geometry problem

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[open fulcrum]

solved yet?

alright

have you seen projective transformations

this is just for convenience

i should probably try to figure out if there's a non coord bash way to solve this actually

i think the first thing to notice is what happens to V_A if P moves very close to A

alright

ok here's the way i did it
D_A, D_B, D_C are similar to ABC
we would like to find the (length) similarity/proportionality factor for each of those
let them be a,b,c respectively
since all that matters is area ratios, we can set ABC's area to be 1
then work out D's and V's in terms of a,b,c

(parallelogram areas and triangle areas can be compared via the former being length * length * sin(angle between), and the latter being half that)

there may also be a more direct way to prove that ||DADBDC/VAVBVC = 1/8|| i'm not sure right now

nah

different lengths, you know, absin(c) but i already used those letters so didn't want to confuse

as an example of this, let V_C be CEPF clockwise
then ||CE = b*CB, by the length similarity with D_B||
||and CF = a*CA||
now ||V_C = a*CA*b*CB*sin(C)||
meanwhile ||the total area is CA*CB*sin(C)/2 = 1, so V_C = 2ab||

[dusky cairn]

you can assume the triangle is equilateral if you want

it won't affect area ratios nor parallel-ness

[dusky cairn]

sudden thought
what happens if you add the six lengths inside the triangle

[dusky cairn]

no, sorry

[open fulcrum]

the proportionality factors

no i meant the similarity factors between triangle D_A and triangle ABC

so D_A = a^2 times the area of ABC

i'm sorry maybe micabo can help you more coherently

[open fulcrum]

.

i basically replaced D_A/V_A with (D_A/ABC) / (V_A/ABC)

oh

so you know how if you take a triangle and make its edges all twice as long, its area is multiplied by 4

if you don't know that, there's probably a better method

i think i do see a better method actually

[open fulcrum]

this helped me

i am failing to help

it can't

we label those 6 lengths like that

then try proving D_A/V_A = ab/2ed

this is unrelated to the previous method so i started over with those letters

formula for area of a triangle with length1 * length2 * sin(angle)/2 is useful here

alright

[open fulcrum]

now you obtain this, the spoiler

well, can you prove that as well

you can do that by rotating the letters, so to speak, since everything is symmetrical

it's so epic

ok now we have 3 things that multiply to 1/8, and we want to prove that they add to at least 3/2

looks familiar?

try proving that for a bit then ask me if you need more info

correct

now there's just one final little thing which is to prove that for any number between 3/2 and infinity, it is possible

yep exactly

and you’re now done

was this really a problem given in a geometry class?

ahh

well good luck with that

no problem

[carmine quiver]

Posting this Problem in here probably wasnt a very good choice. It is from a german competition and the round only ends today.

(even though you have to send the solutions in per post which would be a bit late by now)

Bundeswettbewerb Mathematik

its from Round 2

since only about 400 people qualifyed for the round i doubt that anyone has profited from it

its not my job to make sure that the competition is fair xD i just wanted to prevent anyone from giving out more solutions

i do

but mine is pretty long and the proof is written in german. I didnt read the whole thread so I dont know how much of the solution you already know. If you want I can send you the answer tomorrow