higher dimensional triangle. corner of a cube. 3 vertices, 1 origin. shape is determined by angles formed and length -- or shape is set to fit a cube.. most likely the latter.

3 vertices are always on a 2-dimensional plane Mr. "higher dimensional triangle".

https://en.wikipedia.org/wiki/Trirec...ar_tetrahedron

Now it's my turn, right?

What shape am I thinking about?

Kind of half straight line and half circle.

What triangle is this?

Its perimeter is a prime number. So are all its sides. And so is the sum of the squares of its sides. They are also different.

Who will find it first?

A semi-circle?

I don't think a Pythagorean Triple exists with all 3 numbers prime. Perhaps you are thinking {1,2,sqrt(5)} but the last one isn't considered prime because it isn't a whole number.

Also if all three sides are prime then the perimeter cannot be prime. The sum of any two primes <>2, is always an even number.

I believe he means one of these:

There doesn't appear to be a restriction that this has to be a right triangle.

Odd + Odd + Odd = Odd

Oops.

I was thinking of a...

Yes i was thinking of a special triangle that is related to primes that way and the sums of squares is to get some concept of area involved ie as in the squares formed by the sides, all 3 sides. That is because the triangle itself cannot be prime as stated due to the well known Heron's formula.


I do get solutions of such triplets but they are not satisfying so far the triangular inequalities that secure they form a real triangle. I have tested all triplets of primes up to 2000.

If i relax the all sides different argument then an isosceles solution appears like

3,73,73


It may be impossible to have another solution(s) that is not isosceles and also it may be always needed one side to be 3 for reasons that may be revealed with better number theory analysis of constraints. If one side has to be always 3 for some number theory reason then the isosceles is forced.

There are an infinite number of solutions like this.

Here are in fact many of them . Notice all are using 3 for some reason.


x^3 y^2 z^2 + x^2 y^3 z^2 + x^2 y^2 z^3 + x^5 y^3 z^3 + x^3 y^5 z^3 +
x^5 y^5 z^3 + x^7 y^7 z^3 + x^13 y^13 z^3 + x^17 y^17 z^3 +
x^73 y^73 z^3 + x^137 y^137 z^3 + x^139 y^139 z^3 + x^157 y^157 z^3 +
x^167 y^167 z^3 + x^193 y^193 z^3 + x^227 y^227 z^3 +
x^229 y^229 z^3 + x^269 y^269 z^3 + x^337 y^337 z^3 +
x^379 y^379 z^3 + x^487 y^487 z^3 + x^509 y^509 z^3 +
x^547 y^547 z^3 + x^557 y^557 z^3 + x^643 y^643 z^3 +
x^739 y^739 z^3 + x^743 y^743 z^3 + x^773 y^773 z^3 +
x^853 y^853 z^3 + x^887 y^887 z^3 + x^929 y^929 z^3 +
x^1117 y^1117 z^3 + x^1153 y^1153 z^3 + x^3 y^3 z^5 + x^5 y^3 z^5 +
x^3 y^5 z^5 + x^7 y^3 z^7 + x^3 y^7 z^7 + x^13 y^3 z^13 +
x^3 y^13 z^13 + x^17 y^3 z^17 + x^3 y^17 z^17 + x^73 y^3 z^73 +
x^3 y^73 z^73 + x^137 y^3 z^137 + x^3 y^137 z^137 + x^139 y^3 z^139 +
x^3 y^139 z^139 + x^157 y^3 z^157 + x^3 y^157 z^157 +
x^167 y^3 z^167 + x^3 y^167 z^167 + x^193 y^3 z^193 +
x^3 y^193 z^193 + x^227 y^3 z^227 + x^3 y^227 z^227 +
x^229 y^3 z^229 + x^3 y^229 z^229 + x^269 y^3 z^269 +
x^3 y^269 z^269 + x^337 y^3 z^337 + x^3 y^337 z^337 +
x^379 y^3 z^379 + x^3 y^379 z^379 + x^487 y^3 z^487 +
x^3 y^487 z^487 + x^509 y^3 z^509 + x^3 y^509 z^509 +
x^547 y^3 z^547 + x^3 y^547 z^547 + x^557 y^3 z^557 +
x^3 y^557 z^557 + x^643 y^3 z^643 + x^3 y^643 z^643 +
x^739 y^3 z^739 + x^3 y^739 z^739 + x^743 y^3 z^743 +
x^3 y^743 z^743 + x^773 y^3 z^773 + x^3 y^773 z^773 +
x^853 y^3 z^853 + x^3 y^853 z^853 + x^887 y^3 z^887 +
x^3 y^887 z^887 + x^929 y^3 z^929 + x^3 y^929 z^929 +
x^1117 y^3 z^1117 + x^3 y^1117 z^1117 + x^1153 y^3 z^1153 +
x^3 y^1153 z^1153

That may be true but how do you know there are infinite ?

Also, if the first prime is 3, then the other two must be twin primes p and p+2 to satisfy the triangle inequality. And that forces the perimeter to be divisible by 3.

If 3 < a < b < c are distinct primes, then 3 divides a^2 + b^2 + c^2.

Nice. That forces then the smaller to be 3 and the triangular inequalities that it is isosceles.

The question then becomes if there are infinite solutions ie if there are infinite primes p such that 2p+3 is prime and 2p^2+9 is also prime. A prime congruence kind of problem.

Since you need all p, 2p+3 and 2p^2+9 to be prime and because the probability a number is prime is in a heuristic non strictly trustworthy but very plausible way 1/log(n) and the nth prime goes like n*log(n) then the chance p is prime is

~1/log(n)

the chance 2p+3 is is ~1/log(2n)

and chance 2p^2+9 ~ 1/log(n^2)

suggesting that because 1/(log(x)^3 integral to infinity diverges that there are infinite solutions.

The hours, minutes and seconds hands of a clock are a, b, c (say 3,4,5) units long respectively. At what times during the day the triangle that the tips of the hands form has the highest area?

first time

I have tried to solve the triangle first, with the laws of signs and cosigns, then translate to the clockface with established clockface equations, but failed.

Try using the unit circle (or a circle in this case with radius bigger than c) and describe as function of time the coordinates of the tips (you know like xa=a*sin(w1*t), ya=a*cos(w1*t) describing rotating vectors tips clockwise where eg w1=2pi/3600/12 the angular velocity etc for b and c). Then use this link for how to find the area of a triangle if you know the coordinates of the vertices;

https://en.wikipedia.org/wiki/Triangle (under using coordinates shoelace formula)

New one; (because we are SMP damn it)

Imagine two points A and C that are at the same horizontal level (C slightly lower than A actually by a tiny bit that wont matter here initially) and a distance d horizontally from each other. Now imagine point B lower than A,C and between them forming a triangle ABC.

Find the coordinates of B as functions of coordinates of A,C (basically only the distance between them AC) that makes the time it takes for an object to slide from A to C minimum under the influence of vertical direction gravity g without friction. At B the object has a smooth transition using a very quick curvy part that we can ignore its size compared to the distance between A-C).


Now imagine C is a bit lower than A indeed (say by h) and there is friction coefficient μ. Investigate as function of μ if it is possible to go from A to C and what location for B (ie what triangle ABC) makes the time minimum as functions of h, μ, d.


Lets create problems to challenge this page;

https://www.physics.harvard.edu/acad...rgrad/problems

And have fun with great problems from there here in more organized presentation format;

http://www.people.fas.harvard.edu/~djmorin/Chapter1.pdf (over the years SMP has seen several from this list like the dragon problem, the two envelopes etc)