Brain exercise

[Anonymous No.14639024]

>If the four sides of a quadrilateral are the sides of four squares, and the centers of the opposite squares are connected with lines, the lines will always be the same length

Is /sci/ smart enough to prove this without looking up the proof online?

[Anonymous No.14639061]

>>14639024
Let
A = (x_A, y_A)
B = (x_B, y_B)
C = (x_C, y_C)
D = (x_D, y_D)
Then
P1 = ((x_A + x_B + y_A - y_B)/2 , (-x_A + x_B + y_A + y_B)/2)
P2 = ((x_B + x_C + y_B - y_C)/2 , (-x_B + x_C + y_B + y_C)/2)
P3 = ((x_C + x_D + y_C - y_D)/2 , (-x_C + x_D + y_C + y_D)/2)
P4 = ((x_D + x_A + y_D - y_A)/2 , (-x_D + x_A + y_D + y_A)/2)

Now caluclate the distances with Pythagoras.
P1P3 = 1/2 * sqrt((x_A + x_B - x_C - x_D + y_A - y_B + y_C - y_D)^2 + (-x_A + x_B + x_C - x_D + y_A + y_B - y_C - y_D)^2)
P2P4 = 1/2 * sqrt((-x_A + x_B + x_C - x_D + y_A + y_B - y_C - y_D)^2 + (x_A + x_B - x_C - x_D + y_A - y_B + y_C - y_D)^2)

They are clearly the same. Basically the difference in x-coordinates of P1 and P3 is the same as the difference of the y-coordinates of P2 and P4 and the difference in x-coordinates of P2 and P4 is the same as the difference of the y-coordinates of P1 and P3.

[Anonymous No.14639083]

P_1 = (B+A)/2 + i(B-A)/2
P_2 = (C+B)/2 + i(C-B)/2
P_3 = (D+C)/2 + i(D-C)/2
P_4 = (A+D)/2 + i(A-D)/2

P_3 - P_1 = (D+C-B-A)/2 + i(D-C-B+A)/2
P_4 - P_2 = (A+D-C-B)/2 + i(A-D-C+B)/2 = (D-C-B+A)/2 + i(-D-C+B+A)/2 = -i(P_3 - P_1)

[Anonymous No.14639222]

Is there some way we can see this without coordinates?

I notice
P_3 - P_1 = (D+C-B-A)/2 + i(D-C-B+A)/2
can be rewritten as
(1+i)(D-B)/2 + (1-i)(C-A)/2

Let's try this. We have
[math]\vec{AB} + \vec{BD} + \vec{DC} + \vec{CA} = 0.[/math]

So
[math]\vec{P_1 P_3}[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB}) + \vec{BD} + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{DC})[/math]
[math]= \vec{BD} + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB} + \vec{DC})[/math]
[math]= \vec{BD} - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD} + \vec{CA})[/math]
[math]= \vec{BD} - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{CA})[/math]
[math]= \vec{BD} + \frac{1}{\sqrt{2}} R_{135^\circ}(\vec{BD}) - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{CA})[/math]
[math]= \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{BD}) - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{CA}).[/math]

Similarly
[math]\vec{P_2 P_4}[/math]
[math]= \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{CA}) - \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{DB}).[/math]

Therefore
[math]\vec{P_2 P_4}[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{CA})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) - \frac{1}{\sqrt{2}} R_{-135^\circ}(\vec{CA})[/math]
[math]= R_{-90^\circ}(\vec{P_1 P_3}).[/math]

Ultimately I'd like to see this with just Euclid-style geometry.

[Anonymous No.14639242]

>>14639222
Attempting to simplify:
[math]\vec{P_1 P_3}[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB}) + \vec{BD} + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{DC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB}) + \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{DC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB} + \vec{BD} + \vec{DC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AC})[/math]

Similarly
[math]\vec{P_2 P_4}[/math]
[math]= \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{CA}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}).[/math]

So
[math]\vec{P_2 P_4}[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) - \frac{1}{\sqrt{2}} R_{45^\circ}(\vec{AC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BD}) + \frac{1}{\sqrt{2}} R_{-135^\circ}(\vec{AC})[/math]
[math]= R_{-90^\circ}(\vec{P_1 P_3}).[/math]

[Anonymous No.14639419]

>>14639242
This looks helpful.
[math]\vec{P_1 P_3} = \vec{P_1 B} + \vec{B P_2} + \vec{P_2 C} + \vec{C P_3}[/math]
[math]\vec{P_2 P_4} = \vec{P_2 C} + \vec{C P_3} + \vec{P_3 D} + \vec{D P_4}[/math]

We have
[math]\vec{P_2 C} = R_{-90^\circ}(\vec{B P_2})[/math]
[math]\vec{P_3 D} = R_{-90^\circ}(\vec{C P_3})[/math]

so to conclude
[math]\vec{P_2 P_4} = R_{-90^\circ}(\vec{P_1 P_3})[/math]
we only need to show
[math]\vec{C P_3} + \vec{D P_4} = R_{-90^\circ}(\vec{P_1 B} + \vec{P_2 C}).[/math]

Note that
[math]\vec{C P_3} = R_{-90^\circ}(\vec{D P_3})[/math]
[math]\vec{D P_4} = R_{-90^\circ}(\vec{A P_4})[/math]

and
[math]\vec{A P_4} + \vec{D P_3}[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AD}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{DC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AD} + \vec{DC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB} + \vec{BC})[/math]
[math]= \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{AB}) + \frac{1}{\sqrt{2}} R_{-45^\circ}(\vec{BC})[/math]
[math]= \vec{P_1 B} + \vec{P_2 C}[/math]

Therefore
[math]\vec{C P_3} + \vec{D P_4}[/math]
[math]= R_{-90^\circ}(\vec{A P_4} + \vec{D P_3})[/math]
[math]= R_{-90^\circ}(\vec{P_1 B} + \vec{P_2 C}).[/math]

[Anonymous No.14639504]

>>14639419
New version:

[math]\vec{P_1 P_3}[/math]
[math]= \vec{P_1 B} + \vec{B P_2} + \vec{P_2 C} + \vec{C P_3}[/math]
[math]= \vec{P_1 A} + \vec{A P_4} + \vec{P_4 D} + \vec{D P_3}[/math]

and

[math]\vec{P_2 P_4}[/math]
[math]= \vec{P_2 C} + \vec{C P_3} + \vec{P_3 D} + \vec{D P_4}[/math]
[math]= \vec{P_2 B} + \vec{B P_1} + \vec{P_1 A} + \vec{A P_4}[/math]

so

[math]2 \vec{P_1 P_3}[/math]
[math]= \vec{P_1 B} + \vec{B P_2} + \vec{P_2 C} + \vec{C P_3} + \vec{P_1 A} + \vec{A P_4} + \vec{P_4 D} + \vec{D P_3}[/math]
[math]= R_{90^\circ}(\vec{P_1 A}) + R_{90^\circ}(\vec{P_2 C}) + R_{90^\circ}(\vec{P_2 B}) + R_{90^\circ}(\vec{P_3 D}) + R_{90^\circ}(\vec{B P_1}) + R_{90^\circ}(\vec{D P_4}) + R_{90^\circ}(\vec{A P_4}) + R_{90^\circ}(\vec{C P_3})[/math]
[math]= R_{90^\circ}(\vec{P_1 A} + \vec{P_2 C} + \vec{P_2 B} + \vec{P_3 D} + \vec{B P_1} + \vec{D P_4} + \vec{A P_4} + \vec{C P_3})[/math]
[math]= R_{90^\circ}(\vec{2 P_2 P_4})[/math]

[Anonymous No.14639596]

>>14639504
pictorially

[Anonymous No.14641016]

bumpu

[Anonymous No.14641028]

>>14641016
>oh no muh vanity thread isn't getting the attention i deserve

[Anonymous No.14642169]

>>14641028
bumpu