Post by Flip

Posted: 11/29/2014 3:05 AM

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FractalFusion

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Flip wrote:

-If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?

Suppose that there are n smaller circles of radius x surrounding a central circle of radius 1. The centers of all the smaller circles then lie on a circle of radius 1+x, and the arc between the centers of two consecutive circles has angle 2π/n. So there is an isosceles triangle with sides 1+x, 1+x, 2x, and the angle opposite 2x is 2π/n. By the cosine law, 4x²=(1+x)²+(1+x)²-2(1+x)²cos(2π/n)=(1+x)²(2-2cos(2π/n)) This gives x=(1+x)sqrt(1/2 - (1/2)cos(2π/n)). However, cos(2π/n)=1-2sin(π/n)². So the above equation simplifies to x=(1+x)sin(π/n), or x=sin(π/n)/(1-sin(π/n)). This agrees with r57shell's answer. For n=8, sin(π/8)=sqrt(2-sqrt(2))/2, so we get x=[sqrt(2-sqrt(2))]/[2-sqrt(2-sqrt(2))]=[2sqrt(2-sqrt(2))+2-sqrt(2)]/[2+sqrt(2)], which is about equal to 0.62.

thatguy wrote:

And, AFAIK, the case for spheres is still an open problem today

According to Wikipedia, the sphere packing problem has been solved for over 15 years already.

Posted: 11/29/2014 3:06 AM

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Amaraticando

It/Its

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Flip wrote:

-Prove that a circle can be surrounded by exactly 6 others of its own radius. IE The above picture might actually be incomplete, there may be indistinguishable gaps in between each one; so prove it to make sure. -If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?

My solution is here (click to expand):

So, if n = 6: csc(pi/6) = 1/sin(pi/6) = 2 Then, r = R. If n = 8: csc(pi/8) = sqrt(4 + 2*sqrt(2)) Then, r = R/(sqrt(4 + 2*sqrt(2)) - 1) = R*0.6199.. This problem gets really interesting in higher dimensions: http://en.wikipedia.org/wiki/Kissing_number_problem PS: sorry if my r looks like the symbol for pi. EDIT: whoa, FractalFusion and me posted pretty much the same thing at the same time.

Posted: 11/29/2014 11:21 AM

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Bobo_the_King

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Warp wrote:

Those don't look like circles to me, but ellipses. Which gives me an idea for an additional question: Prove that an ellipse can be surrounded by 6 identical ellipses, each touching an adjacent ellipse only on one point.

Well this part's easy enough. An ellipse is just a linear (stretch and/or skew) transformation of a circle. So here's what we do: 1) Prove that a circle can be surrounded by 6 identical circles with each outer circle mutually tangent to two others. This has been done by other commenters here. 2) Take the set of all points that are on those circle's interiors or perimeters and execute a linear (matrix) transformation on them. Now you have the interiors or perimeters of seven ellipses for which the external ones are each mutually tangent to two others. Provided all the ellipses have the same orientation, we can also work in the opposite direction and do a linear transformation that "compresses" them into circles.

Warp wrote:

Bonus question: Can the ellipses have different orientations, and still fulfill the requirements?

This is much harder. My gut says the answer is no. If we tilt one of our ellipses to the side, it will displace its neighbor and this disturbance will propagate around the exterior ellipses. Once it reaches the original ellipse, I think it's unlikely to be compatible with the original displacement we made. (But I guess my gut instinct was wrong!) On the other hand... Let five of the ellipses be oriented in a "daisy" configuration with their major axes pointing (more or less) radially outward from a given point. This will open up a significant gap in the "petals" where the sixth ellipse should go. (Can anyone prove this elegantly?) Then, you need to make the last ellipse tangent to three others. Three tangencies can be (at best) uniquely filled by an ellipse of given dimensions^*. If the ellipse is too small to fill the gap, then simply tilt all of the "petals" so that they close the gap somewhat until it can be spanned by the last ellipse. Someone can hash out the details of the proof, if they'd like. ^*To prove this, let's examine the number of degrees of freedom. The ellipse has x- and y-coordinates of its center, major and minor axes, and a rotation parameter for a total of five degrees of freedom. We're given the dimensions of the ellipses, so we know their major and minor axes, reducing the degrees of freedom to three. Finally, each tangency "pins down" one degree of freedom. Since there are three tangencies, the ellipse is (at best) uniquely defined.

Posted: 12/4/2014 12:08 AM
(Edited: 12/4/2014 10:21 PM)

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Archanfel

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-How many triangles on this pic? -And how to calculate amount of triangles for arbitrary size of the biggest triangle? (i.e. to find equation for progression a₁=1, a₂=5, a₃=13, a₄=27... a_x=?). Correct answers: * 315 * a_n= (1/8) (2n³+5n²+2n-((1/2)-(1/2)(-1)ⁿ))

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Posted: 12/4/2014 12:26 AM

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Amaraticando

It/Its

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This is the sum of the first n squares, which is a well known progression: http://en.wikipedia.org/wiki/Square_pyramidal_number However, the interesting part of the problem is to show rigorously that we gain n² new triangles in the nth "degree of the pyramid". EDIT: Forget what I said, I made a confusion with the nxn square grid. Gonna take a look at this.

Posted: 12/4/2014 4:23 AM

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Bobo_the_King

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By inspection we see that there are 10 small triangles on the bottom row, 9 on the next two rows, 8 on the next two rows, and so on. So the number of small triangles is 10 + 2*9 + 2*8 + ... + 2*1 = 10 + 2*10*(10-1)/2 = 10 + 10*9 = 100 In general, for an equilateral triangle of side n, there are n^2 little triangles within it. For the 2*2*2 triangles, we see that there are 9 flush with the bottom row, 8 flush with the next row, 7 flush with the next row, and so on. Add 'em up and we get 9 + 8 + 7 + ... + 1 = 9*(9+1)/2 = 45 In general, for a triangle of side n, there are n*(n-1)/2 upward-pointing triangles of side 2 enclosed. Ah! But let's do downward-pointing triangles too! We see 7 in the bottom row, 6 in the next, and so on up the line. The number of downward-pointing 2*2*2 triangles is therefore 7 + 6 + 5 + ... + 1 = 7*(7+1)/2 = 28 For a triangle of side n, there are (n-2)*(n-3)/2 downward-pointing triangles of side 2. These patterns continue, and we see that for a triangle of side n, there are... upward-pointing triangles of side... 1: (n+1)*n/2 2: n*(n-1)/2 3: (n-1)*(n-2)/2 ... n: 2*1/2 and downward-pointing triangles of side... 1: n*(n-1)/2 2: (n-2)*(n-3)/2 3: (n-4)*(n-5)/2 ... n/2: 2*1/2 (This assumes n is even. If n is odd, the series terminates at 3*2/2.) So together we have... ... upward-pointing triangles: sum( (j+1)*j/2, j, 1, n) = 1/2 * sum( j^2 + j, j, 1, n) ... downward-pointing triangles (even side): sum( (2*k)*(2*k-1)/2, k, 1, n/2) = 1/2 * sum( 4k^2 - 2k, k, 1, n/2) ... downward-pointing triangles (odd side): sum( (2*k+1)*(2*k)/2, k, 1, (n-1)/2) = 1/2 * sum( 4k^2 + 2k, k, 1, (n-1)/2) Now we'd like to write these sums in closed form. To do so, we can use induction (which I'll omit): Upward-pointing triangles: 1/6 * ( n^3 + 3*n^2 + 2*n) = 1/24 * ( 4*n^3 + 12*n^2 + 8*n) Downward-pointing triangles (even side): 1/24 * ( 2*n^3 + 3*n^2 + n) Downward-pointing triangles (odd side): 1/24 * ( 2*n^3 + 3*n^2 - 2*n) Adding these together, we find... For triangles with an even length side n: N = 1/24 * ( 6*n^3 + 15*n^2 + 9*n) = 1/8 * ( 2*n^3 + 5*n^2 + 3*n) For triangles with an odd length side n: N = 1/24 * ( 6*n^3 + 15*n^2 + 6*n) = 1/8 * ( 2*n^3 + 5*n^2 + 2*n) where N is the total number of triangles. Let's try these formulas out on the sample data you gave: n=1: Dangit. ^^^ Saved for posterity. Well, that didn't work. But here's a "proof" that will drive the mathematicians here crazy. I recognize that a counting argument like the one above should work. Therefore, I expect a polynomial no higher than 3rd order in n, the length of one side, ~~and with constant coefficient of zero~~. Thus, I just need to find a polynomial that fits the data and I'm done. Let's fit our polynomial to the first three data points (for both even and odd). For even length side, I see that the total number of triangles is... n=2: 5 n=4: 27 n=6: 78 and therefore, N = 1/8 * (2*n^3 + 5*n^2 + 2*n) (n even). And for odds... n=1: 1 n=3: 13 n=5: 48 n=7: 118 and therefore, N = 1/8 * (2*n^3 + 5*n^2 + 2*n - 1) (n odd). Hooray for cheating! Is there a name for what I just did? Basically, I assume induction should work and then I just fit the polynomial to the data. Edit: Messed up a calculation. More corrections not unlikely... Edit 2: Apparently, I can't assume the constant coefficient is zero. No matter! Just take more data! The formulas should both be correct now.

Posted: 12/4/2014 2:23 PM

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Scepheo

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Joined: 7/16/2009
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Bobo the King wrote:

Is there a name for what I just did? Basically, I assume induction should work and then I just fit the polynomial to the data.

I think the scientific term is "guessing". EDIT: It seems your polynomials hold up, at least for the first 10,000 values.

Posted: 12/4/2014 6:10 PM

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FractalFusion

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Bobo the King is right (and I can see where he got that guess; in each case of odd or even, the count is nothing more than sequential sums of polynomials of degree 2, which are polynomials of degree 3), but let's actually go through with a proof. Mere counting is overrated; let's go find a recurrence first thing. Let T_n be the big triangle of size n and a_n be the count of triangles (of any size and orientation) in it. Let n≥3. Let T_a, T_b, T_c be the subtriangles of T_n formed by removing the bottom row, top right diagonal, and top left diagonal, respectively; note that each one forms a T_n-1, the intersections of two of them form a T_n-2, and the intersection of all three forms a T_n-3. Obviously this calls for inclusion-exclusion. Let C_a, C_b, C_c be the conditions that a subtriangle of T_n is contained in T_a, T_b, T_c, respectively. The notation N(condition) means "number of objects in sample space satisfying condition". Then, taking the sample space to be all subtriangles of T_n, a_n = M + N(C_a) + N(C_b) + N(C_c) - N(C_aC_b) - N(C_aC_c) - N(C_bC_c) + N(C_aC_bC_c) = M + 3a_n-1 - 3a_n-2 + a_n-3, or, in other words, a_n - 3a_n-1 + 3a_n-2 - a_n-3 = M, where M = N(not (C_a or C_b or C_c)). A subtriangle of T_n is contained in none of T_a, T_b, T_c if and only if it touches all sides of T_n. The only up-pointing triangle satisfying this is the whole T_n. The only down-pointing triangle satisfying this is when n is even, and it is the triangle joining the midpoints of each side of T_n. Thus, M=1 if n is odd, and M=2 if n is even. So M = 3/2 + (1/2)(-1)ⁿ. The recurrence is thus: a_n - 3a_n-1 + 3a_n-2 - a_n-3 = 3/2 + (1/2)(-1)ⁿ (n≥3), a₀=0, a₁=1, a₂=5. The characteristic equation of this recurrence is r³-3r²+3r-1=(r-1)³. Therefore there exists constants A,B,C,D,E such that a_n=A+Bn+Cn²+Dn³+E(-1)ⁿ, where Dn³+E(-1)ⁿ is a particular solution of the recurrence. Substituting Dn³+E(-1)ⁿ for a_n in the recurrence gives 6D + 8E(-1)ⁿ = 3/2 + 1/2 (-1)ⁿ for all n≥3, so 6D=3/2 (D=1/4) and 8E=1/2 (E=1/16). So a_n=A+Bn+Cn²+(1/4)n³+(1/16)(-1)ⁿ, a₀=0, a₁=1, a₂=5. Substituting the initial conditions gives A+(1/16)=0 (A=-1/16), A+B+C+(1/4)-(1/16)=1 (B+C=7/8), and A+2B+4C+2+(1/16)= 5 (B+2C=3/2). The last two give C=5/8, B=1/4. So a_n= (1/4)n³ + (5/8)n² + (1/4)n - 1/16 + (1/16)(-1)ⁿ = (1/8) (2n³+5n²+2n-((1/2)-(1/2)(-1)ⁿ)), and (1/2)-(1/2)(-1)ⁿ is 1 if n is odd, and 0 if n is even. This agrees with Bobo the King's result. I think it should be possible to derive this by a straight counting argument like Bobo the King has tried, but I'll leave it for someone else.

Posted: 12/4/2014 9:29 PM

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NxCy

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I went ahead and tried a counting argument. I find the number of up-pointing triangles N_up and down-pointing triangles N_down separately. Let the big triangle have side length n. N_up: Let's count the number of triangles with size s. Clearly 1 <= s <= n. Suppose I am on row j of the big triangle where j = 1 at the bottom. For fixed s I can only increase j until the remaining height left above me is equal to s, i.e. until s > n - j + 1, so j_max = n - s + 1. On row j there are n - (j-1) - s + 1 = n - j - s + 2 triangles of size s, so N_up = sum from s=1 to s=n of [sum from j=1 to j=j_max of (n-j-s+2)] = 1/6 * n(n+1)(n+2) N_down: Here I flipped the picture upside down and labelled the 'tip', which is now at the bottom, as row 0, the base of that small 'tip triangle' as row 1 etc. What's the largest size s I can have? Well when I am on row s I could at best have a triangle of size s, and the room above me is n-s, so I must stop when s > n-s, or s > n/2. So s_max = (n-1)/2 for n odd and n/2 for n even. I only start counting triangles of size s from row s onward, so I label my rows as s+j where 0 <= j <= j_max. Again I stop when the room above me is less than my triangle size, i.e. n - (s+j) < s, giving j_max = n - 2s. Clearly on row s+j there are j+1 triangles. Hence N_down = sum from s=1 to s=s_max of [ sum from j=0 to j=j_max of (j+1) ] = 1/24 * (2n^3 + 3n^2 - 2n - 3) for n odd and 1/24 * (2n^3 + 3n^2 -2n) for n even. Finally N = N_up + N_down = 1/8 * (2n^3 + 5n^2 + 2n -1) for n odd and 1/8 ( 2n^3 + 5n^2 + 2n) for n even, in agreement with the above.

Posted: 12/4/2014 10:33 PM

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Archanfel

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Well done guys, now one more easy challenge: Rectangle was cut into 9 squares as shown in the picture. Size of the smallest white square is 1. -Find sides of this rectangle.

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Posted: 12/4/2014 11:09 PM

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Bobo_the_King

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There may be a nice geometric way to do this, but I favor the good ol' brute force equations method. Let the widths of the squares be given by orange red yellow magenta green cyan (kind of a stretch) blue dark green. Also, let the width of the rectangle be w and the height be h. Then from the figure, we see o + r + g = w o + 1 + m + g = w y + m + g = w y + c + d = w b + d = w o + y + b = h r + 1 + y + b = h r + m + c + b = h r + m + d = h g + d = h That's ten equations and ten unknowns. I'll solve it later.

Posted: 12/5/2014 12:08 AM

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AntyMew

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I got 34x34, but knowing myself, I could have made a number of silly mistakes :P r=m+1 o=r+1 y=o+1 y=m+c-1 y=m+3 m+c-1=m+3 c-1=3 c=4 b=y+4 d=b+4 d=(y+4)+4 d=y+8 d=g+m-4 d=y+8 g+m-4=y+8 g=y-m+12 g=(m+3)-m+12 g=15 r+m=15 (m-1)+m=15 2m=16 m=8 r=9 o=10 y=11 d=y+8 d=11+8 d=19 w=o+r+g w=10+9+15 w=34 h=d+g h=19+15 h=34

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Posted: 12/5/2014 12:24 AM
(Edited: 12/5/2014 12:45 AM)

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Archanfel

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Bobo the King wrote:

I favor the good ol' brute force equations method. ... That's ten equations and ten unknowns.

Method of solving proposed by Bobo the King is absolutely correct.

Anty-Lemon wrote:

I got 34x34, but knowing myself, I could have made a number of silly mistakes :P

You are good at know yourself :P This answer is wrong. Scepheo's calculations (before his edit) were wrong too :] --- Using Bobo the King's notation, сorrect answer is w = 32 h = 33 Anty-Lemon_v2 is сorrect.

I show you how deep the rabbit hole goes. Current projects: NES: Tetris "fastest 999999" (improvement, with r57shell) Genesis: Adventures of Batman & Robin (with Truncated); Pocahontas; Comix Zone (improvement); Mickey Mania (improvement); RoboCop versus The Terminator (improvement); Gargoyles (with feos)

Posted: 12/5/2014 12:28 AM
(Edited: 12/5/2014 12:38 AM)

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Scepheo

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Using Bobo the King's notation, here's some equations: r = m + 1 o = r + 1 = m + 2 y = o + 1 = m + 3 c = y + 1 - m = (m + 3) + 1 - m = 4 b = y + c = (m + 3) + 4 = m + 7 d = b + c = (m + 7) + 4 = m + 11 g = d + c - m = (m + 11) + 4 - m = 15 m = g - r = 15 - (m + 1) 2m = 14 m = 7 As we just expressed everything in m, we now know everything. Namely: w = b + d = (m + 7) + (m + 11) = (7 + 7) + (7 + 11) = 32 h = g + d = 15 + (m + 11) = 15 + (7 + 11) = 33

Posted: 12/5/2014 12:34 AM

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AntyMew

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I found one error changing the answer to 32x33, hopefully it's the only one r+m=15 (m+1)+m=15 2m=14 m=7 r=8 o=9 y=10 d=y+8 d=10+8 d=18 w=o+r+g w=9+8+15 w=32 h=d+g h=18+15 h=33

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Posted: 12/5/2014 10:05 PM
(Edited: 12/6/2014 5:44 PM)

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Archanfel

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Ok, problems with triangles and with squares were easily solved. Now time for pentagon! -Radiuses of all small circles = 1. Find radius of big circle. ---

Aktan wrote:

My wild guess is 5

Nope. Answer is irrational number, value somewhere between 4.5 and 5; A lot of square roots are needed to exact calculation.

I show you how deep the rabbit hole goes. Current projects: NES: Tetris "fastest 999999" (improvement, with r57shell) Genesis: Adventures of Batman & Robin (with Truncated); Pocahontas; Comix Zone (improvement); Mickey Mania (improvement); RoboCop versus The Terminator (improvement); Gargoyles (with feos)

Posted: 12/6/2014 3:36 AM

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FractalFusion

Editor, Expert player (2080)

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By the way, here is a solution to the squares problem without words and using only one variable:

Posted: 12/6/2014 4:55 AM

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Aktan

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My wild guess is 5

Posted: 12/6/2014 3:53 PM

Post subject: Assuming previous questions correct...

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Flip

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By previous question, at the origin we have a smaller circle of radius r, touching the surrounding five of radius R=1. This gives the interior circle a radius of r = R*(1/(sin(pi/n))-1=1/sin(pi/5)-1= ~0.7 (Yellow) Distance from origin to inner circle origin D(O,O_i)=r+R=r+1 (Red) Distance from inner to outer circle origin D(O_i,O_o)=2R=2 (Green) Distance from outer origin to outer edge D(O_o,E)=R=1 (Green) But these are obviously not in the same direction, so to find the overall distance we need a bit of trig. First consider the closest triangle, formed by 2 reds and a green. This is isosceles, and we know the inner angle, thus: A(O_i,O,O_i)=2pi/5 A(O_i,O_i,O)=(pi-2pi/5)/2=(5pi/5-2pi/5)/2=3pi/10 We know the four inner/outer circles create a square, giving A(O_i,O_i,O_o)=pi/2 A(O,O_i,O_o)=A(O_i,O_i,O)+A(O_i,O_i,O_o)=pi/2+3pi/10=5pi/10+3pi/10=8pi/10=4pi/5 Now we know this angle, we can use it for the larger triangle O_oO_iO (Green Red Blue). From a²=b²+c²-2bcCos(A), we have D(O, O_o)²=2²+(1+r)^2-2(2)(1+r)Cos(4Pi/5)= ~12.4 D(O, O_o)=L= ~3.52 (Blue) Next, we have consider triangle O_oOO_o (blue blue green), being isosceles with lengths L, L, and 2R=2. We need the interior angle, so therefore we have A(O_o,O,O_o)=ArcCos ((b²+c²-a²)/(2bc))=ArcCos ((L²+L²-2²)/(2(L)(L)))=ArcCos ((2L²-4)/(2L²)=ArcCos (1-2/L²) =~ArcCos(0.84) =~ 0.58 radians. This gives the other triangles' angles as: A(O,O_O,O_O)=(Pi-0.58)/2=~1.28 radians. A(O,O_O,E)=1.28+Pi/2=~2.86 radians. Finally, consider the triangle going from OO_oE (Blue Green Purple). We know 2 lengths, and one angle (2.86), giving our overall radius D (Purple) to be a²=b²+c²-2bcCos(A) D²=3.52²+1²-2(3.52)(1)Cos(2.86)=~20.1 D=~4.49

Posted: 12/7/2014 7:07 AM

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FractalFusion

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For the last question (pentagon and square arrangement of unit circles problem): Let the center be the origin. Take a point where the inner circles touch and rotate the diagram so it is on the positive x-axis. Then the inner and outer circle in quadrant I touching the x-axis form the following diagram (not perfectly to scale):

Then x=tan 54° and so the distance from the origin to the center of the outer circle is sqrt(x²+4x+5)=sqrt((tan 54°)²+4(tan 54°)+5) Therefore, the radius of the big circle (enclosing the outer circles) is one unit more; that is, sqrt((tan 54°)²+4(tan 54°)+5)+1. You can substitute tan 54°=φ/sqrt(3-φ) if you like (φ is the golden ratio (1+sqrt(5))/2), but I won't do that here. The number comes out to be about 4.521, close to Flip's estimate.

Posted: 12/7/2014 12:01 PM

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Scepheo

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Joined: 7/16/2009
Posts: 686

FractalFusion's answer is correct (search for "circle packing in a circle"), but I'm still curious as to how he found the 36 degrees. I'm guessing 360/5/2, but why is that correct?

Posted: 12/7/2014 3:48 PM
(Edited: 12/7/2014 5:33 PM)

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Archanfel

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Joined: 11/26/2011
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Scepheo wrote:

FractalFusion's answer is correct (search for "circle packing in a circle")

Using google search to find correct answer is a kind of cheating! But, true it is case for "15 units" from circle packing in a circle. http://en.wikipedia.org/wiki/Circle_packing_in_a_circle Correct answer is

= 4.5213569647061642409073640084

Scepheo wrote:

but I'm still curious as to how he found the 36 degrees.

Obviously, because it is pentagon in basis.

I show you how deep the rabbit hole goes. Current projects: NES: Tetris "fastest 999999" (improvement, with r57shell) Genesis: Adventures of Batman & Robin (with Truncated); Pocahontas; Comix Zone (improvement); Mickey Mania (improvement); RoboCop versus The Terminator (improvement); Gargoyles (with feos)

Posted: 12/7/2014 5:07 PM

Post subject: Reverse formula of a linear congruential generator

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Masterjun

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Posts: 1185
Location: Germany

I always wondered how it is possible to find a reverse formula of a linear congruential generator. For example we have the formula: X(n+1) = ( ( 0x41C64E6D * X(n) ) + 0x3039 ) mod 0x100000000 which, starting with 0, gives you values: 0, 12345, 3554416254, 2802067423... So I wonder how it is possible to calculate this reverse formula: Y(n+1) = ( ( 0xEEB9EB65 * Y(n) ) + 0xFC77A683 ) mod 0x100000000 which gives the expected values: ...2802067423, 3554416254, 12345, 0...

Warning: Might glitch to credits I will finish this ACE soon as possible (or will I?)

Posted: 12/7/2014 5:20 PM

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Aktan

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Joined: 4/23/2009
Posts: 1283

Dang, I was way off, oh well, you guys know math a lot better than me! Archanfel: you should re hide those words in your quote.

Forum Off topic some math challenges.

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some math challenges.