Trivia Contest 2021

[Ferrus]

𑁜
James K. Polk

[Ferrus]

𑁜𑁓
Israel

[Utisz]

𑁗

Curve is catenary (thanks to the physics of hanging chains) shifted to origin so c*cosh(x) - c.

Arch length of c*cosh(x) - c = I_a^b cosh(x) = sinh(x)

(derivative in arclength term cancels out constant)

Therefore c*(sinh(1) -sinh(-1)) = 2

Therefore c*2.35 ≈ 2
c ≈ 2.35/2 ≈ 0.85

So curve is approximately 0.85*cosh(x) - 0.85

0.85*cosh(1) - 0.85 or 0.85*cosh(-1) - 0.85 ≈ 0.46

Not what I have! You're on the right track though. (I used a different method, but sinh(1) -sinh(-1) = 0? Also c ≈ 2/2.35? I think the latter is just a typo though.)

𑁔

At Oklo a site was flooded with groundwater then that acted as a natural neutron moderator to the uranium deposits and therefore formed a natural nuclear reactor.

Correct! 10 points.

𑁛𑁖

Clallam county, Washington State. Has been bellwether (neutered ram, a wether, with bell used to lead flock of sheep) since 1980.

Correct! 10 points.

𑁚

Ossip Bernstein, died 80. Was arrested by the Cheka in Odessa in 1918 due to supposed counterrevolutionary activities and due to be executed by firing squad. However at the last minute the commanding office upon finding out his identity as a chess player offered his life if he could win a game of chess against him, which he did, thus saving his life.

Correct! 10 points.

𑁓

L'Atelier Rouge by Henri Matisse.

Correct! 10 points.

𑁜
James K. Polk

Correct! 10 points.

𑁜𑁓
Israel

Not the one. You're on a roll though!

[Utisz]

0.46

Just on this, I think it's okay to mention that it cannot be 0.46 m. A reasonable upper bound would be if it formed (three sides of) a rectangle, in which case it would be 2*0.46 m + 1 m, which is less than 2 m.

So it must be >0.5 m.

[Ferrus]

0.46

Just on this, I think it's okay to mention that it cannot be 0.46 m. A reasonable upper bound would be if it formed (three sides of) a rectangle, in which case it would be 2*0.46 m + 1 m, which is less than 2 m.

So it must be >0.5 m.

sinh is odd function (https://images.app.goo.gl/jsgPaysUfMuX8TDHA), so it is possible to get a length from that calculation (cosh cancels). Also the integration was between 1 and -1 which was incorrect it should be -0.5 and 0.5. sinh(0.5) - sinh(-0.5) is 2sinh(0.5)

However I realised the actual equation for a catenary is a*cosh(x/a) not a*cosh(x).

if you do the arclength integral calculations the arclength of a catenary is asinh(x/a)

My workings:

a(sinh(0.5/a) - sinh(-0.5/a)) = 2

2a(sinh(0.5/a)) = 2

a*sinh(1/2a) = 1

So...

I will just graph a*sinh(1/2a) - 1, graphing that I get 0.23 as the minimum.

So the equation is 0.23cosh(x/0.23) - 0.23

(As acosh(0) = a)

at 0.5 and -0.5 the height is:

0.79

it strikes me there is an upper bound by pythagoras of 0.87

and a lower bound of 0.5 as you mentioned.

So after a sanity check this result seems at least plausible.

graphs here:

https://www.desmos.com/calculator/9xhchsv4o2

[Ferrus]

𑁜𑁓

Jordan

[Ferrus]

𑁒

William Faulkner

[Ferrus]

𑁛𑁕

A compass. These areas in Ontario, Belgograd and the Central African Republic are all within magnetic anomalies.

[Ferrus]

𑁕

Senseye, 1916

[Ferrus]

𑁜𑁒

Ötzi the Iceman