Someone Else
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Well, let's see. The triangle is like this:

    B
 -------  A1: H^2 + (B-b)H/2
H|..../   P1: H+B+(B-b)+sqrt(H^2+(B-b)^2)
 -----
   b
 ----     A2: bh/2
 |./      P2: b+h+sqrt(b^2+h^2)
h|/
 |

We know that the second triangle is isomorphic to the original triangle. Without loss of generality, we can set both B and b. So let B=2, b=1. Then

2(H^2+H/2)=h

b+h+sqrt(b^2+h^2)=H+B+(B-b)+sqrt(H^2+(B-b)^2) 
=> 1+2(H^2+H/2)+sqrt(1+4(H^2+H/2)^2)=H+3+sqrt(H^2+1)

Wolfram Alpha helpfully tells us that this means

H=3/4 => h=15/8.
sin(2/(3/4+15/8)) = 69 degrees

So 21, 69, 90!
I don't know what you mean by "no abstruse formulae are needed".

Nuntar
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Well, for starters, the area of a trapezium is not H^2 + (B-b)H/2

A more significant error, which I'll put in secret tags as it contains a hint to the correct solution:


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[Last edited by Nuntar at 12-18-2013 10:29 PM]

Lucky Luc
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File: triangle.png (2.1 KB)
Downloaded 741 times.
License: Public Domain

Hmm, maybe I'm missing something, but I think it's impossible to achieve this:


Feel free to prove me wrong

Nuntar
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a^2 = b^2 + c^2 is only true if "a" is the hypotenuse

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Lucky Luc
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Oops, I knew I missed something obvious

EDIT: Okay, my next offer is 45°, 45° and 90° (how boring!). My solution isn't very pretty and I'm almost certain I'm missing a trick somewhere, but it should be correct this time:

Up to

a=(sqrt(2)-1)*(b+c)    (I)

everything was fine, so let's try some trigonometry now. Let's call the bottom angle A.

sin(A)=a/c
cos(A)=b/c
(I)/c:
sin(A)=(sqrt(2)-1)^2*(cos(A)+1)^2             | ^2
1-cos^2(A)=(sqrt(2)-1)^2*(cos(A)+1)^2         | x:=cos(A)
1-x^2=(3-sqrt(2))*(x+1)^2
(1+x)(1-x)=(3-2*sqrt(2))*(x+1)(x+1)
1-x=(3-2*sqrt(2))*(1+x)
(4-2*sqrt(2))*x=(2*sqrt(2)-2)
x=1/sqrt(2) (That's what my calculator says, at least)
=> A=45°


[Last edited by Lucky Luc at 12-18-2013 11:12 PM]

RyTracer
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I agree: 45°, 45°, and 90°. Though I don't follow Lucky Luc's logic, but I get the same answer.

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Nuntar
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You've got it

Your logic is substantially correct, so I'll see if I can make it clearer:

a = (sqrt(2)-1)*(b+c)
a/c = (sqrt(2)-1)*b/c + (sqrt(2)-1)*c/c         | divide through by c
sin(A) = (sqrt(2)-1)*cos(A) + (sqrt(2)-1)       | using sin(A)=a/c and cos(A)=b/c
sin^2(A) = (sqrt(2)-1)^2 * (cos(A)+1)^2         | group RHS and square both sides
1 - cos^2(A) = (sqrt(2)-1)^2 * (cos(A)+1)^2     | using sin^2 + cos^2 = 1
1 - x^2 = (3 - 2 sqrt(2)) * (x+1)^2             | let x = cos(A), and multiply out first term of RHS
(1+x)(1-x) = (3 - 2 sqrt(2)) * (x+1) * (x+1)    | factorise diff. of two squares
1 - x = (3 - 2 sqrt(2)) * (1+x)                 | divide by 1+x
(4 - 2 sqrt(2)) * x = 2 sqrt(2) - 2             | multiply out RHS and rearrange
(sqrt(16) - sqrt(8)) * x = sqrt(8) - sqrt(4)
x = 1/sqrt(2)
A = 45°

As for my own solution, well... it's a bit of a cheat.

a = (sqrt(2)-1)*(b+c)
tan A = (sqrt(2)-1)*(sec A + 1)                 | scale triangle so b = 1
tan A (sqrt(2)+1) = sec A + 1                   | using the handy fact that (sqrt(2)-1) * (sqrt(2)+1) = 1

from which the solution of 45° is "spottable".

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[Last edited by Nuntar at 12-21-2013 10:00 PM]

Lucky Luc
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Yay!

Hm, you probably know this puzzle, but I can't think of any better right now:

You have two pieces of string and a lighter. You know that if you light a string from one side, it will take exactly 60 minutes to burn down; however it doesn't burn down consistently, the first half of it might only take ten minutes while the second half of it needs 50. How can you measure 45 minutes exactly?

RyTracer
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I'll go ahead and solve this. Light three of the four ends at the same time. When the double-lit string burns completely away, light the last end. When the second string is gone, 45 minutes have elapsed since the beginning.

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Lucky Luc
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Thanks

Not very surprisingly, that's correct. Your turn!

RyTracer
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The following notation represents 3-digit numbers, not three terms multiplied together. What are the digits?

XYZ+XZY=YZX

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Nuntar
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Okay, this to earn my 1000th rank point!

Any number is equivalent to the sum of its digits, mod 9. Therefore, if R is the digital root of XYZ, R obeys R + R = R (mod 9), therefore R = 9.

From the units place, Z + Y = X or X + 10, but the former is impossible as X < Y.

Therefore 2(Z + Y) = X + Y + Z + 10 = 19 or 28. The former is ruled out as 2(Z + Y) is even.

Therefore Z + Y = 14, X = 4, and there is a carry in the tens place, so Y = 2X + 1 = 9.

The solution is 495 + 459 = 954.

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Penumbra
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edit: dagnabbit!

Answer X = 0, Y = 0, Z = 0

Or, if you don't like that one . . .

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[Last edited by Penumbra at 01-14-2014 12:33 AM]

RyTracer
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Well done both of you, but a bit like killing spiders with a shotgun. Nuntar is up.

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TripleM
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RyTracer wrote:
Well done both of you, but a bit like killing spiders with a shotgun.

Heh, indeed. The tens digits mean Y=0 or 9; the former gives the normally-invalid 000 solution, while Y=9 means, based on the hundreds digits, X=4 and therefore Z=5.

Nuntar
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A firm manufactures tiles that are all regular polygons with side length 10cm. There are a finite number of ways to arrange these tiles around a point so as to fit exactly (for example, four squares, or two squares and three triangles). My question is: what is the largest polygon (greatest number of sides) that can be used in one of these arrangements?

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Someone Else
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It's simple: a hexagon.
Or at least, that's true if the tiling only uses one shape... I know that octagons can be used, but I don't know for 12-gons or 16-gons. :/

Nuntar
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Dodecagons can tile the plane in combination with triangles, squares and hexagons. However, the question is about the possible arrangements of polygons around a single point, including ones that cannnot be extended to tile the plane.

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Maurog
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Just use two infinigons.

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Paint the sky in crimson red!

TripleM
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stigant
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I think you the biggest is


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TripleM
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Ooh, you're right. My mistake.

Nuntar
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Over to you, stigant

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stigant
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Ok, Grover the Grocer is stacking oranges in a pyramid. The base of the pyramid is a square of oranges, and each layer of oranges is a square, with sides one orange smaller than the layer below it, and there is a single orange at the top of the pyramid. So for example, when the pyramid has 4 layers, the layers consist of 1, 4, 9, and 16 oranges. After stacking for a while (ie his pyramid has several layers), Grover realizes that he has completed a pyramid in which the total number of oranges is a perfect square. How many oranges are in his pyramid?



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RyTracer
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When the pyramid is at its 24th level, there will be 4900 oranges in it, which is 70 squared.

The progression is x^3/3+3x^2/2+13x/6+1, which I put into my calculator table and then scanned the list by sight. Thankfully I only had to go to x=23 before I recognized a square.

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RyTracer
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If I recall correctly, the number of dimensions in string theory is 11. How are these two problems related?

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stigant
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the number of dimensions in string theory is 11

Yeah, clearly your answer is wrong ;-) j/k

I went looking for the article that I was reading, and couldn't find it. I think it was a particular flavor of String Theory (Bosonic or something) which predicts that there are actually 24 + 2 = 26 dimensions, and that has something to do with the fact that 1^2 + 2^2 + 3^2 + ... + 24^2 is the only series of squares that sums to a square.

At any rate, you've found the correct answer, so you're up.

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stigant
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Well, this isn't the article I was talking about, but it mentions the result:
Cannon Ball Problem


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RyTracer
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Alrighty then. To my knowledge, this is my own original variation to a classic. The classic is Cannibals and Missionaries, and some people have even made flash games of it. You should google them.

EDIT: This is not how I posted this originally. This is how I should have posted it. Try this one. It is harder.

There are 5 Lions, 3 Cannibals, and 3 Missionaries that want to cross a river to the other side. They have a boat that will hold up to 3 Passengers (a combination of Lions, Cannibals, and Missionaries; Lions take up 2 human seats), and must have at least 1 Passenger in it to guide it across (yes, the Lions can do this; they're very talented). The trouble is that Cannibals eat Missionaries if they ever outnumber them, and Lions eat Cannibals and Missionaries if they ever outnumber the combined Human total. Consumption can happen just after the boat has left among the remaining population, while in transit among the sailors, or when the boat arrives and everyone disembarks with the new population and all passengers must disembark every time. How can you shuttle everyone over without anyone getting eaten?

I can do it in this many one-way boat trips, but that may not necessarily be the most efficient.


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[Last edited by RyTracer at 01-23-2014 06:55 AM]

Penumbra
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I feel oddly compelled to post this.


Ok, now to write a script to solve this one!