taat's Profile

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SteamID64	76561198041848219
SteamID3	[U:1:81582491]
SteamID32	STEAM_0:1:40791245
Country	Finland
Signed Up	August 2, 2016
Last Posted	October 5, 2026 at 7:17 PM
Posts	116 (0 per day)

Account Details

SteamID64

76561198041848219

SteamID3

[U:1:81582491]

SteamID32

STEAM_0:1:40791245

Country

Finland

Signed Up

August 2, 2016

Last Posted

October 5, 2026 at 7:17 PM

Posts

116 (0 per day)

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In-game Sensitivity	6
Windows Sensitivity	9/11
Raw Input	1
DPI	32,000
Resolution	1600x1200
Refresh Rate	60

Game Settings

In-game Sensitivity

Windows Sensitivity

9/11

Raw Input

DPI

32,000

Resolution

1600x1200

Refresh Rate

Hardware Peripherals
Mouse	Razer Death Adder RGB
Keyboard	full size acer oem
Mousepad	none
Headphones	Beats by dre
Monitor	vintage lcd

Hardware Peripherals

Mouse

Razer Death Adder RGB

Keyboard

full size acer oem

Mousepad

none

Headphones

Beats by dre

Monitor

vintage lcd

toads_tfnegative area square?

negative area makes sense if you consider it as the "area under a function" (so a defined integral of a function)
and you could just handwave it anyways.

This looks like a "find the overlapping area of three circles" question. It's extremely tedious, but pretty braindead with a symbolic calculator

Here's a very rough outline of how to find A3 (not wasting any more time on this):

Show Content

let's start with an A_s (assuming it's the outer square) of 1 and multiply by the actual A_s at the very end
https://i.imgur.com/2GSGwQ1.png
I guessed that the vague curves are sections of circle with center at a corner of the large square and radius the same as the square's side length
https://i.imgur.com/qny2VWS.png
the program gives the equations for the circles easily
https://i.imgur.com/mzP36RE.png
change the relevant half of the circle into a y=f(x) format
https://i.imgur.com/y15qo1k.png
https://i.imgur.com/53bLMWX.png
find the intersections of the relevant curves, and use them as the bounds of the defined integrals..
https://i.imgur.com/VqDoIGT.png
and just smack the integrals into a calculator, and you add them all together:

+ integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) dx

- integral_(3/8 - sqrt(7)/8)^(1 - sqrt(3)/2) (sqrt(2 x - x^2) + 1/2 (-1 + 2 sqrt(x - x^2))) dx

+ integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx

- integral_(1 - sqrt(3)/2)^(5/8 - sqrt(7)/8) 1/2 (1 - 2 sqrt(x - x^2)) dx

+ integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(2 x - x^2)) dx

- integral_(5/8 - sqrt(7)/8)^(1/2) (1 - sqrt(1 - x^2)) dx
and multiply the result by the side length
and the final answer for A3 (unless I made mistakes....)

surprise, it's a massive clusterfuck:
(-47)*(9/64 - sqrt(3)/8 - (5 π)/12 + tan^(-1)(4 + sqrt(7)) + 1/64 (-17 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)((2 sqrt(5 - sqrt(7)))/3)) + (1/64 (sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + 8 (-3 + 4 sqrt(3) - sqrt(7) + 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (5 π)/12) - (1/64 (-9 + 16 sqrt(3) - 4 sqrt(7) - 8 sqrt(14 sqrt(3) - 24) - 16 sin^(-1)(3^(1/4)/sqrt(2)) + 16 csc^(-1)(2 sqrt(3 - sqrt(7))))) + (-1/64 sqrt(321 + 48 sqrt(7) - 48 sqrt(43 + 16 sqrt(7))) + π/3 + 1/8 (-1 + sqrt(7) - 8 cot^(-1)(3/sqrt(31 + 8 sqrt(7))))) - (1/192 (3 - 24 sqrt(3) + 24 sqrt(7) + 32 π - 96 cos^(-1)(1/8 (5 - sqrt(7))))))

= ~0.036566432691702221271670 - 0.019630731925696712432468 + 0.061918112844524846486709 - 0.0150274870976263361702352 + 0.04230790314076436373846 - 0.0173897792991522791591702

= 0.0887444503545161037349656 * (-47)
= ~-4.1709891666622568755433832..

and then just do that two more times for the final answer. It's not going to get much shorter.
Of course the question is pretty vague so this could be the solution to the wrong problem.