Hewwo

I got nothing done at work today because I have been obsessed with this sink. This is very much a me problem, nevertheless I have Findings.

Some astute tumblrinas (to whom I am indebted for the waste of an afternoon) noted the knob had the name of the Italian design company Ritmonio. My wild goose chase through all the products on their website yielded nothing like this sink.

I had more success with a reverse image search. This led me to this article by Leicht Los Angeles (a kitchen cabinetry company).

A basic web search for Sonobath turned up these articles by Kitchen and Bath Business.

However, the Sonobath website seems defunct. So naturally, I plugged it into the Wayback Machine anddddd

Bumdada da! Dumb sinks galore! The sink in your photo seems to be a version of the Titus sink.

Yes, all the models of sinks have names like that. The model Hydros has the tagline "vessel sink or interactive sculpture? you decide." It looks like this.

Which, I'll be honest, looks exactly like the Ethos sink, but rotated 90 degrees and rocks added. Or perhaps it's the inverse. Who knows!

The genius? behind these "sinks" seems to be one Jae Omar. He seems to have transitioned from bathrooms to whole houses. He has a new website.

I have no point and no final conclusion (at least, not beyond "rich people have remarkably poor taste"). But I spent all day figuring this out and none of my friends have been appreciative of my efforts. So now I'm dumping it on to tumblr.

From the second equation we get that cheeseburger plus fries-squared is five. Subtracting cheeseburger, which is six, from both sides, we get that fries-squared is negative-one. Math fans will know that there are two solutions to this; either fries are the "imaginary unit" 𝒾 or they are its negative, -𝒾. We'll do the rest of the problem with 𝒾, keeping in mind that at the end we should also take the complex conjugates as solutions.

Finally, we have that cup to the power of fries, minus cup, equals three. Replacing fries with 𝒾, and moving a cup to the other side, we get that cup-to-the-𝒾 is equal to cup-plus-three.

Now, the weird part about this is the cup-to-the-i. The problem with this is that complex exponentiation is technically not a thing. That is to say, there is no one function which is mathematically equal to "input-to-the-power-of-𝒾". In fact, there are infinitely many such functions.

Fortunately, due to reasons that take about six pages to explain (trust me I've done it), there is one particular function that many people have agreed is "the most reasonable one". This is not a mathematical notion, but a human preference. Seeing as this question was presumably written by a human, I am comfortable with using this function.

So, what function is this? Well, given a complex number r∠θ written in polar form (if you don't know what that means don't worry), where -π < θ ≤ π, then (r∠θ)^𝒾 = e^(-θ)∠ln(r).

Applying this to our problem a value r∠θ will be a possible solution for cup if e^(-θ)∠ln(r) = r∠θ + 3. Splitting this into real and imaginary parts, we get two equations: e^(-θ) cos(ln(r)) = r cos(θ) + 3 and e^(-θ) sin(ln(r)) = r sin(θ). We can graph these equations on Desmos:

let the value of the glass = z, for z ∈ ℂ:

z^(±i) = 3+z

let z = r·e^(iθ) for r,θ ∈ ℝ, -π < θ ≤ π

∴ z = r·e^i(θ+2πn) for all n ∈ ℤ

∴ (r·e^i(θ+2πn))^(±i) = 3+r·e^i(θ+2πn)

distribute powers (apologies for the use of ∓):

r^(±i)·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

convert to the same base:

e^i(±ln(r))·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

split into real and imaginary components:

re: cos(±ln(r))·e^(∓(θ+2πn)) = 3+r·cos(θ)

im: sin(±ln(r))·e^(∓(θ+2πn)) = r·sin(θ)