Can SBOT solve this math problem?

[Soccerplayer2233]

I posted my answer on OT and saw you mention SBOT. I see that there's some reward so I thought I'd post it here as well:
Points:
p=(u+v)/2 - i * (v-u)/2
q = (v+w)/2 - i*(w-v)/2
r = (w+z)/2 - i *(z-w)/2
s = (z+u)/2 - i * (u-z)/2

Line segments:
p-r = (u+v-w-z)/2 - i (-u+v+w-z)/2
q-s = (-u+v+w-z)/2 - i (-u -v+w+z)/2

Line segments are equal length and perpendicular:
p-r = (u+v-w-z)/2 - i (-u+v+w-z)/2 =- i*[i*(u+v-w-z)/2+(-u+v+w-z)/2]=-i*[(-u+v+w-z)/2-i*(-u-v+w+z)/2] = -i*(q-s)
(p-r)/(q-s) = -i, and are therefore perpendicular.
As well, |p-r|=|-i|*|q-s| = |q-s| and therefore have equal length.

The problem is pretty fun and geometric if you know your complex geometry.

Edit: Fixed some negative signs.

 

[GiantWizard]

I posted my answer on OT and saw you mention SBOT. I see that there's some reward so I thought I'd post it here as well:
Points:
p=(u+v)/2 - i * (v-u)/2
q = (v+w)/2 - i*(w-v)/2
r = (w+z)/2 - i *(z-w)/2
s = (z+u)/2 - i * (u-z)/2

Line segments:
p-r = (u+v-w-z)/2 - i (-u+v+w-z)/2
q-s = (-u+v+w-z)/2 - i (-u -v+w+z)/2

Line segments are equal length and perpendicular:
p-r = (u+v-w-z)/2 - i (-u+v+w-z)/2 =- i*[i*(u+v-w-z)/2+(-u+v+w-z)/2]=-i*[(-u+v+w-z)/2-i*(-u-v+w+z)/2] = -i*(q-s)
(p-r)/(q-s) = -i, and are therefore perpendicular.
As well, |p-r|=|-i|*|q-s| = |q-s| and therefore have equal length.

The problem is pretty fun and geometric if you know your complex geometry.

Edit: Fixed some negative signs.

Man you didn’t need to reply to both threads

 

[Soccerplayer2233]

Man you didn’t need to reply to both threads

Mb, I saw a reward on this one after the first one. I can delete the other one.

 

[GiantWizard]

Mb, I saw a reward on this one after the first one. I can delete the other one.

there's a reward?

 

[qFalcon_]

bruh
you're not online

yo why do u have like 112 rod of legends worth about 200m???

 

[xov]

View attachment 3414638
Also posted in OT. Y'all wanna give it a shot? I've already solved it if no one figures it out.

The skyblock related part is that I'm doing math for bazaar so close enough.

2 plus 2 is 4
minus 1 is 3 quick maths

 

[DehJoker]

We can solve this elegant geometry problem using complex numbers. Let’s interpret each point u,v,w,zu, v, w, zu,v,w,z as complex numbers on the plane.

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Part a)​

Let’s consider the center of the square constructed on segment uv‾\overline{uv}uv, lying outside the quadrilateral. If two points uuu and vvv are interpreted as complex numbers, then a square on segment uv‾\overline{uv}uv, constructed outwardly (counterclockwise turn), will have its center at:

Center=u+v2+(v−u)i2\text{Center} = \frac{u + v}{2} + \frac{(v - u)i}{2}Center=2u+v+2(v−u)i
Here’s why:

• u+v2\frac{u+v}{2}2u+v: midpoint of uuu and vvv
• (v−u)i2\frac{(v-u)i}{2}2(v−u)i: half the vector uv→\overrightarrow{uv}uv

LaTeX syntax spotted.