General: ∈ ∉ ∩ ⋃ ⊂ ⊃ Ø ⋀ ⋁ ∃ ∀ ⇒ ← ↑ → ↓ ↔ ± ∓ × ≈ ≅ ≠ ≡ ≢ ≤ ≥ ∞ √ ∛ ∜ ° ∡ ∑ ∏ ∂ ∫ ∤ ∥ ∦ ∆ ∇ ⊙ Subscripts/superscripts: ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ⁿ Greek: Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ Σ Τ Υ Φ Χ Ψ Ω α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω Fractions: ½ ⅓ ⅔ ¼ ¾ ⅕ ⅖ ⅗ ⅘ ⅙ ⅚ ⅛ ⅜ ⅝ ⅞
At an evening party the clock on the wall stroke 8 o'clock. Several seconds later one of the guests commented: 'Look everybody, the time is now 08:00:20 and the clock hands form a "Mercedes" configuration: the hour hand points at 8, the minute hand - at 12, the second hand - at 4; the angles between the hands are equal!' I objected immediately, because in that moment the minute hand had moved ⅓ of a minute past 12, the hour hand was also little after 8 (assuming of course that each hand rotates at uniform speed - for example at 08:30:00 the hour hand is half-way between 8 and 9), so that the hands formed approximately, but not exactly, three 120° angles in that moment. The company agreed, but a question arose is there a moment when the exact configuration is possible, like 3 vectors, connecting the centroid of an equilateral triangle with its vertices?
I recalled another well-known problem: to prove that the 3 clock hands point at the same direction only at 12:00:00 noon and 12:00:00 midnight. The proof uses that 11 = 12 - 1 and 59 = 60 - 1 are relatively prime. The same consideration can help here too, but I would like to award the best answer to the easiest and shortest possible solution, submitted by our contributors.
Cars & Transportation > Car Makes > Mercedes-Benz7 AnswersMathematics7 years ago
You have sufficiently many congruent regular tetrahedra - #1 on the following picture:
You begin joining them face-to-face: joining the second tetrahedron to the first, you get a regular triangular bipyramid (#2); then you join the third tetrahedron to one of its faces (#3) and so on. Let the joined tetrahedra form a chain so that each of them (except first and last) shares exactly 2 faces with its neighbors (unlike #4). Can the chain close in a frame-like construction, i.e. the first and last to be joined precisely face-to-face too?
For example 8 congruent cubes make an obvious frame (#5) with a cubic hole (how convenient are these right dihedral angles!), but somebody might be surprised that 8 congruent regular octahedra do it also (#6)! The success of the latter construction is due to the fact that the dihedral angle between 2 faces, sharing a common vertex, but not a common edge, is θ = arccos(1/3); the dihedral angle between 2 faces, sharing a common edge is exactly π - θ. But in the regular tetrahedron the only dihedral angle is θ.
Many irregular congruent polyhedra also make frames, remind the prisms in the famous Rubik's Snake:
But no cubes. No octahedra. Regular tetrahedra only. Their centroids need not be coplanar (like the centers of the 8 cubes above), but remember, not edge-to-edge or vertex-to-vertex joinings as these kaleidocycles:
All to be joined face-to-face, or "cheek-to-cheek" - this may inspire you:
This question may turn out difficult and the web search I made (I am not sure I did it thoroughly) didn't help - such rigid frame seems highly unlikely with only dihedral angle θ available, but I can neither prove, nor disprove it is possible.
To get into a good shape, try another much easier problem: prove that the only convex polyhedra - unions of joined face-to-face congruent regular tetrahedra - are the tetrahedron itself and the triangular bipyramid (#1 and #2).7 AnswersMathematics7 years ago
Let below "g-parallel lines/planes" means "parallel in generalized sense", i.e. objects are parallel or some of them may coincide. Given 3 g-parallel lines in the plane, distance between 1st and 2nd a ≥ 0, distance between 2nd and 3rd b ≥ 0, there exists always an equilateral triangle, whose vertices are one by one on each line:
The original problem (#1) - "Triangle on the rails" - is an old one and reminds some mountain railways in Switzerland with 3 rails, the middle one jagged for the cogwheel, allowing the trains to overcome unusually steep slopes.
Now some generalizations (see the diagram):
#2: Same as #1, but the 3 lines not necessary co-planar, with given distances a, b, c between them, also an old one. The equation we arrive about the side length x according Intermediate Value Theorem always has a real root, i.e. an equilateral triangle again always exists, or, say it otherwise, every trihedral prismatic surface has regular triangular cross-sections.
Further generalizations involve regular tetrahedron instead of regular triangle.
#3: Given 4 g-parallel planes at given distances between them, Zo Maar proved that a regular tetrahedron with vertices on each plane always exists, and found its edge length - read his excellent answer here:
Now the most difficult problem of this kind in my opinion, not always having a solution.
#4: What conditions must satisfy 4 g-parallel lines in 3D space, so that a regular tetrahedron, having vertices on each of them, exists? For example, if the common points with a perpendicular plane form a square, or an equilateral triangle with its centroid, solution obviously exists, but there are many other configurations of these common points with the same property. Suggest a description for them.2 AnswersMathematics7 years ago
(inspired by: http://answers.yahoo.com/question/index;_ylt=AjB.E...
Mathematics is a difficult thing. Here are some questions that have always intrigued me, I would like to know the answers before my Christmas vacation:
1) How many are Euclid's elements?
2) What crime was prosecuted and punished in Bernoulli Trial?
3) Why did Lagrange use Multipliers only, but neglected Divisors?
4) What can be wrapped up in the Folium of Descartes?
5) Why Euler Bricks are not used in the construction?
6) What material are Dedekind, Artinian and Noetherian Rings made of?
7) What is the difference between Right Ideal and Wrong Ideal?
8) What crop grows best on the Galois Fields?
9) In the good old days of DOS there was a mathematical software 'Eureka - the Solver'. Did Archimedes (as far as I know he has lived also some time ago) use 'Eureka' to discover the Law of Archimedes?
10) French mathematician Fermat is known for his 'Fermat's Little Theorem' - does it mean he has not been capable of proving something big?
11) The most famous apple in the history of science has fallen on Sir Isaac's Newton's head. These days I heard about Newton's Method - is it a method for painless picking of apples?
12) German mathematician Gauss has calculated the sum
1 + 2 + 3 + .. + 99 + 100 being 6-7 years old.
Very impressive! Recently I tried to do the same and found a remarkable identity:
n = n!/(n-1)!
So I'll have to calculate 0!, 1!, 2!, . . , 99! and 100!, then I'll have to divide each by the previous, and the expression
1!/0! + 2!/1! + 3!/2! + . . + 99!/98! + 100!/99!
should produce the required result. Do you like that approach or can you suggest something better? I need to finish it by all means till Christmas!
Thanks to everyone who will enlighten me!7 AnswersMathematics8 years ago
Recently I answered a question on the visibility of sides of a polygon and that recalled me a problem I have read long ago in a book: imagine a tourist in Washington D.C., looking from a random point at one of the most remarkable buildings in the city - the Pentagon. What are the probabilities he can see 0, 1, 2, 3, 4, 5 sides of the building?
See the picture:
On the left the regions, from which 1, 2, 3 and 5 sides are visible, are shown. Now the generalization: given a regular planar n-gon (n = 3, 4, 5, ..) and a random point in its plane, find the probabilities P(0), P(1), P(2), . . , P(n), exactly 0, 1, 2, . . , n sides of the polygon to be entirely visible from that point (cases n=3 and n=4 are shown on the right).
To avoid any misunderstanding what exactly "random point" should mean, imagine the regular polygon inscribed in the unit circle, centered in the origin (so that the vertices are, let's say, the roots of 1); the coordinates (x, y) of the viewpoint are uniformly distributed in a circle with radius R > 1, centered in the origin. What are the values of the aforementioned probabilities when R → ∞ (some of them are obviously 0)?
Prove or disprove the following:
1) if n is even, then P(n/2) = 1, the rest probabilities are 0;
2) if n is odd, then P((n-1)/2) = P((n+1)/2) = 0.5, the rest probabilities are 0.5 AnswersMathematics8 years ago
It's a Christmas time, let's postpone the difficult mathematical questions for later. Most of us, more or less regular contributors in this Forum, know each other already quite well, we've accustomed to our nicknames and avatars, so let's have a little holiday fun taking a survey on them.
Please share your opinion:
# 1: Whose user name sounds most "mathematically"?
# 2: Whose user name is most difficult to spell/pronounce (the harder, the better)?
# 3: Whose user name is most impressive (e.g. long palindrome, or sounds best to the ear, or otherwise remarkable)?
/for ## 1-3 please translate if the name uses non-Latin or non-Greek characters/
# 4: Who has the most ''mathematically looking" avatar?
# 5: Whose avatar is most picturesque?
# 6: Whose avatar is otherwise most remarkable?
Some of my choices:
# 1: Very difficult choice between Mathematishan, Mathemagician, MathMan TG, Mathmanretired, MathNerd, Divide By Zero and several others... Finally I decided for
# 4: Also very difficult to decide, final decision:
# 5: Definitely: http://answers.yahoo.com/activity?show=5cc0d7cf7f3...
# 6: Definitely: http://answers.yahoo.com/activity?show=1c25f077392...
I haven't decided yet, but depending on the answers I may send the question to public vote.
&feature=related7 AnswersMathematics9 years ago
Here are some:
Feel free to submit more as many as you wish!21 AnswersMathematics9 years ago
12 years ago when I was 50, I had to undergo a cornea transplantation on my left eye (very successful, Thank God!) because I suffered from keratoconus and that was the last thing to help. It helped indeed, the eye is doing well, but developed astigmatism and now I use glasses to read, my last prescription (abut a year ago - O.S. only) is:
SPH CYL Ax
+2.5 +2.5 150°
Several days ago I noticed that when I turn the lens at a right angle I see even better. Is it possible or is it normal the degrees of my astigmatism to have changed so rapidly?3 AnswersOptical10 years ago
When Yahoo! introduced some time ago the new default icon (currently in use) my friend Alberich commented it in a question as an "anemic Halloween pumpkin (Jack-o'-lantern)" or some awful thing "anyone with even the slightest tinge of paranoia, will probably run for the hills, screaming 'help, help, it wants to eat me' ". Agree?
If we add teeth to the monster, it may look like this:
or even (CAUTION: Scary! Only brave persons click the link on the next text line!):
Anyway, here is a problem. It isn't difficult, just for fun. Enjoy!
ABCD is a square with side length 2a (a > 0). The eyes are circles with radius r < a/2, tangent to the sides AD and DC, BC and CD respectively. A circle with radius R (R > a) is tangent to AB and internally to both eyes as shown, the lower arc along with the horizontal chord forming the mouth.
1) Derive a relationship between a, r and R;
2) Derive expressions, yielding all solutions - integer triples (a, r, R), satisfying the above relationship and
0 < r < a/2 < a < R;
3) Find the least value of a for which there are 2 distinct triples (a, r', R') and (a, r", R");
4) Find a Pythagorean triple (a, r, R) if any;
5) What is Your Favorite Triple? The icon with most "aesthetic dimensions"? Please help me to decide how to carve my Halloween Pumpkin!
This is inspired by Manjyomesando1's very interesting question:
As shown there, infinitely many pairs of non-congruent polyhedrons with the same volume (V) and surface area(S) exist, I repeat here some of them for convenience:
/both #2 and #3 can be obtained truncating Kepler's Stella Octangula:
removing 6 out of its 8 thorns/
Dragan K's cuboids: 1 x 1 x (p(p+1)/2) and p x p x ((p+1)/(2p)) /p≠1/
Now the question: what is the combined minimal number of faces of such pair?
I began to play with a right triangular prism with right isosceles bases and lateral faces in planes x=0, y=0, x+y=1 as shown here:
Taking red triangles instead of black ones as bases, we obtain a pair of pentahedrons (#1 and #2) with the desired property - we can even join the rightmost vertexes in #2 to obtain a pyramid for the following result:
Prism: vertexes: (1, 0, 0), (1, 0, -2), (0, 1, 0), (0, 1, 2), (0, 0, ±1);
faces x = 0, y = 0, x + y = 1, x - y + z - 1 = 0, x - y + z + 1 = 0;
Pyramid: vertexes: (1, 0, 0), (0, 1, ±2), (0, 0, ±1);
faces x = 0, y = 0, x + y = 1, x - y + z - 1 = 0, x - y - z - 1 = 0
V = 1, S = 4 + 2√2 + √3 for both of them.
Finally, is 10 faces the combined minimum?2 AnswersMathematics1 decade ago
Answering a question in Y!A, I tried to paste a link to my previous answer to a similar question, but it was not accepted and I got a 'Error 999' message. Neither of the reasons in the message is relevant to the problem (my system is virus clean and the situation with the provider is OK), nor contacting Yahoo! Customer Care helped, so I suspect an internal bug in Y!A.
Moreover, about a year ago I had the same problem, trying to paste a link to a Wiki Article - then I was advised to change manually in the link 'wikipedia.com' to 'wikipedia.org' (or vice versa, I don't remember already) and it worked!
But this time it is an internal link and I don't know what to do. Every reasonable advice is welcome, thanks in advance.8 AnswersYahoo Answers1 decade ago
Let R = const > 0 and x₁, x₂, x₃, x₄, y₁, y₂, y₃, y₄ are 8 independent uniformly distributed in [0, R] random variables. Let us consider the quadrilateral with vertexes
P₁(x₁, y₁), P₂(x₂, y₂), P₃(x₃, y₃) and P₄(x₄, y₄) /following in that order, not the convex envelope of these 4 points!/. What are the probabilities P₁P₂P₃P₄ to be:
/of course the probability of degenerated quadrilateral with at least 3 collinear vertexes is 0/
See the picture:
What is the behavior of the above probabilities when R → ∞?
Browsing back through Q & A I found this:
It reminded me another similar question, asked some time ago (see links below) but I haven't encountered much more interesting generalized question: if A, B, C are random real coefficients (not necessarily digits!), what is the probability the equation:
At² + Bt + C = 0 to have real roots?
To avoid any misunderstanding what exactly "random real coefficients" should mean, I suggest the following:
Approach 1) Let the point (A,B,C) is uniformly distributed in the cube [-R,R; -R,R; -R,R] /R=const>0/, or, equivalently, A, B, C are uniformly distributed independent random variables in [-R,R] each. Express the probability in terms of R, then let R → ∞.
Approach 2) Same as above, but the point (A,B,C) to be uniformly distributed in a sphere with radius R.
I consider Approach 1) most natural, but feel free to choose 2) instead (that could led to more complicated integrations), or to suggest another eventual approach (and solution) You find reasonable enough. It may be useful to read the following comments and discussions:
and Pascal's excellent explanation:
Of course P(Real Roots) =
= P[(A≠0 and B²-4AC ≥ 0) or (A=0 and B≠0) or (A=0 and B=0 and C=0)]=
= P(B² ≥ 4AC) since P(A=0) = 0. So what is P(B² ≥ 4AC)?5 AnswersMathematics1 decade ago
A question about some problems with elegant solutions was asked recently:
Encouraging further discussion on this fascinating subject, started by Ana, I decided to repost 2 problems despite the fact I know the solutions - they fully deserve it, I'm sure Y!A Math Community will like them very much. Enjoy!
1) Let P is an arbitrary point in space. How many rays, starting at P exist, with the property: the angle between any 2 of them is the same? Prove that the maximal number is 4.
2) Consider all octahedrons, circumscribed around the unit sphere. Prove that the regular octahedron (one of the Platonian Solids) IS NOT the polyhedron with the minimal volume among them.
The most elegant (or most neatly presented in my opinion) answer will be chosen as best, I am not going to put this into vote. Additional Details immediately follow.
What do you think about him and his legacy these days?3 AnswersBooks & Authors1 decade ago
Well, sooner or later it had to happen. Revisiting today one of my answers:
I noticed surprisingly a second answer by another Duke, so in Math category there are already at least 2 of us!
Has anybody heard of 3rd, 4th, etc. Duke in Y!A community?
Well, we would hardly be able to compose a Duke football team, but I'm still curious about the number of solutions. I know, the answerers in this category are extremely active and competent on serious problems, so I decided to suggest this one before the summer vacation time.
Any witty suggestions how to determine the number of Dukes are welcome.
As far as I can remember (may have read long ago), the Russian mathematician B. Delaunay has proven that x=3, y=5 is the only solution of the Diophantine equation
y² = x³ - 2 in natural numbers.
A question about rational solutions of this interesting equation was asked recently:
Using the common tangent-chord method:
I found x = 1290/1000 = 1.29, y = 383/1000 = 0.383
Next rational points I found on this elliptic curve have enormously huge denominators, so I'm extremely curious whether a solution with a denominator less than 1000 exists (x and y to be expressed as positive fractions, at least one of them irreducible, with common denominator d, 1<d<1000).
Any analytical or computer-aided approach to do eventually the work is acceptable.
The following question, asked recently:
reminded me an old similar problem I have almost forgotten about: in triangle ABC the angle A is 3 times greater than angle B and all side lengths are integers (unlike the most well-known example with angles 30°, 60° and 90°, all of its sides can not be integers).
Find all such triangles (or, at least , some of them).
The solution involves a 3rd degree Diophantine equation - the most interesting point.
Probably similar questions about how LCD monitors affect the eyesight were asked many times.
From time to time I feel my eyes somewhat tired. I am not young already, have a best possible quality large LCD monitor at home, but nevertheless try not to stay for too long in front of it and make pauses for rest as ophthalmologists recommend. I'm not even online every day (and inevitably miss many interesting questions). I have also accessories like umbral glasses, eye drops etc.
Despite all that I'll be much grateful for any advices or recommendations on the subject. Is there any day limit?5 AnswersOptical1 decade ago
Do You remember the old problem about the equilateral triangle on the rails? Given are 3 co-planar parallel lines, the distance between 1st and 2nd is a, between 2nd and 3rd - b (and a + b between 1st and 3rd of course).
Please follow the link below to see a picture:
Easily follows that for all a ≥ 0 and b ≥ 0 there exists an equilateral triangle, having one vertex on each line with side length:
(2/√3)*√(a² + ab + b²)
Now the 3D analog: imagine 4 parallel planes at distances a, b and c between them. Does a regular tetrahedron exist, having one vertex on each plane? If so, find its edge length.