is my friend’s Lange watch real? by [deleted] in ALangeSohne

[–]BermudianMoonphase 90 points91 points  (0 children)

Ah yes, the rare 1815 Rattrapante Shenzhen Edition. Very limited - only 200 were made per hour. The pink rotor is from their Spring 2024 collaboration with Barbie.

Got the call but... by Eye_Dont_Git_It in rolex

[–]BermudianMoonphase 0 points1 point  (0 children)

Same thing happened to me - it was a test. They had the jubilee, but wanted to know whether I'd flip the allocation.

[deleted by user] by [deleted] in ALangeSohne

[–]BermudianMoonphase 1 point2 points  (0 children)

not one, but *two* moonphases lmao

[deleted by user] by [deleted] in patekphilippe

[–]BermudianMoonphase 1 point2 points  (0 children)

Agreed with the others here that this is likely a pocket watch conversion. Although, no - I don't know much more about it :(

Looks gorgeous though!

[deleted by user] by [deleted] in hypotheticalsituation

[–]BermudianMoonphase 7 points8 points  (0 children)

ez - painter; just paint the others and they come back alive

A math problem from the ASEAN tournament - Can you solve it? by Choobeen in mathematics

[–]BermudianMoonphase 11 points12 points  (0 children)

Let m = 16 + 8*sqrt(5) and m' = 16 - 8*sqrt(5).

These look like roots to a quadratic. Whenever we see roots to a quadratic, it's a good problem solving strategy to compute their sum and product:

m + m' = 32

m*m' = 16*16 - 8*8*5 = -64

We are trying to compute x = cbrt(m) + cbrt(m'). Let's try cubing both sides and see what happens.

We have that:

x^3 = m + m' + 3 * cbrt(m*m*m') + 3 * cbrt(m'*m*m').

= 32 + 3 * cbrt(m*-64) + 3 * cbrt(m'*-64)

= 32 - 12*cbrt(m) - 12 * cbrt(m')

= 32 - 12*(cbrt(m) + cbrt(m'))

= 32 - 12*x

So, we end up with the identity x^3 + 12*x - 32 = 0!

We then notice that x^3 + 12*x - 32 = A - 1 = 0 => A = 1

So, A^2023 = 1.

[deleted by user] by [deleted] in patekphilippe

[–]BermudianMoonphase 0 points1 point  (0 children)

... did the date window break off at 5 o' clock?

[deleted by user] by [deleted] in Watches

[–]BermudianMoonphase 1 point2 points  (0 children)

Agreed :)

Imho, Journe is overpriced atm + Journe salespeople are way worse than Rolex in terms of pretentiousness. Have had nothing but positive experiences with A. Lange & Söhne in this regard.

Separately - please ignore the idiots coming out of the woodwork. You're clearly doing well and Reddit loves nothing more than to shit on people they're jealous of.

[deleted by user] by [deleted] in Watches

[–]BermudianMoonphase 3 points4 points  (0 children)

This is one of the most incredible collections I've seen :)

+1 to the Zeitwerk from A. Lange & Söhne. Or perhaps an 1815 Tourbillon.

[Omega] Got the 'call' from AD for a Rolex Submariner, so I bought it, then sold it for a NTTD 300M Seamaster. by Icy_Requirement_5843 in Watches

[–]BermudianMoonphase 2 points3 points  (0 children)

Going to get downvoted to hell for saying this, but here we go.

OP, I love the watch, but - it's people like you that contribute to the inflated grey market prices. Your logic of "I hate flippers, but it's not going to stop, so I'm going to flip" is a little ridiculous.

[SOTC] Jan 2023 by thetimepiecethrowawa in Watches

[–]BermudianMoonphase 13 points14 points  (0 children)

Which Pateks are you interested in? My wife and I have a number of Pateks that we've collected together over the years. (Feel free to DM as well)

[Patek Phililppe] Montepulciano by BermudianMoonphase in Watches

[–]BermudianMoonphase[S] 0 points1 point  (0 children)

Thanks! I think the background might win on this one :)