be cubes. The only numbers which satisfy this are $x$ = 0. However, this means $B^3 = 0$.  $x$ cannot be irrational because then $C^3$ is irrational. Therefore $x$ is a fraction.  We can represent this now as: $$A^3 + \frac{p}{q}B^3 = C^3 (p,q \in \mathbb {Z})(GCD(p,q) = 1)$$ This means $\frac{p}{q}$ is actually a cube over a cube. Therefore $p$ and $q$ are cube numbers.  However, $1 + \frac{p}{q}$ is also a cube. Therefore $\frac{p+q}{q}$ is a cube over a cube. Hence: $$\sqrt[3]{p} ,\sqrt[3]{q}, \sqrt[3]{p+q} \in \mathbb {Z}$$  We can now create another variable $v$ for $p+q$. Therefore, $$p+q = v$$ (Where $\sqrt[3]{v} \in \mathbb {Z}$)  Now we must deal with $\frac{p}{q}+1$. Since this is a cube we can write this as: $$\frac{p}{q} + 1 = \lambda^3$$ Since $p$ and $q$ are co-prime and in simplest form, $\lambda$ and $\lambda^3$ are also fractions in simplest form. Rearranging, we get: $$p = q(\lambda^3 - 1)$$  Substituting this into $p+q=v$, we get: $$q+q(\lambda^3 - 1) = v$$ $$q\lambda^3 = v$$  Since $\lambda^3$ is a fraction, we can express it as $\frac{D^3}{E^3}$. Hence: $$qD^3 = vE^3$$  However, because $v$ is a cube, $vE^3$ is also a cube. Therefore $\lambda^3$ doesn't have to be a fraction. This is a contradiction, and so there can be no integer solutions for $p$ and $q$. Subsequently, there can be no integer solutions for A,B,C in: $$A^3 + B^3 = C^3$$.  If anyone could point out the problem in this and what you think of it it would be greatly appreciated. Thank you.
110739
I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine). I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here?
40745
Consider a tetrahedron with a right-angle corner (like the corner of a cube) at $D$. Prove that the foot of the height from $D$ to the face ABC is the orthocenter of $ABC$.  Is this well-known?
108948
I am trying to solve question 3.7 (b) from Chapter 15 in Artin's book "Algebra".  The problem is:     Is it true that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$?   It is clear by Eisenstein's criterion that both $\sqrt[3]{2}$ and $\sqrt[3]{5}$ have their minimal polynomails of egree $3$ over $\mathbb Q$.  Thus if we assume that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$, then we must have $\mathbb Q(\sqrt[3]{5})=\mathbb Q(\sqrt[3]{2})$.  This didn't get me anywhere.  Then I tried the most simple minded approach.  Say $\sqrt[3]{5}=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, for some $a,b,c\in \mathbb Q$.  By cubing and rearranging, it is likely that we will arrive at some absurdity.  But clearly this shouldn't be what the author must have intended.  Can somebody point me towards a slick solution?
64423
What is the probability that a particular set of integer edge lengths selected from an interval $[1, N]$, can form a triangle?  How might this extend to the case where one selects real number edge lengths from the unit interval?   I'd also be very interested in any comments on what the probability distribution would look like for the distances between the triangles vertices.
114697
Consider a rectangle whose length is m and whose height is n. A path from bottom left corner to top right corner is called valid, if in each step, it either goes one unit to the right or one unit upwards. if m = 5 and n = 3, how many valid paths are there?  This rectangle is basically a table with 15 cells(3 vertical x 5 horizontal). I need to find the total number of paths that I can take from bottom left corner to top right corner. I get the answer 16 paths. Can anyone verify this?
73885
Can anyone give me a vectorial solution to the following problem:     Prove that if each pair of opposite edges of the tetrahedron $ABCD$ is perpendicular (that is, $AB \perp CD$ and $AC \perp BD$ and $AD \perp BC$), then the heights of the tetrahedron are concurrent.   Here, the heights (also known as the altitudes) of tetrahedron $ABCD$ are the perpendicular from $A$ to the plane $BCD$, and three other similarly defined perpendiculars.  Tetrahedra satisfying the condition of this problem are called , and this appears to be a known result.
16362
I drew a parallelepiped that is spanned by three vectors, and we know the volume is given by the absolute value of the determinant of the matrix - with the three vectors arranged in rows (or columns, since $detA = detA^T$).  How can I draw the tetrahedron from this parallelepiped to convince myself it is indeed 1/6th of the volume, i.e., 1/6 * |detA|?  Thanks,
66140
I am looking for an algorithm that can return the number of size of n squares that fit into a a rectangle of a given width and height, maximizing the use of space (thus, leaving the least amount of leftover space for squares that do not fit). Neither the rectangle nor the squares can be rotated.  For example, let's say I have a rectangle that is 5 inches by 7 inches, and I need it to fit 35 squares. The algorithm needs to tell me that the 35 squares fit if they are 1 inch wide/tall (as they could be laid out inside the rectangle in a 5 x 7 grid). Another example is if I need to divide a rectangle 35 inches by 1 inch into 35 squares. It should still tell me that the squares will fit if they are 1-inch wide/tall (as they could be laid out in the rectangle in a 35 x 1 grid).  The tricky part is that sometimes there may be leftover space, as the squares cannot be divided into partial squares. Let's say for either of the two examples above I need to lay out 34 squares and not 35 (in which case the answers would still be 1 inch), or maybe 33, or 7 squares. Or, perhaps the rectangle width and height aren't whole numbers. With the number of squares being a variable I need an algorithm that can tell me the size of the squares for a given rectangle width and height.  Thanks for your help!
119776
I suspect I'm missing something obvious, but I'm coming up blank. I've gotten pretty comfortable with so-called natural units over the years: in doing quantum mechanics/QFT, it's common to set $c = \hbar = 1$ and in GR, it's common to set $c = G = 1$. I'm told that quantum gravity researchers are fond of , where (according to that Wiki page) we set $c = \hbar = G = 1$, along with a couple other fundamental constants. I don't know what to make of this, though: it's dimensionally inconsistent in 3+1 dimensions. That same article gives the dimensions of each constant:  $$\begin{align} [c] &amp;= L T^{-1} \\ [G] &amp;= L^3 M^{−1} T^{−2} \\ [\hbar] &amp;= L^2 M T^{−1} \end{align}$$  Setting $c$ and $G$ to 1 implies $L=T=M$. Setting $c$ and $\hbar$ to 1 implies $L=T=M^{-1}$. Hence, making all three unity is internally inconsistent: mass cannot simultaneously have dimensions of length and reciprocal length. So, how do Planck units work?
41544
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.   I have tried completing square by several method but all failed. Can anyone help me please? Thank you.  p.s. I'm a poor question-tagger.
63126
How to find the number of ways in which six digits $1,2,..,6$ can be assigned to six faces of a cube (without repetition of digits) so that one arrangement cannot be obtained from another by a rotation of the cube?  I tried to find the number of unique $4-$adjacencies of the faces of the cube. I drew a simple undirected graph $A,B,...,F$ having $6$ vertices with each of them having degrees equal to $4$. There were $12$ edges. Considering the choice between the top and the bottom of the cube, I found the result to be $12 \times 2=24.$  Am I correct? Please suggest better approach approach or bijections(if there are any).
4903
That's my problem: Can every triangle be divided into five isosceles triangles?  I've got to give evidence why this is true or not true...  (sorry for possible language mistakes - I'm from Germany)  Thanks in advance  Markus
96969
Problem: Let n be a positive integer greater than four and let $P_1P_2P_3..P_n$ be a convex n-sided polygon. Zachary wants to draw $n-3$ diagonals that partition the region enclosed by the polygon into $n-2$ triangular regions and that may intersect only at vertices of the polygon. In addition, he wants each triangular region to have at least one side that is a side of the polygon. In how many different ways can Zachary do this?   My attempt: I began with the case $n=5.$ I observed that the number of ways in which $3$ triangular regions can be constructed is $5$. I then went on to consider a hexagon and observed that the number of ways to partition was $(6-1)*5.$ Furthermore, I deduced for $n$ sided polygons we must have $(n-1)(n-2)...5*5$ ways to partition the polygon. I would like to know whether this answer is correct and if not, then a proof or a sketch would be much appreciated.