geometry problem

[peak edge]

hello, i need some help with this problem, if anything is unclear please tell me. merry christmas and many thanks!

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[solemn moon]

Hmm... Well, it shouldn't be too hard to solve this algebraically. Purely geometrically, though... Don't know.

[peak edge]

what do you mean?

i just need a solution, i dont think i have to use any specific method

[ember pumice]

since you have a 45 degree angle there try drawing a perpendicular from AB to C to make a 45-45-90 trinagle which might help

and consider if Q is necesarily on that line or not

also ADC and idk lets call the perpendicular point E will be a rectangle, so use that

[peak edge]

that is what the book says

[ember pumice]

if i draw a perpendicular from C to AB then Q is on the line?

[peak edge]

and ik it kinda sounds stupid but it looks like pq is a 3rd of mn

and yes

wait

wait

no sorry

[ember pumice]

i mean i would fakesolve by assuming CD=AD because why not

seems to give you too much flexibility

actually idek if theres one solution

[peak edge]

the solution should be 12

but i cant reach the answer

[ember pumice]

hmm okay try making the whole thing one rectangle

like that

[peak edge]

cb will be CE√2

but i dont know further

[ember pumice]

looks like Q doesnt have to be on CE since we dont know what CD is it could just be super tiny

you can always do a fakesolve by assuming CD=AD

currently trying to break the problem

[peak edge]

previous guy said this could also be solved algebrically, may you explain if you dont mind?

[ember pumice]

idk what the algebraic solution is lol

oh i proved its always 6 regardless of AD :D

hmmm

[peak edge]

?

[ember pumice]

how do you solve without the fakesolve tho

[peak edge]

my thought process might be wrong but i was thinking that (B-b)/2 = 6 and B=b+12 and that MN = DC+PQ

[ember pumice]

whats lowercase b

[peak edge]

DC

and B is AB

[ember pumice]

uhhh

(AB+DC)/2 is MN

cuz thats what midpoint line is

[peak edge]

i meant (B-b)/2 sorry i am tired

[ember pumice]

that should be QN

[peak edge]

(AB-DC)/2

=6

[ember pumice]

QN is half of that

wait that is kinda useful

okay can you think about how MP relates to PQ

hint AECD is a rectangle and AC is a diagonal

wait nvm

[peak edge]

wait

i think i found it

[ember pumice]

Q isnt necessarily on CE

i got a geometry solution 😭 its so weird

the issue with geometry problems is that everyone has different thought processes and teaching them usually makes 0 sense

[peak edge]

if q was always on ce then (ab-cd)/2 would have been equal to both qn and pq

[ember pumice]

yea

sadly its not always

actually

you can prove q is on CE if and only if AD=CD

i got it for both if CD>AD and CD<AD

okay lets start with CD>AD

ive called stuff x and y for convenience

wait let me draw the letters

okay nice

[peak edge]

this is an exercise from the 2nd subject of a practice test and they are usually not that hard

😭

[ember pumice]

i think i overcomplicated it lol

i think youre supposed to assume CD=AD but i got annoyed by loose definitions

anyways

first off, CEB is a 45-45-90 triangle

since CE=AD=y, then EB=CE=y

now we consider rectangle ABFD

BD is a diagonal

MN is the middle line

so their intersection, Q, is the middle of the rectangle

now MQ is then half AB

which is $\frac{x+y}{2}$

[cerulean inlet]

!𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪!

[ember pumice]

then AECD is another rectangle with AC diagonal and MN mid line, so MP is x/2 and

MQ=PQ+MP
$\\\frac{x+y}{2}=6+\frac x2\\$
$\frac y2=6\$

[cerulean inlet]

!𒐪 ɹɐupoɯ⇂ㄥ8𝟝 𒐪!

[ember pumice]

@peak edge does that make sense or not

[peak edge]

im reading 1 sec

[ember pumice]

gotcha

the issue is theres probably better solutions and this isnt really good for your learning hmm

[peak edge]

it makes

it makes sense

but i would have never thought of this

thank you so much

[ember pumice]

yea i think theres a better solution

anyways this is the proof for CD<AD

and then CD=AD you should be able to do

let me do CD>AD

[peak edge]

it only said ab>cd

[ember pumice]

notably this time, Q is now after CE

[ember pumice]

yea which annoys me :P

wait hold up

i think 6+x/2 is still half of AB

yup

it is alright nm

i originally had a more complex solution for that

okay so what should you take away from this

you know the rectangle diagonal and midline rule now right

[peak edge]

i ll make sure to put it in practice

again, thank you

didnt find a solution anywhere

[ember pumice]

yea and i guess you should be careful of where certain points lie i guess

and to think of the extremes

like in this case they didnt specify that CD has to be greater, less, or equal than AD

so you want to think of the extremems

when its really small or really large

also think that if it can be anything then pick the most convenient one

usually in tests you can get away with assuming that a certain variable that could be anything is something thats most convenient as in this case we find that the answer is always 12 no matter whatn CD is

[peak edge]

i guess im gonna close this thread, thanks again and merry christmas!

+close