It's our old buddy correlation with population size at work.
Geometry puzzle

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Re: Geometry puzzle
People who believe God or History are on their side provide the chaos.

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Re: Geometry puzzle
Abdul Alhazred wrote: ↑Thu Sep 17, 2020 3:13 pmIt's our old buddy correlation with population size at work.
More high school geometry students available?

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Re: Geometry puzzle
Math is hard
still working on Sophrosyne, but I will no doubt end up with Hubris

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Re: Geometry puzzle
Geometry even harder
still working on Sophrosyne, but I will no doubt end up with Hubris

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Re: Geometry puzzle
A back of the envelope calculation gives me 9. If that's correct I'll write up my solution. (If not I'll just hang my head in shame.)

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Re: Geometry puzzle
I’m an idiot
Last edited by robinson on Sat Sep 19, 2020 6:05 am, edited 1 time in total.
still working on Sophrosyne, but I will no doubt end up with Hubris

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Re: Geometry puzzle
One of you please explain how to get there.
People who believe God or History are on their side provide the chaos.

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Re: Geometry puzzle
Jesus fucking christ almighty
still working on Sophrosyne, but I will no doubt end up with Hubris

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Re: Geometry puzzle
It's like posting the Riemann Hypothesis in a puzzle forum. Then laughing like a maniac later that night.
still working on Sophrosyne, but I will no doubt end up with Hubris

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Re: Geometry puzzle
Here's a simple geometric proof by Pan Narrans (a poster at the Freethought Forum). Pan was actually the first poster to solve my posting of the puzzle, but I posted the puzzle there about the same time as here, and before I did at ISF, so the ISF guys were the fastest. I didn't invent the puzzle by the way  I first saw it posed in a YouTube "recommended for you" still  but I didn't watch the video and solved it myself instead.
AD = AE = AG = AK (the radius of the small circle)
similarly
BD = BF = BH = BL (the radius of the large circle)
Triangle ADK is an isosceles triangle, with A as its apex. The same goes for triangle BDL, with apex B. The "bases" of the two isosceles triangles (the non equal sides) are KD and DL. Note that the line AC is the same length as half the base of the small triangle plus half the base of the large triangle
AC = (KD + DL)/2, and we know that KD + DL = 6, so AC = 3
Now AG + AC + BH = 8 (the height of the rectangle)
Subtracting AC = 3 gives AG + BH = 5 (the sum of the radii)
Since AD + BD = AG + BH = 5, we have a 3/4/5 right triangle ABC with a hypotenuse (AB) 5 and one side (AC) 3, the other side (BC) must be 4.
X = AE + BF + BC = AG + BH + BC = 5 + 4 = 9
Pan did a clever proof using trigonometry and simultaneous equations, similar to Witness's, first, but then followed it up with this simpler geometric one.
The way I did it was something of a cheat. I reasoned that the puzzle wouldn't have been set unless it had a neat constant answer  not some horrible formula depending on the sizes of the circles drawn. So I figured I could make the smaller circle as small as I wanted without changing the answer. If you draw it with the small circle diameter zero, then you just have a rectangle with short side 8, and the large circle touching the bottom and right hand side of the rectangle and passing through the top left hand corner. The left hand edge of rectangle cuts a chord length 6 across the circle. Then the 3.4.5 triangle is immediately obvious, and you get the solution in a flash.
AD = AE = AG = AK (the radius of the small circle)
similarly
BD = BF = BH = BL (the radius of the large circle)
Triangle ADK is an isosceles triangle, with A as its apex. The same goes for triangle BDL, with apex B. The "bases" of the two isosceles triangles (the non equal sides) are KD and DL. Note that the line AC is the same length as half the base of the small triangle plus half the base of the large triangle
AC = (KD + DL)/2, and we know that KD + DL = 6, so AC = 3
Now AG + AC + BH = 8 (the height of the rectangle)
Subtracting AC = 3 gives AG + BH = 5 (the sum of the radii)
Since AD + BD = AG + BH = 5, we have a 3/4/5 right triangle ABC with a hypotenuse (AB) 5 and one side (AC) 3, the other side (BC) must be 4.
X = AE + BF + BC = AG + BH + BC = 5 + 4 = 9
Pan did a clever proof using trigonometry and simultaneous equations, similar to Witness's, first, but then followed it up with this simpler geometric one.
The way I did it was something of a cheat. I reasoned that the puzzle wouldn't have been set unless it had a neat constant answer  not some horrible formula depending on the sizes of the circles drawn. So I figured I could make the smaller circle as small as I wanted without changing the answer. If you draw it with the small circle diameter zero, then you just have a rectangle with short side 8, and the large circle touching the bottom and right hand side of the rectangle and passing through the top left hand corner. The left hand edge of rectangle cuts a chord length 6 across the circle. Then the 3.4.5 triangle is immediately obvious, and you get the solution in a flash.
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Re: Geometry puzzle
Finally got it. Messy and convoluted, but it works. That was fun.
Such potential!

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Re: Geometry puzzle
It was quite satisfying to see a 3,4,5 right triangle emerge from my mess.
Such potential!

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Re: Geometry puzzle
People who believe God or History are on their side provide the chaos.

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Re: Geometry puzzle
Yeah, sometimes they emerge from my ass when I'm dropping a log.
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Freedom of choice
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Freedom from choice
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"what dicking deep shit i produce"  pillory
Freedom of choice
Is what you got
Freedom from choice
Is what you want
People are shitting themselves to death
Crap so much they fail to take a breath
But even when their kids are starvin'
They thought Trump would throw them Charmin.