Right, I tried expressing the sides of the smaller isosceles using the fact that the medians split eachother with a ratio of 2:1 but it didn't seem to bring anything up
#help-13
noble trench
worldly chasm
So from properties of medians we know the area of SQP is 1/2 RQP, and from your observation, we know POQ is 2/3 SPQ, or in other words POQ is 1/3 RQP
I don't know if this helps though
noble trench
I understand, I'll try using that
worldly chasm
if I mental mathed this right, the area of isosceles triangle is ab cos(A/2) for an isosceles triangle with base length a, pair of sides length b, and opposite angle A.
and cos(A/2) = sqrt((1+cos(A))/2)
It might be half of that, actually: ab cos(A/2)/2
cedar kilnBOT
@noble trench Has your question been resolved?
cedar kilnBOT
@noble trench Has your question been resolved?
uneven quartz
yup
steel heart
Centroid divides median in 2:1
uneven quartz
you can find OQ and OP in terms of 'b' and 'a' and then use the law of cosines on the angle indicated in the image to get a relation that will be solely in terms of a and b.
steel heart
I don't see how that is different
hollow kiln
yours work only if pm is median
steel heart
The question specifically mentions it
hollow kiln
simplified ver. of this
steel heart
simplified ver. of this
uyeah
hollow kiln
from this formula
and using it here
you could find cos
if you know double angle formulas too
hollow kiln
nice catch
uneven quartz
||AO:OR = 2:1||
||Use the pythagoras theorem on triangle ABR to find x in terms of 'a' and 'b'.||
||AR passes through O since the median will be the same as the perpendicular bisector and also because the medians of a triangle are concurrent||
Use the definition of cosine(x) and use it in triangle BOR.
cedar kilnBOT
@noble trench Has your question been resolved?
noble trench
I'm seeing this only now, I'll try out your suggestions. Thank you
cedar kilnBOT
noble trench
.reopen
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noble trench
Alright I got an answer but its incorrect. Can someone help me find where I had a mistake?
Sorry for the messy arrows, they're supposed to indicate which values are used where
cedar kilnBOT
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cedar kilnBOT
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Available help channel!
wild pagoda
!help
cedar kilnBOT
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wild pagoda
x^3.dy-(xy-1)dx=0. Find y(0)
wraith daggerBOT
Pro_Hecker
random shale
What have you tried
dreamy void
You can try to bring this into the form [ y' + a(x)y = s(x) ] and use variations of parameters.
wraith daggerBOT
𝔸dωn𝓲²s
dreamy void
First you can derive if $s(x)$ = 0 the homogeneous solution $y_h = Ce^{-\int a(x) : \dd x} = Ce^{-A(x)}.\$
Then the particular solution is $y_p = C(x)e^{-A(x)}.\$
You would next calculate $y'_p$ and plug in both $y_p$ and $y'_p$ into the differential equation to solve for $C(x).\$ Then you get the solution $y = y_h+y_p.$
wraith daggerBOT
𝔸dωn𝓲²s
wild pagoda
oh
dreamy void
**𝔸dωn𝓲²s**
Bringing the equation into this form is just algebra before you really start solving the differential equation yet
dreamy void
**Pro\_Hecker**
You can start by dividing by dx and x³ both sides and then group the terms
honest zephyr
Or it's possible to like make it in the form of d/dx
wild pagoda
Oh
dire cove
Can you solve question 4
dreamy void
Can you solve question 4
!redir
cedar kilnBOT
Can you solve question 4
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
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dreamy void
this is occupied pick another channel
dire cove
Ok sorry
winter path
cedar kilnBOT
winter path
I am half stuck on 1 and 3.
Can't quite describe where I am struggling tho, I see some information that's useful but can't see the framework to solve the answer
fair oracle
for the first one you can look at this shape
to find the area of the weird thingy
the rest should be straightforward
winter path
ah I see now
fair oracle
third one you can look at these shapes instead
and should be pretty straightforward too
you just need to be able to visualize each single part
simplify into simpler shapes you can work with
winter path
I see, I need to work on being able to see the simpler shapes
let me try solving it now
I got 2 unit squared for No. 1 and 1.14 unit squared for No. 3
winter path
ah ok, also could you share me your method, to see if I could have done it faster
dapper raven
For the first one you can move the two small bits to the larger bits to get a triangle
For no 3 you can add up the quarter circle to the leftover bit in the bottom right to get a unit square
winter path
For the first one you can move the two small bits to the larger bits to get a tr...
wow I never thought to see it that way
cedar kilnBOT
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digital bear
How am I gonna find g-1(9)'s value?
cedar kilnBOT
How am I gonna find g-1(9)'s value?
earnest idol
f(x) = 9 ie x =2
earnest idol
nope, but you can see that y = g(x) and y = f(x) intersect at a point where the output is 9
if we call that point x1, f(x1) = 9, g(x1) = 9
digital bear
oh like the point where they collide
earnest idol
intersect yes
earnest idol
no problem
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crimson sedge
might be a bit obivous but just trying to figure out why this doesn't work:
I get how this works:
crimson sedge
But of course this involves making x^2 - 9 = (x+3)(x-3).
if we don't do that and just evaluate it as is
with this:
so I just put 3 in place of x..
I get 0
why is not the same thing? why is making it (x+3)(x-3) necessary?
oh nvm ofc
because obviously can't divide by 0
been a long day
thanks anyway
.close
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ashen shard
Prove that there does not exist two positive integers $m,n$ such that $\2 \cdot 5^{6m+1}-1 = 9^{6n+1}$
wraith daggerBOT
Dork9399
ashen shard
mod 7 won't reveal anything interesting
hollow kiln
rearrange as $2 \cdot 5^{6m+1}=9^{6n+1}+1$
pastel vault
What have you tried then
wraith daggerBOT
haygiya
ashen shard
What have you tried then
idk what to try
all mods ive tried don't do anything
maybe we can mess with parity using like 5=2^2+1, 9=2^3+1 and binomial thm
but i doubt that will give a result
hollow kiln
$10 \cdot 5^{6m} = 9 \cdot 9^{6n}+ 1$
wraith daggerBOT
haygiya
hollow kiln
$5^{6m}=\frac{9 \cdot 9^{6n}+1}{10}$
wraith daggerBOT
haygiya
hollow kiln
for 9^something to be divisible by 10, its last digit should be 9
9
729
59049
…
but this only happens if the exponent is even! since 6n+1 isnt even, this equation cant be true
hollow kiln
i still feel like the solution must lay somewhere close
i kinda cheated
9^6n+1 will only end with either 69, 89, 29, 49 or 09
ashen shard
and 2*5^(6n+1)-1 will only end in 49
hollow kiln
i just proved myself wrong nevermind…
hollow kiln
9^6n+1 will only end with either 69, 89, 29, 49 or 09
n=1 69
n=2 29
89
49
09
alternates like this
n=4, 9, 14… will interest us
wraith daggerBOT
Dork9399
ashen shard
<@&286206848099549185>
steel crest
2*5^(6m+7)-1=(5^6)(2*5^(6m+1)-1)+5^6-1
ashen shard
how would I use that?
ashen shard
yes but it wont show anything
twilit bison
doesn't it show the left side is 2 and the right side is 4?
oh that 5 isn't meant to be there in the exponent of 9, is it?
steel crest
what is 5^6 - 1 mod 9
cedar kilnBOT
@ashen shard Has your question been resolved?
ashen shard
what is 5^6 - 1 mod 9
0
steel crest
what is v3(5^6-1)
ashen shard
v3?
cedar kilnBOT
@ashen shard Has your question been resolved?
cedar kilnBOT
@ashen shard Has your question been resolved?
obtuse mango
Prove that there does not exist two positive integers $m,n$ such that $\\2 \cdot...
if you know what Zsigmondy's theorem is, Zsigmondy's theorem kills the problem
alternatively lifting the exponent seems useful too
(if you know what LTE is)
obtuse mango
**haygiya**
if you don't, this is a helpful rearrangement
then factor the RHS normally
see if you can conclude something
wait hang on it's not immediately obvious how to conclude LTE from that
obtuse mango
Prove that there does not exist two positive integers $m,n$ such that $\\2 \cdot...
alternatively it might be possible to pell's equation this asw
ashen shard
alternatively it might be possible to pell's equation this asw
isnt pells for second degree only?
obtuse mango
well as in it's an equation of the form x^2 - 10y^2 = -1
ashen shard
ah
obtuse mango
wait hang on it's not immediately obvious how to conclude LTE from that
it's not too hard to finish the problem using LTE
i haven't found a non LTE soln, there probably will be one
but yeah LTE is very useful
ashen shard
@obtuse mango can u give me a hint on how to start lte
this is what i know abt lte
but idk how to start to apply
ashen shard
**haygiya**
ill probably start with this form right
obtuse mango
yh
another thing you can do is replace b -> -b to get
v_p (a^n + b^n) = v_p(a+b) + v_p(n)
what prime shouldwe use it with?
ashen shard
the RHS is LTE-able
can you explain why
ashen shard
$2 \cdot 5^{6m+1}=9^{6n+1}+1$
wraith daggerBOT
Dork9399
ashen shard
v_p (a^n + b^n) = v_p(a+b) + v_p(n)
using this
inland jackal
👀
obtuse mango
using this
well ok using that it'll be b=1
ashen shard
ah ok
obtuse mango
but yeah so what prime should we LTE with and what do we get?
ashen shard
but yeah so what prime should we LTE with and what do we get?
5?
obtuse mango
so v5(9^6n+1) = 1 + v5(6n+1)
yh
ashen shard
now we do lhs?
obtuse mango
yh
ashen shard
do we just write as 5^6n+1 + 5^6n+1
ashen shard
**Dork9399**
are we doing lhs of this?
ashen shard
so v5(9^6n+1) = 1 + v5(6n+1)
or this
obtuse mango
are we doing lhs of this?
yh
obtuse mango
so v5(5^6n+1 + 5^6n+1) = v5(10) + v5(6n+1)?
what is v_5?
as in what is the definition?
ashen shard
the number of times 5 divides a number
ashen shard
6n+1
ashen shard
6n+1 = 1 + v_5(6n+1)?
obtuse mango
well one of them's meant to be an m
6m+1 = 1 + v_5(6n+1)
then think about what that means
(we want to set up an inequality to finish the problem)
ashen shard
so 6m = v_5(6n+1)
so there exists some number 6n+1 such that 6n+1 = 5^(6m)*k, where k is an integer?
obtuse mango
so there exists some number 6n+1 such that 6n+1 = 5^(6m)*k, where k is an intege...
^
ashen shard
$6n+1 = 5^{6m} \cdot k$
wraith daggerBOT
Dork9399
wraith daggerBOT
Dork9399
obtuse mango
if we plug this back into our original thing, we get?
ashen shard
$2 \cdot 5^{6m+1}=9^{6n+1}+1$
wraith daggerBOT
Dork9399
ashen shard
how do I plug an ineq into an eq?
waitwaiataiwat
i think i see it
one sec
i dont see it 😭
obtuse mango
$2 \cdot 5^{6m+1} \geq 9^{5^{6m}} + 1$
wraith daggerBOT
LY
obtuse mango
contradiction for m sufficiently large
ashen shard
wait im confused
ashen shard
**LY**
howd u get this
you can get 9^6n+1 >= 9^5^6m+1
oh and it follows
got it
but how is it a contradiction?
obtuse mango
i suspect this won't be the intended solution
because we don't use 6m vs m
(haven't used the 6 at all)
obtuse mango
but how is it a contradiction?
intuitively, LHS is small, RHS is very big
ashen shard
but rigorously how would i say that
obtuse mango
you need to prove it though but it is probable by i.e. induction, calculus, taking logs etc. etc.
obtuse mango
should be relatively straightforward right?
yh
ashen shard
.close
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silent island
how do i know which values are 'appropriate', is it just values that make the expansion converge or is there something more?
worldly chasm
These are some random values of x. Honestly, none of these seem appropriate, because none of them have the property that 4 + 5x ≈ 5
crimson sedge
how do i know which values are 'appropriate', is it just values that make the ex...
we use the binomial expansion rule right?
silent island
we use the binomial expansion rule right?
wdym? the expansion is already done for us anyway
silent island
These are some random values of x. Honestly, none of these seem appropriate, bec...
yeah i thought so too, surely they should be around 0.2
worldly chasm
So there is also another reason why you would have a problem and that has to do with regions of convergence
So even if we rewrote this to be the expansion of sqrt(-11 + 5x) and then x=3.2 did turn turn out to correspond to sqrt(5) this would still be bad, because the series would be divergent
And finally negative numbers are undesirable because they converge more slowly
Because it wrecks the alternating series we have
But that isn't a huge deal or anything
silent island
So there is also another reason why you would have a problem and that has to do ...
|x|<4/5 right?
worldly chasm
Not sure! I'd have to know the exact form of the series
In fact, I'm not 100% on 3.2 being divergent.
It just seemed suspicious to me
silent island
Not sure! I'd have to know the exact form of the series
i was taught that if you have a binomial expansion in the form (1+x)^n and n isnt a positive integer, then you need |x|<1 for it to converge
not exactly sure why though
worldly chasm
Regions of convergence have to do with poles of the function in the complex plane
Or other undefined regions
silent island
oh i thought you just meant where the series converges
worldly chasm
I guess nonholomorphic regions in general
worldly chasm
oh i thought you just meant where the series converges
Yup, this is what I mean
Essentially, the region of convergence is the largest circle on the complex plane where the interior is holomorphic
At least that is my possibly flawed understanding
silent island
i have no clue what that means 😭
worldly chasm
Well, that's ok, I also only have a tenuous grasp of it, so it's probably best to disregard it
silent island
Not sure! I'd have to know the exact form of the series
well wolfram alpha seems to agree at least
so i can get rid of x=3.2
not sure about the other 2 though
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Available help channel!
pliant pine
k is supposed to be the curvature vector at p and x is supposed to be a tangent vector
pliant pine
i thought k is always orthogonal to x but in this diagram it doesnt look remotely close?
dusk goblet
curvature vector
🤔🤔
curvature is a scalar
i assume N is the normal vector
so maybe you mean binormal vector?
dusk goblet
i thought k is always orthogonal to x but in this diagram it doesnt look remotel...
and idk it looks pretty orthogonal to me
cedar kilnBOT
@pliant pine Has your question been resolved?
odd seal
So my question is asking me to use the gram-shmidt process to orthonormalize the set of vectors
odd seal
This is what I have
I think I did something wrong though. just asking for a double check on my work.
odd seal
yeah looks alright to me
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willow dock
We have a triangle, height 2 meters, every angle is 60°, what's the length of every side?
keen compass
are you familiar with "sohcahtoa"
willow dock
No
prisma gull
trignometry
keen compass
you can probably also do this without trig maybe
if every angle is 60 degrees, it is an equilateral
have you drawn the triangle?
prisma gull
@keen compass have you learned pythagoras theorem?
keen compass
well I hope so cause I am in Year 13
cedar kilnBOT
@willow dock Has your question been resolved?
karmic timber
cedar kilnBOT
karmic timber
!status
cedar kilnBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
im pretty sure you're meant to turn the integral into the tan(1/2pi...) form
but im not sure how
fervent snow
hi
vestal bluff
half angle formulas
karmic timber
half angle formulas
which equality are you talking about
if its these, ive done them
im not sure how to deal with the cos alpha here
karmic timber
you want to get the integral into the form 1+cos/sin
yeah, how do i do that
<@&286206848099549185>
vestal bluff
intuitively i would multiply top and bottom by 1 - cos alpha sin theta
but i domt think that transforms into thr form we want
though i could be wrong
karmic timber
let me try that
inland stirrup
intuitively i would multiply top and bottom by 1 - cos alpha sin theta
I think that would probably work
Ive done similar before
karmic timber
I think that would probably work
what kind of
identity are we using then
after wards
inland stirrup
Idk.. Show pic of ur work?
vestal bluff
show us what weve got
vestal bluff
am i stupid or did that do nothing
karmic timber
am i stupid or did that do nothing
it did nothing
because i dont know what to do next
vestal bluff
sin(alpha + theta) - cos alpha sin theta?
no that looks like a bad idea
karmic timber
i dont know how to get it in 1+cosx/sinx form
karmic timber
no that looks like a bad idea
then what do we do next
karmic timber
i mean it wants u to weierstrauss it?
vestal bluff
but it seems like the question wants you to use weierstrauss in a specific form
inland stirrup
Maybe try u = cos alpha sin thata?
vestal bluff
yeah but in that form
karmic timber
but i dont know how to deal with that cos alpha thing
karmic timber
Maybe try u = cos alpha sin thata?
hm
karmic timber
Maybe try u = cos alpha sin thata?
but then du = cos alpha cos theta d theta
karmic timber
they want it in 1+cos/sin form
what about
sec (pi/2 - alpha) cos (pi/2 - theta)
then sec = 1/cos
nv
i misread it
man
how do u even start
vestal bluff
just weierstrauss it in the current form
cedar kilnBOT
@karmic timber Has your question been resolved?
pallid kelp
@cedar kiln
karmic timber
?
Got to here but now too sure how to continue
i want to make sure this is correct so far
wait are you meant to
do king's first
like
sin(pi/2 -theta)
thats cosine
then what
2I =
1/1+cos sinx + 1/1+ cos cosx?
this might wokr
karmic timber
heres what i got t
to
2/sin alpha arctan ((t + cos alpha) / sin alpha)
now what
is there some
property to split arctan
<@&286206848099549185>
trail dune
Is this correct?
desert nova
Is this correct?
trail dune
Is this correct?
What I did here was find start and end of period, by doing 2(x-pi/3) = -+ pi/2, then did the phase shift to the right
desert nova
What I did here was find start and end of period, by doing 2(x-pi/3) = -+ pi/2, ...
hmm
can u explain specifically how u phase shifted it to the right?
trail dune
can u explain specifically how u phase shifted it to the right?
I drew it poorly but yk how da middle of da tan lands on 0
I tried to make da middle of tan on pi/3
Do you get me?
desert nova
ahhh okay
desert nova
Is this correct?
yes this is correct
i was a bit confused at first by the graph but all the information is correct
cedar kilnBOT
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crimson sedge
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rigid schooner
You can calculate the distance of a trip by multiplying the time you drive and the speed driven
carmine coyote
minus the time it takes
for the first
one time would equal to 45
i believe
from a quick glance
wait no
105
crimson sedge
I’m still lost
rigid schooner
You can calculate the distance of a trip by multiplying the time you drive and t...
So for the drive back, they multiple the speed driven back and the time it took to drive back
hexed vortex
Why is trip 2
27(1.75-T) ?
distance is speed * time, we let the time it took to drive to her job be represented as t
if we she drives for t hours towards her job
but she drives a total of 1.75 hours to and from her job
the remaning time driven back must be 1.75 - t
since we denoted the time driving to her job as t
she drove a distance of 36t to her job
driving back she must have drove a distance of 27(1.75-t)
does that make sense?
or are you still a bit confused
crimson sedge
or are you still a bit confused
Confused. T= trip 1’s time?
hexed vortex
yeah t is the time it took her to drive to her job
hexed vortex
ok thats good
crimson sedge
yeah t is the time it took her to drive to her job
Can you also do 36(1.75-t) if t equals trip 2 time?
hexed vortex
yep
crimson sedge
.reopen
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twilit bison
are these arguments using inner automorphisms valid?
cedar kilnBOT
are these arguments using inner automorphisms valid?
twilit bison
i'm not sure if the book said
for a subgroup H, i_g[H] forms a subgroup
but i think it's true
cedar kilnBOT
@twilit bison Has your question been resolved?
dawn junco
i_g is conjugation by g right ? @twilit bison
twilit bison
yeah
dawn junco
yeah your proofs look fine
dawn junco
i'm not sure if the book said
> for a subgroup H, i_g[H] forms a subgroup
but i ...
and indeed the image of a subgroup by a homomorphism is a subgroup
mellow yacht
I'm trying to integrate this. I'm trying to use triangle method.
mellow yacht
When I draw the triangle like this
I'm ending up with 3 * tanx * sinx integral
but when I draw the triangle like this:
I get -9 * cosx^2 integral
but how ? Teach always said it doesn't matter when you draw triangle
is there a trigo trick I don't know like cosx^2 = sinx * tanx or something?
cedar kilnBOT
@mellow yacht Has your question been resolved?
crimson sedge
Why is cosine = x and sine = y ?
cedar kilnBOT
Why is cosine = x and sine = y ?
covert spoke
Why is cosine = x and sine = y ?
It's by definition, for convenience
It's just a name for the x and y of a unit circle
cedar kilnBOT
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frigid heath
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frigid heath
Can someone tell me if this has any solution?
ancient lodge
,w y’=(-y+sqrt((d-x)^2+y^2))/(d-x)
wraith daggerBOT
cedar kilnBOT
@frigid heath Has your question been resolved?
frigid heath
,w y’=(-y+sqrt((d-x)^2+y^2))/(d-x)
Thank you
@ancient lodge do you know if the solution is a parabola ? That was the exercise to show that this is a parabola, but I dont know how to find the vertex and the semi latus rectum of this parabola
I think this is the closest form to a parabola
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trim wagon
HUHH
cedar kilnBOT
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honest urchin
What are functions
cedar kilnBOT
What are functions
cedar kilnBOT
crimson sedge
You can watch a bunch of videos on Khan Academy to get an adequate understanding of functions
use this to get you started
and then start exploring more complex topics
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Available help channel!
polar magnet
If I take the Union and Intersection over a family of sets but only to a certain range rather than the whole set of families. How do I describe that range? On a finite family of sets
rocky harness
Just write the members of the range individually? Its a finite family 
I mean the real answer lies in this question: how are you taking the union over a range without having described the range first?
There are plenty ways to do this
cedar kilnBOT
@polar magnet Has your question been resolved?
radiant fiber
sinful stag
Help i dont get question 20-21-22 on this conditional probability lesson , how do i know which one comes first (I/M) or (M/I) ?
cedar kilnBOT
@sinful stag Has your question been resolved?
cedar kilnBOT
@sinful stag Has your question been resolved?
atomic saddle
it has "see example 1,2 and 3", to get a better understnading you may look there
scenic brook
i dont get it why isnt it 53/100
How you gott 100
sinful stag
1,2,3 examples are way simpler than this
sinful stag
How you gott 100
53 + 47
sinful stag
yea
sinful stag
but then 53 +65 is with no improved
atomic saddle
for 22 compare their fractions
see which one is larger
21 shouldn't be a problem, is it?
sinful stag
independent says that the events doesnt affect the probability and dependent it does
atomic saddle
yea , so you know 21
atomic saddle
compare the ratios from both placebo and medication
do the same thing for placebo
like on 20
atomic saddle
no, yes... You have to say that medication dosent work
instead of placebo works
any confusions, or is this finished?
atomic saddle
does medication affect your health?
sinful stag
yeah
atomic saddle
so....?
sinful stag
so dependent
atomic saddle
yes, but no
sinful stag
huh
atomic saddle
placebo means fake medication
that dosent have effect
you have to say that medication is worse than juist leaving them
and not reccomend the medication
sinful stag
ohh
Ok one more thing how do i know when to use permutation or combination equations
cedar kilnBOT
cedar kilnBOT
Available help channel!
placid nacelle
$M_1 = {(x,y) ; |x| + |y| < 1}$
cedar kilnBOT
$M_1 = {(x,y) ; |x| + |y| < 1}$
wraith daggerBOT
Merineth 🇸🇪
placid nacelle
If i'm asked to do a rough sketch of this
how would i go about it?
I understand that if y = 0 then x = 1 giving (0,1).
Same applies to when x = 0 then x = 1 giving (1,0)
that gives me two points
the green and purple is that i concluded but i'm not entirely sure how i should come ot the conclusion that it's rotated 45 degrees
placid nacelle
Do do mean, what is y if x = 0,5?
in that case y would be 0,5
Yeah that might work
dawn junco
Do do mean, what is y if x = 0,5?
no
if x is positive |x| is just x right
same for y
so the condition becomes a plain old line x+y<1 i.e y < 1-x
placid nacelle
Well in that case the highest x and y can be if they are both positive would be 0,5
oh i see
y = x
dawn junco
so the points in the 1st quadrant that are in M1 are such that y < 1-x
ie all these points
y=1-x is the dotted red line
repeat that for all quadrants and you're done
dawn junco
it's like when you're solving absolute value equations
you're trying to see when the things in the absolute value are positive or negative to simplify your work
that's all I'm doing
placid nacelle
So to get the 3rd quadrant (bottom right?) i would say that y is negative and x is positive
$x - y < 1 \implies y < x - 1$
wraith daggerBOT
Merineth 🇸🇪
dawn junco
yeah
placid nacelle
yeah oki that makes sense
dawn junco
wait 3rd quadrant is bottom left tho
placid nacelle
Do you know how i'd interpret
dawn junco
but 4th quadrant yeah
placid nacelle
$max(|x|,|y|) \le 1$
wraith daggerBOT
Merineth 🇸🇪
placid nacelle
what does the max mean?
dawn junco
the biggest of the two values inside
you can rephrase that as 2 simultaneous inequalities
placid nacelle
so i choose the one that's largest?
dawn junco
|x| <= 1 and |y| <= 1
placid nacelle
aaah so essentially we get the square
dawn junco
yea
placid nacelle
a compact square to be even more precise
placid nacelle
No, i meant only the last square
with the max()
Since it's both closed and bound then it's compact
I'm not sure how to tackle these problems
normally when i'm asked to sketch something
i choose some point that seems appropriate
and then it comes together
dawn junco
well they aren't closed to begin with
placid nacelle
$|xy| < \frac{1}{4}$
wraith daggerBOT
Merineth 🇸🇪
placid nacelle
This for example, here i would assume that
$y < \frac{1}{4x}$ and $x < \frac{1}{4y}$
or is that also wrong?
wraith daggerBOT
Merineth 🇸🇪
mental trail
**Merineth 🇸🇪**
if x,y are both positive, sure
but they're the same thing
they both mean xy < 1/4
placid nacelle
yeah i get that
i'm just not sure how to apply the method of sketching the area
I'm used to finding the max/min points
and then of that try some random points to get a rough sketch
y < 1/4x would give me points such as
(0,0)
mental trail
well
y < 1/4x is a good start
it means when x,y are positive
(x,y) is below the curve y = 1/4x
placid nacelle
Well yeah
but we can conclude that the start point would be (0,0)
since both when x = 0 and y = 0 we get (0,0)
ooooh do we apply the same method as we did with the cube?
So we check the first quadrant (top right iirc)
then x and y are positive
dawn junco
you can yeah
dawn junco
not necessarily, there's a ton of positive points which don't satisfy that condition
placid nacelle
$y < 1/4x,\y < -1/4x, \ -y <-1/4x, \-y < 1/4x$
wraith daggerBOT
Merineth 🇸🇪
placid nacelle
don't these 4 equations satisfy the 4 quadrants?
dawn junco
it's unclear what you mean there
one equation is valid in one quadrant
the points in Q1 that are in |xy| < 1/4 also satisfy y<1/4x
the points in Q2 that are in |xy| < 1/4 also satisfy y<-1/4x
etc...
it's 4 conditions, 1 for each quadrant
dawn junco
yeah
mental trail
Let's try plotting the area on the first quadrant
we know that when x > 0, y > 0
we have y < 1/(4x)
so plotting y = 1/(4x)
the points in this quadrant that are in the area
placid nacelle
Yeah but how did we conclude that?
mental trail
are the ones below
placid nacelle
just choose randomg points and draw?
mental trail
?
no
if y < 1/(4x)
then (x,y) is below (x,1/(4x))
since the y coordinate is smaller
so the point (x,y) is below the curve where (x,1/(4x)) is
placid nacelle
I had no idea y = 1/4x is graphed like that
mental trail
I had no idea y = 1/4x is graphed like that
never seen the plot of inverse function? 1/x
placid nacelle
i have to choose x points to determine y and then draw some sort of curve between the point
Well they are equal to each other when x = 0,5
that gives me the point (1/2, 1/2)
if x = 1 i get y = 1/4 and if x = 1/4 then y = 1
Took a while but i guess i can draw it now
Would i apply the same method for the 2nd quadrant?
y < 1/(-4x)
.close
ripe lance
Let quadrilateral ABEC be inscribed in circle with center O with $AB<AC$. Let the intersection of diagonals be D and $\angle CAO=\angle ABD$ Let the midpoints of BE and OD be M and N. How to prove line MN intersects AC at its midpoint?
wraith daggerBOT
babario
ripe lance
<@&286206848099549185>
pallid trout
Let quadrilateral ABEC be inscribed in circle with center O with $AB<AC$. Let th...
We are given that quadrilateral (ABEC) is inscribed in a circle with center (O), and (AB < AC). The diagonals (AC) and (BE) intersect at point (D). Additionally, we are provided the condition that ( \angle CAO = \angle ABD ). We are tasked with proving that the line segment (MN), where (M) and (N) are the midpoints of segments (BE) and (OD) respectively, intersects (AC) at its midpoint.
Step 1: Understand the configuration
- Quadrilateral (ABEC) is inscribed in a circle, so (A), (B), (C), and (E) all lie on the circle with center (O).
- The diagonals (AC) and (BE) intersect at point (D).
- The condition ( \angle CAO = \angle ABD ) is provided.
- (M) is the midpoint of (BE), and (N) is the midpoint of (OD).
Our goal is to prove that line (MN) intersects (AC) at its midpoint.
Step 2: Utilize properties of cyclic quadrilaterals
Since quadrilateral (ABEC) is inscribed in a circle, it has several important properties:
- Opposite angles of cyclic quadrilaterals are supplementary (by the Inscribed Angle Theorem).
- The diagonals of a cyclic quadrilateral may intersect at points that satisfy specific angle conditions.
Step 3: Investigate the condition ( \angle CAO = \angle ABD )
The condition ( \angle CAO = \angle ABD ) suggests a relationship between the points (A), (B), (C), and (D). This is a key element in understanding the symmetry of the figure. It implies a certain geometric harmony between the diagonals and the center of the circle, which is likely to have implications for the midpoint properties.
Step 4: Apply the midpoint theorem
The midpoint theorem tells us that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length. We will use this idea to establish a relationship between the line (MN) and (AC).
Step 5: Consider the quadrilateral formed by points (A), (B), (C), and (E)
In cyclic quadrilaterals, the diagonals and the lines joining midpoints of certain segments often exhibit symmetry. Given the specific conditions about the angles and midpoints, it suggests that line (MN) intersects (AC) at its midpoint.
wraith daggerBOT
King Hassan
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pallid trout
To prove this rigorously, one could use projective geometry or vector geometry, but the symmetry of the situation strongly suggests that (MN) indeed intersects (AC) at its midpoint. Specifically, the configuration of midpoints (M) and (N), and the fact that the diagonals intersect at (D) in such a symmetric way, implies that the line (MN) must pass through the midpoint of (AC).
Step 6: Conclusion
Thus, by leveraging the properties of cyclic quadrilaterals, the condition ( \angle CAO = \angle ABD ), and the symmetry of the midpoints, we conclude that line (MN) intersects (AC) at its midpoint. Therefore, the statement is proved.
wraith daggerBOT
King Hassan
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neon prism
We are given that quadrilateral \(ABEC\) is inscribed in a circle with center \(...
is this ai lol
solemn bobcat
is this ai lol
gpt
ripe lance
did you try to do a sketch
looks smth like tiz
cedar kilnBOT
@ripe lance Has your question been resolved?
ripe lance
<@&286206848099549185>
steel crest
am just copy pasting the qn for reference
Let quadrilateral ABEC be inscribed in circle with center O with $AB<AC$. Let the intersection of diagonals be D and $\angle CAO=\angle ABD$ Let the midpoints of BE and OD be M and N. How to prove line MN intersects AC at its midpoint?
wraith daggerBOT
shreyanjha
steel crest
CAO=ABC
lets ABC = CAO=OCA = k
=>AEC = 90-k
so we have k = 45*
so in your diagram C will be in the x axis, and we know that B is in the 2nd quadrant
steel crest
why?
cedar kilnBOT
ripe lance
.reopen
cedar kilnBOT
✅
steel crest
oh
ripe lance
yea btw my observations are that $\Delta BDE \cong \Delta ADC$ and both are right angled
wraith daggerBOT
babario
ripe lance
i think it also suffices to prove that PDC or PDA are isoceles
ripe lance
yea btw my observations are that $\Delta BDE \cong \Delta ADC$ and both are righ...
ah wait congruent only when CAO=45 so no
so basically its just that BDE=ADC=90
ripe lance
90 - BAE = AEC
yep
steel crest
yes
steel crest
so basically its just that BDE=ADC=90
this is correct
ripe lance
and hence since we want to prove that P is the midpoint of AC it suffices to show that AP=AC=DP since ADC is right angled tri
and we can do so by angle chase to prove either PDA or PDC is isoceles
steel crest
AP is not AC
ripe lance
which i cant
steel crest
hm
ripe lance
hmm if we want to prove PDA is isoceles we just have to smh prove EAC=180-EDP
ripe lance
hmm if we want to prove PDA is isoceles we just have to smh prove EAC=180-EDP
from ggb MN=NP too btw, which means $\Delta DNP\cong \Delta ONM$
wraith daggerBOT
babario
ripe lance
if we can prove this its gonna be helpful in the angle chase i think
ripe lance
try showing that BA parralel to EC
actually tats only the case when CAO=45
steel crest
but not the one you are syaing
steel crest
actually tats only the case when CAO=45
no, that is always true (i believe i have proven that)
ripe lance
which means AC=BE
but the question is let ABEC be a cyclic quad of any sort
steel crest
ik
ripe lance
which means AC=BE
hence this is not always the case
steel crest
but you are given enough information to prove that is the case
ripe lance
how so
ripe lance
looks smth like tiz
the diagram alone is a counterexample
steel crest
sry that is wrong
ripe lance
nvm
ripe lance
**babario**
can you help?
steel crest
perpendicular bisector
ripe lance
which means 180-ADP=180-(QDO+DOQ)=DQO
steel crest
one minute
ripe lance
=MQE=EBD=EBC=EAC
so... we just need to prove 1. MN=NP and 2. OM perp to BE
then we are done
ripe lance
so... we just need to prove 1. MN=NP and 2. OM perp to BE
because that would imply $\Delta NDP \cong \Delta NOM$ and we can do the angle chase 180-ADP=180-(QDO+DOQ)=DQO=MQE=EBD=EBC=EAC to show that ADP is isoceles
wraith daggerBOT
babario
ripe lance
oh and we also gotta prove BDE=ADC=90
then ADP is isoceles would imply P is the midpoint of AC
ripe lance
one minute
hmm at least any idea why ADC=90?
steel crest
am back
steel crest
hmm at least any idea why ADC=90?
yes
BAD=CAO
=>
BAE = 90-AEC
=>
BAE=90-ABC
=>BAD=90-ABD
=BDA=90
@ripe lance
are you there?
ripe lance
BAE = 90-AEC
this requires us to prove that ADC=90 first unfortunately
ripe lance
ahh rightt
ripe lance
so... we just need to prove 1. MN=NP and 2. OM perp to BE
oh right so now this remains
steel crest
which means 180-ADP=180-(QDO+DOQ)=DQO
180-ADP is not QDO
ripe lance
or honestly we can congruence test by MNO=DNP, DN=NO and NOM=NDP
yep i think this settles 2. right?
steel crest
how did you get
ripe lance
180-ADP is not QDO
yep its QDP
steel crest
NOM = NDP?
ripe lance
yep its QDP
but the rest still holds i just mistyped
steel crest
QDP is nor MQE
ripe lance
wait i rewrite
steel crest
ah right but that requires them to be parallel
if what we are trying to prove is true that is true
ripe lance
so if MN=NP is true, 180-QDP=180-(QDO+DOQ) (since MN=NP implies NDP and NOM is cong)
180-(QDO+DOQ)=DQO=MQE=90-OEM=EBD=EBC=EAC
ripe lance
QDP is nor MQE
Q is just intersection of MO and DE
so its true cuz er opposite angles i think its called
yep
ripe lance
so if MN=NP is true, 180-QDP=180-(QDO+DOQ) (since MN=NP implies NDP and NOM is c...
here i wrote DQO=MQE
which is right
the entire angle chase is done we just need to prove MN=NP now
@steel crest any ideas?
steel crest
hm
ripe lance
BAE = 90-AEC
actually how is this true?
steel crest
CAO = 90-AEC
ripe lance
you agree this is quivalent to proving
wdym
steel crest
i have to go now
cedar kilnBOT
ripe lance
.reopen
cedar kilnBOT
✅
steel crest
So I was saying
This is equivalent to proving, the line connecting midpoints of AC and BE bisected OQ
right?
ripe lance
proving MN=NP is equivalent to proving P is midpoint of AC yea
steel crest
Am not saying that
ripe lance
This is equivalent to proving, the line connecting midpoints of AC and BE bisect...
eh wait i misunderstood
actually u mean OD?
(just so i dont hv to scroll)
steel crest
Ok
ripe lance
honestly we can angle chase in a lot of ways, but since we have gone with MN bisects AC and trying to prove specifically PDA is isoceles i think we should just keep on with it, its just left proving MN=NP basically