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Amaraticando wrote:
It's possible to get the sin/cos/tan of all integers angles (in degrees), because sin(2x) is a 2nd degree function of sin(x) and sin(3x) is a 3rd degree function of sin(x).
sin/cos/tan of all integer degree angles are solvable by radicals, but not because sin(2x) is a 2nd degree function of sin(x) and sin(3x) is a 3rd degree function of sin(x). (A now-deleted post by OmnipotentEntity pointed out you need sin(5x), a 5th degree function. In general, 5th degree polynomials are not solvable by radicals.) It is because all roots of unity (of angle θ which is a rational multiple of π) are solvable by radicals, and so are their real part (cos θ), imaginary part (sin θ) and the ratio of the two (tan θ).
Even though every root of unity is solvable by radicals, some of them involve things like radicals of complex numbers, as you can see from the posted table. (There is no rule against taking radicals of complex numbers, however mind-bending that is.) If you want only solvability by radicals using only real numbers, this restricts things. Correct me if I am wrong, but I think the following applies (not necessarily exhaustive):
- You are allowed to solve any 2nd degree equations.
- You are allowed to solve irreducible 3rd degree equations as long as it does not have three distinct real roots, which is the "casus irreducibilis" (but the equation for sin(3x) in terms of sin(x) is always casus irreducibilis, since you can always get three different angles from the equation).
I believe that in order for the sin/cos/tan of an nth root of unity to be solvable by radicals in real numbers, the Euler phi number φ(n), which is how many numbers between 1 and n are relatively prime to n, must be a power of 2. This occurs whenever n is a product of zero or more distinct Fermat primes times a power of 2. (Ex. from the table, sin 3° is represented in radicals using only real numbers, n=120=23*3*5. However sin 1° fails because n=360=23*32*5, and 3 occurs as a factor twice.) However, I can't prove it.
Edit: Fixed some numbers.
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Warp wrote:
Exact solutions preferred.
Let g(x) be the inverse function of (x-1)ln(x). Then the second solution is x=g(ln(2)).
Couldn't think of any better way to state the exact solution (even WolframAlpha doesn't give me anything). An approximation for x is x≈0.34632336227858092206485655.
Edit: Technically, to be well-defined, g(x) should be defined as the inverse function of (x-1)ln(x) on the interval (0,1). (By itself, (x-1)ln(x) has two branches, one on (0,1) and one on (1,∞).)
Yes, problem 1 uses an idea that is extremely similar to how Hamming codes are formed: parity. An explanation is here: http://datagenetics.com/blog/december12014/index.html (section "Parity"). If we eliminate the top-left corner of the board (which is not involved in any of the 6 parity filters), and take all 257 possible boards on the 63 remaining squares that all have the same parity pattern, these boards form a Hamming code on 63 bits.
BrunoVisnadi wrote:
Also, is there an algorithm to find the 2^(2^N - N - 1) numbers that cover every binary number with 2^N - 1 digits, for higher values of N?
There is a "general algorithm" based on parity and data bits that can be used to generate Hamming codes: https://en.wikipedia.org/wiki/Hamming_code (section "General algorithm"). Note that the Hamming code generated is a (2N-N-1)-dimensional subspace of Z2 ^ (2N-1). The columns of the parity check matrix consist of all possible combinations of nonzero vectors of length N; the ones corresponding to powers of two represent parity bits (N of them) and the others represent data bits (2N-N-1 of them). Since the parity bits are uniquely determined from the data bits, the Hamming code is obtained by plugging in all 2^(2N-N-1) possible sequences for the data bits. To get just a basis of size 2N-N-1, for each of the 2N-N-1 data bits, plug in 1 for that bit and 0 for all other data bits to get its corresponding bitword. The case N=3 is illustrated on Wikipedia.
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To finish the Riddler question, BrunoVisnadi indeed guessed correctly on a previous post that the strategy is improvable to 7/8. This is actually a coding theory problem in disguise:
Let 0=white, 1=black and assign an order to the 7 friends. Map each of the 27=128 possible hat distributions to a binary sequence of 0's and 1's of length 7, e.g. 0110101. This problem then becomes: can at least 1 of the 7 friends correctly guess the sequence given that each one sees every symbol (0 or 1) except their own?
Consider a particular sequence (say 0110101). Say we take a chance and assume that 0110101 is not the sequence. If 0110101 is the actual sequence, then we lose. However, if 0110101 is not the sequence, but 1110101 is, then we can win; friend 1 sees ?110101 and so correctly guesses that it is 1110101. Likewise, we can correctly guess 0010101, 0100101, ..., 0110100, that is, any of the 7 sequences that are one symbol away from 0110101. So for 1 losing sequence, we can generate up to 7 winners (maximum winning rate is thus 7/8).
The strategy is thus: Choose a subset S of the 128 binary sequences, such that the number of sequences that are not in S and are one away from an element of S (and thus sure winners) is maximized. It turns out for n=7 that there is such an S (of size 128/8=16) that exactly achieves the maximum; that is, every element not in S is one away from exactly one element in S. For example:
S={0000000,1101000,0110100,0011010,0001101,1000110,0100011,1010001, 1111111,0010111,1001011,1100101,1110010,0111001,1011100,0101110}
(I.e. all 0's and 1101000 and all its cycles, and then all their complements.)
The strategy: For each friend, if one of the two possibilities is an element of S, guess the one which is not in S; otherwise pass. The strategy wins if the sequence is not in S and loses if the sequence is in S. Win rate: 7/8.
S is what is known as a Hamming code of 7 bits. The Hamming codes are the "optimal" solutions for the S in the strategy above; they occur whenever n=2N-1 and the win rate is (2N-1)/2N. E.g. for n=3, S={000,111}; this is the "guess the other color if you see two hats of the same color and pass otherwise" strategy already described in one of the previous posts and has a win rate of 3/4.
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About the Scaravich totems, the first time you enter totem 1, it randomly sets the totem 1 and 2 areas (from a choice of four areas). Likewise, the first time you enter totem 3, it randomly sets the totem 3 and 4 areas (from a choice of four areas).
So it may take a while to get the right manipulation each time. Not to mention the random layouts within each area.
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OK, I finally understand the problem and why exactly the Riddler said 2N−1. BrunoVisnadi had the right idea; each person is allowed to see both the other hats that the friends are wearing as well as the friends who wear them. This means that the group of friends can set up rules for guessing or passing (based on who they see wears what) to maximize their chances of winning. Remember that they cannot communicate during the game, so effectively, everyone makes their choice at the same time. To win, at least one person needs to guess right, and nobody else is permitted to guess wrong.
There is indeed something special about a group of 3, 7, 15, 31, ... that requires the generalization to be 2N−1. Well, it could be generalized to any n, but if it is not in the form 2N−1, the question becomes far more difficult and requires computer search; in the form 2N−1 it is still difficult but at least there is some theory out there that may be helpful. 7 can be done by hand in any case.
I actually like the hat-only version of the problem (where you can't see your friends so you can only base your choice on the number of hats of each color). The strategy I mentioned above can be generalized to all n; the win probability in this version approaches 2/3 as n approaches infinity. (The strategy sums two of every three binomial coefficients. Using complex numbers, you can show that summing every third binomial coefficient gives one of the two integers nearest 2n/3.)
Warp wrote:
Bobo the King wrote:
sin(18) = sqrt(3 - sqrt(5))/8
That doesn't seem to be correct. Note that sin(18°) is approximately 0.309, while your result is approximately 0.109.
What Bobo the King clearly meant was sqrt((3 - sqrt(5))/8).
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Nice to see X beat this game (instead of Zero) for a change. Though the WARNING sequences and fighting Dynamo over and over is entirely the game at fault. If it would just let you exit from a secondary area...
Oh, and nice to see you back, Rolanmen1.
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I found a slightly better strategy, at 85/128 (~66.4%). Strategy is likewise based solely on number of hats of each color seen (first number is black, second is white):
- 0/6: Guess white (aim for 0/7)
- 1/5: Guess black (aim for 2/5)
- 2/4: Pass
- 3/3: Guess white (aim for 3/4)
- 4/2: Guess black (aim for 5/2)
- 5/1: Pass
- 6/0: Guess white (aim for 6/1)
The strategy wins when the distribution is 0/7, 2/5, 3/4, 5/2, or 6/1, and loses otherwise. The chance of winning is then (1+21+35+21+7)/128 = 85/128.
Can this type of strategy be generalized to n people? Strangely, Riddler's generalization specifically says 2N−1 instead of just n. Since there is nothing we talked about so far that makes 2N−1 special, it makes me wonder whether we missed anything.
Riddler wrote:
then you win a fabulous, all-expenses-paid trip to see the next eclipse
How oddly non-specific this "eclipse" is. I can see the next eclipse just by walking outside at night. :)
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Here's a playaround that was uploaded to nicovideo a few days ago:
Link to video
It beats one of the cups with the smallest number of points (in the standings) possible. Most of this playaround consists of trolling the CPU players.
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Bobo the King wrote:
I believe your error lies where you say, "So the true proportion of 2s in the new sequence is x*(1/3) + (1-x)*(1/4)." Because the strings 3332 and 332 are not of equal length, you can't just average the ratios together to find the the combined ratio. At least I think that's correct.
You're right. I used wrong logic to declare that I can just average the ratios together.
Actually, there is an even more fundamental problem in that I never bothered to find out if it converges at all. Even if the strings were of equal length, it would be flawed to assume that the ratio converges. For example, the ratio of 0s to 1s in 01010011000011110000000011111111... doesn't converge, even though if you looked only up to the 2nd, 4th, 8th, 16th, ... , the ratio is always 1:1.
But, assuming that the ratio of 2s to 3s in the sequence 333233323332332... does converge, we can calculate the ratio. If, at some point of the sequence, there are x 2s and y 3s, then the current ratio is x:y. After replacing 2→332 and 3→3332, the new ratio is x+y:2x+3y. Since we assumed that the ratio converges, then we can take as the limit x:y = x+y:2x+3y. Letting y=1 and x=α, we get:
α:1 = α+1:2α+3
α(2α+3) = α+1
2α2+2α-1 = 0
α = (-1+sqrt(3))/2.
So, assuming that it converges, the ratio of 2s to 3s is (-1+sqrt(3))/2:1 = -1+sqrt(3):2 ≈ 0.366:1, so the ratio of 3s to 2s is 2:-1+sqrt(3) = 1+sqrt(3):1 ≈ 2.732:1. (The proportion of 2s is 2-sqrt(3) ≈ 0.268; the proportion of 3s is sqrt(3)-1 ≈ 0.732).
Proving that it does converge is harder (and judging from past Riddler posts, I don't think Riddler cares). It would go something like this (won't go into the details):
1) If there are x 2s up to the nth symbol, then by replacing 2→332 and 3→3332, there are n 2s up to the (4n-x)th symbol.
2) The ratio n/(4n-x) approximates β=2-sqrt(3) better than x/n does. The key is that (x-βn)*(x-(2+sqrt(3))n) = x2-4xn+n2 is a fixed number (it gives the same value if you replace x→n and n→4n-x).
3) So repeatedly doing the replacement 2→332 and 3→3332 causes x/n to converge to β, for all starting n.
4) Only the numbers n such that the nth symbol is 2 can be targeted directly by 2→332 and 3→3332, but every n is within 3 of such a number.
5) Limits then show that the error between x/n and β goes to 0 as n goes to infinity.
As an aside, x is exactly the integer part of βn (that is, floor(βn)) for all n, and the nth symbol is 2 if and only if the integer part of βn is one more than the integer part of β(n-1).
Edit: Fixed another thing where I switched the roles of 2s and 3s.
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Bobo the King wrote:
Did anyone else get it? I'm looking at you, FractalFusion.
How the sequence is defined (borrowing from Riddler):
The sequence has the property that, if you replace every 3 with 3332 and 2 with 332, the resulting sequence will be the same.
Let x be the long-term proportion of 2s in the sequence. Replace every 3 with 3332 and 2 with 332. If we look at only the 2s being replaced with 332, the proportion of 2s in the new sequence is 1/3, whereas if we look at only the 3s being replaced with 3332, the proportion of 2s in the new sequence is 1/4. So the true proportion of 2s in the new sequence is x*(1/3) + (1-x)*(1/4). Since this replacement does not change the sequence:
x = x*(1/3) + (1-x)*(1/4)
x = (1/12)*x + 1/4
(11/12)*x = 1/4
x = 3/11
So the proportion of 2s in the sequence is 3/11, and the proportion of 3s in the sequence is 8/11. The ratio of 3s to 2s is 8/11 : 3/11, or 8:3.Edit: This post is wrong. See http://tasvideos.org/forum/viewtopic.php?p=457755#457755 for the correction.
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Hm, I haven't seen the word "surd" used a lot. It looks weird.
Anyway, before I look at the video, I'll post two other methods:
Method 1 (along the same lines as arflech):
Let x=(7+sqrt(50))1/3, y=(7-sqrt(50))1/3. We want to find z=x+y.
Now
z3 = x3+3x2y+3xy2+y3 = x3+y3+3xy(x+y) = x3+y3+3xyz
= 7+sqrt(50) + 7-sqrt(50) + 3(7+sqrt(50))(7-sqrt(50))z = 14 - 3z.
So (7+sqrt(50))1/3+(7-sqrt(50))1/3 is a solution to z3+3z-14=0. We see that 2 is a root of z3+3z-14, and it factors as (z-2)(z2+2z+7). Since z2+2z+7 only has complex roots, then 2 is the only real root of z3+3z-14, and so (7+sqrt(50))1/3+(7-sqrt(50))1/3=2.
Method 2 (if you are familiar with the Pell equation x2-2y2=±1):
Note that 7+sqrt(50)=7+5sqrt(2)=(1+sqrt(2))3. (This is recognizable if you already know that 7+5sqrt(2) is the third power of the fundamental solution x+y*sqrt(2) to x2-2y2=±1.) Similarly 7-sqrt(50)=7-5sqrt(2)=(1-sqrt(2))3. Then we immediately get:
(7+sqrt(50))1/3 + (7-sqrt(50))1/3 = 1+sqrt(2) + 1-sqrt(2) = 2.
Edit: Video uses Method 1.
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Russian Slot Machines getting TASed IRL
Just pointing this out, but "Russian Slot Machine Hack" in the wired.com title bar here means "Russian hack of slot machines", not "hack of Russian slot machines". The slot machines mentioned in the article are in other parts of the world, and there isn't a lot of gambling in Russia in the first place.
News headlines can be quite silly.
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Bobo the King wrote:
Oooh, so sorry! The correct answer is 1/29. The initial holder of the potato has zero probability of winning. :P
Based on the statement of the question, I'm getting 1/30, if I'm understanding this right.
Riddler Express wrote:
A class of 30 children is playing a game where they all stand in a circle along with their teacher. The teacher is holding two things: a coin and a potato. The game progresses like this: The teacher tosses the coin. Whoever holds the potato passes it to the left if the coin comes up heads and to the right if the coin comes up tails. The game ends when every child except one has held the potato, and the one who hasn’t is declared the winner.
By my understanding, the circle is 31 people large, and none of the 30 children start with the potato, because the teacher does.
Riddler Express: It's one of those questions where either you get it or you don't.
Every child has the same chance (1/30) of winning. At some point, the potato is first held by either of the two people beside you; let's call them A and B, and without loss of generality, A has the potato; thus B never had it. To win, the potato must be passed from A to B the long way around (i.e. not through you). This applies to every child regardless of how the potato was passed previously; therefore every child has the same chance of winning.
Riddler Classic: I assume likewise as Bobo the King did: whoever attempts to use the washroom leaves immediately if the bathroom is occupied (i.e. the only person modifying the door sign is the current user of the bathroom).
First, in the long run, the state of the door sign when vacant tends to probability 1/2 for both "occupied" and "vacant" (this is because only 2/3 of the users affect the sign: half of them set it to "occupied" and half of them set it to "vacant").
Given that, we can calculate the probability of cases. I call user behavior T1 if they ignore the sign, T2 if they set it to "occupied" but not "vacant" at the end, and T3 if they set it to "occupied", then "vacant" at the end. Each behavior occurs with equal probability (1/3).
occupied, user is T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
occupied, user is T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, user is T3, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, sign="occupied": 5/12
occupied, sign="vacant": 1/2 - 5/12 = 1/12
vacant, last user was T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
vacant, last user was T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
vacant, last user was T3, sign="occupied": 1/2 * 1/3 * 0 = 0
vacant, sign="occupied": 1/4
vacant, sign="vacant": 1/2 - 1/4 = 1/4
sign="occupied": 5/12 + 1/4 = 2/3
sign="vacant": 1/3
occupied given sign="occupied": (5/12) / (2/3) = 5/8
vacant given sign="vacant": (1/4) / (1/3) = 3/4
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Bobo the King wrote:
What I think is interesting is that for k<<n, ak(n) settles pretty closely to (n/k)^k for the reason I outlined above.
If indeed ak(n) is well-approximated by (n/k)k, then the maximum would occur at k=n/e (by setting the derivative of ln((n/k)k) to be 0 and solving for k). However, I'm not sure that the behavior of ak(n) is well-approximated by (n/k)k when k is not significantly small compared to n.
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What I find most interesting about this game is that it presents itself as a Super Mario World port at first (although a bad one at that), and then it slowly deteriorates over time until it no longer resembles SMW. Which I guess is a little more interesting than just being a bad but faithful SMW port all the way.
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Oh nice. You managed to use just 3 deck edits!
It would indeed be difficult to improve on this. Better luck manipulation maybe, but as is typical of TASes, the amount of effort required per second gained tends to grow exponentially with the revision number of the TAS.
What's your next TAS project? Is it Pokemon Card GB2?
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OmnipotentEntity wrote:
(1) n is the smallest m for which p divides Fm, and p2 also divides Fn.
Does this imply that n = m? That seems nonsensical, so I must be misunderstanding. What is the relationship between n and m?
What I should have said was:
"n is the smallest positive integer i for which for which p divides Fi"
or
"n is the smallest {i is a natural number: p divides Fi}"
or
"p divides Fn and p does not divide Fi for 0<i<n."
I have edited the above post.
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OmnipotentEntity wrote:
With the exception of O6 = 4 these numbers seem to be square free. Are they?
In fact, prime powers dividing Fibonacci numbers has been studied before. What is known so far may surprise you.
There are two ways in which On can be non-square-free. For a prime p:
(1) n is the smallest positive integer i for which p divides Fi, and p2 also divides Fn.
(2) n is the smallest positive integer i for which pk divides Fi (k≥2), and pk+1 also divides Fn.
Start with case (2) first.
Case (2):
We have the identity (when m≥0)
Fmn = Sum［i=1 to m］(C(m,i) Fni Fn-1m-i Fi) (*)
(Click for better mathematical formatting; note that C(m,i) is binomial coefficient .) See bottom of post for details. I won't prove (*) unless someone asks.
Let pk (k≥1) be the highest power of p dividing Fn. Take (*) mod pk+1:
Fmn ≡ mFnFn-1m-1.
Since consecutive Fibonacci numbers are relatively prime, the smallest m for which pk+1 divides the RHS is p. Let m=p and take (*) mod pk+2:
Fpn ≡ pFnFn-1p-1 + C(p,2)Fn2Fn-1p-2.
If p is an odd prime (so that p divides C(p,2)), or if k>1, then C(p,2)Fn2Fn-1p-2 ≡ 0 (mod pk+2) and so
Fpn ≡ pFnFn-1p-1 ≢ 0 (mod pk+2).
So whenever Fn is the smallest Fibonacci number divisible by pk (k≥1), and pk+1 does not divide it, then the smallest Fibonacci number divisible by pk+1 is Fpn, and pk+2 does not divide it. The only exception is when p=2, k=1 and that corresponds to O6 = 4, a non-square-free number.
Case (1):
Here's the part which is up in the air and nobody knows.
If n is the smallest m for which p divides Fm, and p2 divides Fn, then p is a Wall-Sun-Sun prime. (Can be shown π(p)=n or 2n or 4n, where π(p) is the Pisano period as defined on that page.) No Wall-Sun-Sun primes are known to exist and there are none below 9.7×1014.
So all this confirms that O6 = 4 is very likely the only exception within a reasonable range.
------------------------------------
About (*): (*) is generalized as (when m≥0):
Fmn+r = Sum［i=0 to m］(C(m,i) Fni Fn-1m-i Fi+r) (**)
which is identity (7) in this paper. (**) can be proved using Fm+n = Fn+1Fm + FnFm-1 along with some clever induction.
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Warp wrote:
Famously, lim(x->∞) (1 + a/x)bx = eab
This got me thinking what happens if that 1 were something else.
Not that hard. Consider lim(x->∞) (c + a/x)bx. If c>0, then:
lim(x->∞) (c + a/x)bx = lim(x->∞) cbx lim(x->∞) (1 + a/(cx))bx = eab/c lim(x->∞) cbx,
which results in:
● ∞, if c>1 and b>0, or 0<c<1 and b<0,
● 0, if 0<c<1 and b>0, or c>1 and b<0,
● eab, if c=1,
● 1, if b=0.
If c=0, then:
lim(x->∞) (a/x)bx = e^(lim(x->∞) bx ln(a/x)),
which results in:
● 0, if a≥0 and b>0,
● ∞, if a≥0 and b<0,
● 1, if b=0.
All other cases result in real powers of negative numbers or 00, which are undefined.
Warp wrote:
The challenge is not finding the answer. The challenge is to find a geometric proof (or, could it perhaps be more accurate to say "geometric argument"?) that requires no math at all.
Here's an image I made which shows why the green square is twice the area of the red square:
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Let's suppose that, instead of meeting at x=0.5, the segments meet at x=d where d is in (0,1). Then we write the segments as:
y=A(x/d)m, x in [0,d]
y=1-B((1-x)/(1-d))n, x in (d,1]
The function must be continuous and differentiable at d. This gives:
A = 1-B
Am/d = Bn/(1-d)
Solving the equations for A and B gives A=dn/(dn+(1-d)m) and B=(1-d)m/(dn+(1-d)m). The segments meet at x=d, y=dn/(dn+(1-d)m).
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Found the mistake. I put A=n2n(m+n) when it should have been A=n2m(m+n). (Apologies, that was a really dumb error on my part.) That gives y=n/(m+n) at x=0.5. So now y is clearly between 0 and 1.
Warp wrote:
Edit: By the way, your functions are not correct. They should be:
y = A(2x)m, x=[0, 0.5]
y = 1 - B(2(1-x))n, x=[0.5, 1]
It doesn't really matter; y = A(2x)m can be rewritten as y = Cxm where C = A2m. Similarly y = 1 - B(2(1-x))n can be rewritten as y = 1 - D(1-x)n where D = B2n. Edit: But now I see that it would probably have been better to solve it that way.
The sequence has the property that, if you replace every 3 with 3332 and 2 with 332, the resulting sequence will be the same.
.) See bottom of post for details. I won't prove (*) unless someone asks.
Let pk (k≥1) be the highest power of p dividing Fn. Take (*) mod pk+1:
Fmn ≡ mFnFn-1m-1.
Since consecutive Fibonacci numbers are relatively prime, the smallest m for which pk+1 divides the RHS is p. Let m=p and take (*) mod pk+2:
Fpn ≡ pFnFn-1p-1 + C(p,2)Fn2Fn-1p-2.
If p is an odd prime (so that p divides C(p,2)), or if k>1, then C(p,2)Fn2Fn-1p-2 ≡ 0 (mod pk+2) and so
Fpn ≡ pFnFn-1p-1 ≢ 0 (mod pk+2).
So whenever Fn is the smallest Fibonacci number divisible by pk (k≥1), and pk+1 does not divide it, then the smallest Fibonacci number divisible by pk+1 is Fpn, and pk+2 does not divide it. The only exception is when p=2, k=1 and that corresponds to O6 = 4, a non-square-free number.
Case (1):
Here's the part which is up in the air and nobody knows.
If n is the smallest m for which p divides Fm, and p2 divides Fn, then p is a Wall-Sun-Sun prime. (Can be shown π(p)=n or 2n or 4n, where π(p) is the Pisano period as defined on that page.) No Wall-Sun-Sun primes are known to exist and there are none below 9.7×1014.
So all this confirms that O6 = 4 is very likely the only exception within a reasonable range.
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About (*): (*) is generalized as (when m≥0):
Fmn+r = Sum［i=0 to m］(C(m,i) Fni Fn-1m-i Fi+r) (**)
which is identity (7) in this paper. (**) can be proved using Fm+n = Fn+1Fm + FnFm-1 along with some clever induction.The challenge is not finding the answer. The challenge is to find a geometric proof (or, could it perhaps be more accurate to say "geometric argument"?) that requires no math at all.
