Help Geo again

[scarlet aurora]

Hello Micabot

[bitter latch]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
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   +close
   

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[elfin scroll]

send drawing pls

wait what???

[serene tapir]

I think this means that this would work for any triangle.

Which is very surprising!

[elfin scroll]

bisects the perimeter and area?

oh wait nvm

lol

[serene tapir]

To be honest, I'm not even sure how to prove that such a line even always exists 😅

[elfin scroll]

such an interesting question

[elfin palm]

You can always find a line that bisects the perimeter, and then if it doesn't bisect the area that means the larger portion of area is on one side. From there, presumably you can find a perimeter bisector where the larger area is on the other side, and then continuously transform one into the other, then it's just the intermediate value theorem.

[serene tapir]

Ohh, I see, yeah.

[elfin palm]

Actually, you don't need the "presumably". You can just rotate the line 180 degrees, and then the larger area is "on the other side" from the perspective of the line.

But there is exactly one perimeter bisector passing through any point on the perimeter of a triangle.

[serene tapir]

Ah, yeah. Kinda reminds me of Borsuk-Ulam theorem.

[scarlet aurora]

@elfin scroll

[elfin scroll]

oh

[elfin palm]

As I said, there's exactly one perimeter bisector that passes through any point on the perimeter of a triangle, so you might try to prove that any specific one that also bisects the area must pass through the incenter.

If it passes through the incenter, then it bisects the incircle.

There's also exactly one area bisector that passes through any given point on the triangle's perimeter.

[scarlet aurora]

@dry thicket

[scarlet aurora]

Could you please give a solution. I have given this problem a good amount of time. Couldn't solve it tho

[elfin palm]

I don't have one, I was just speculating.

[dry thicket]

what?

!

💀

[serene tapir]

If you have a function on a sphere, then it maps at least one pair of antipodal points to the same value.

[dry thicket]

yeah i know just how here

[safe nacelle]

i’m lost here, sorry

try using heron’s formula?

[scarlet aurora]

No worries, everyone lost on this one

Unless Generating function comes and sweeps away the victory.

[scarlet aurora]

Or perhaps stayhomedomath

[safe nacelle]

i proved the converse

proving uniqueness shouldn’t be too hard; i’ll try for a reverse reconstruction

[safe nacelle]

||Let the line l intersect (WLOG) AB and AC at X, Y respectively.

AX+AY + XY = XB + BC + CY + XY, so AX+AY = XB+BC+CY. this means AX+AY = (AB+BC+CA)/2 = s.

let there be a point J on XY lying on the angle bisector of ABC. this allows the distance from J to AB and AC to be equal.* set that equal to j.
let AX = x and AY = y.

the area of AXY is equal to AJX + AJY. area of AJX is jx/2 and area of AJY is jy/2 (since j is the height and x, y are the bases)

area AXY = j(x+y)/2 = js/2. however, area AXY = (area ABC)/2 = rs/2.

this means j=r, which implies I=J.**
therefore I lies on XY. qed, i don’t write proofs, please ping for more clarification||

||* such a J existing is evident via IVT, as J=X gives (distance from J to AB) < (distance from J to AC) and J=Y gives the opposite.

** distance AJ can be found as j/sin(A/2). it may help to draw a diagram for this. since J is on the ray AI, this uniquely determines J (and clearly J is inside the triangle ABC as it lies on segment XY).||

should i have spoilered those?

[scarlet aurora]

Nah you are fine

[scarlet aurora]

Imma look at this after some tiwm

[safe nacelle]

Yeah, that is much easier than this..

[safe nacelle]

is it?

[safe nacelle]

I think so

have you proven both?