hello?
#geometry-and-trigonometry
trail minnow
lmao
dark sparrow
ok i got ghosted great
vale bobcat
|
|
|(7) 60° \14
|
|■____\30°
(12.12)
I got it
to hell with this, im going to the other triangle lol
vale bobcat
I just remembered my geometry last year
vale bobcat
its art
dark sparrow
-_-
trail minnow
the ascii lmao
queen python
Sad
molten oracle
Anyone here fancy giving a second opinion on a maths olution before i code it?
Its to do with aligning two congruent triangles in 3d.
worthy root
@rigid wharf Ok the question is poorly worded but you can always see the diagram .
rigid wharf
Diagram doesnt help if the question isnt clicking
mighty narwhal
Ohkay so
lets figure out when is the sum going to be maximum
~~sin x_1 cos x_2
This would be obviously at its maximum when x_1=90 deg, and x_2=0 deg.
But this would result in the second term being 0
The more you increase one term's value, the more the term after or before it decreases
So lets say that all x's are equal
sin(x)cos(x) is maximum when x=45 deg, where it would be 1/2
We have n terms in total, each being 1/2 at most
so the total is at most n/2~~
pretty much thats it
of course this is way far away from being formal, mostly because you cant just say "all of them should be equal so its maximum" without proving why it would give the maximum
try your best to read my handwriting
tidal rune
Given AB > AC (thus, z > y by side angle inequality) and AD is an internal angle bisector of A, how can I show that DB > DC without using Angle Bisector Theorem on AD?
Using the angle bisector theorem on AD makes the problem rather trivial. But that's not what the question asks. It asks to use side angle inequality to solve this problem.
By external angles, <ADC = x+z, and <ADC = x+y.
I can then use side angle inequality to determine that AB > DB, AB > AD, and AC > CD, but that's all I can get
upper karma
@tidal rune you have yourself written the solution
tidal rune
How so?
upper karma
If AB > AC , And AB > DB ,AC>CD it implies AB>DB>AC>CD .
tidal rune
Nothing says DB > AC
tidal rune
It's okay. We all do that sometimes.
mental wharf
yes
it says (not pictured), so they havent drawn it in
you just draw the lines in if youre going to use them
upper karma
You can always draw dotted lines.
It kind of clears up the matter.
You can tell which of the lines were there originally and which are part of your own construction.
Like this.
restive wren
@mighty narwhal thnx for the help man.......
upper karma
@upper karma did you solve it? The answer is ||30 degrees||, isn't it?
tidal rune
@restive wren You can't subtract inequalities like that.
upper karma
Great.
restive wren
Nvm maybe i got confused
tidal rune
I solved my own question
Just overlap angle z with a line segment of angle y from AC
It'll intersect AD at a new point, say, H
Then there are two similar triangles ABD and ACH
and DHC is isoceles
Using ratios, AB/AC = DB/HC (by similarity) = DB/DC (by the isoceles)
And since AB > AC, therefore DB > DC because AB/AC = DB/DC > 1
BTW, my previous problem still stands.
I need to prove that a simple quadrilateral with one pair of opposite sides congruent and one pair of opposite angles congruent does not determine that the quadrilateral is a parallelogram.
i.e. I need an example of a simple quadrilateral with the given properties but is not a parallelogram.
tidal rune
Supposedly related to SSA =/=> congruency in triangles.
tidal rune
Hmm
I got something like that before but wasn't able to draw one to scale
I'll try again though
upper karma
Hello home
I'm a bit confused on doing number 2
I know about the inequality theorems
fringe dirge
Then how are you having trouble applying them?
thin tartan
Can Anyone tell me what this is asking me to do?
umbral snow
What values of x make that true?
stuck torrent
$2\cos^{-1}(0.5)$
somber coyoteBOT
Whoever:
stuck torrent
vale bobcat
can someone teach me how to solve this
f(x)=2x+3 with domain: 0<x<10
(< has an underline)
royal crest
what are you trying to solve?
vale bobcat
plotting the graph
royal crest
what can you tell about that function?
royal crest
wait, if i understand right, you need to plot that function for x between 0 and 10 right?
vale bobcat
yes
royal crest
are you asking for a function plotter or about how that function should be drawn?
do you recognize that it's a linear function?
if you need to plot the function for screenshotting it or something just use desmos
online and free
vale bobcat
really ok ill try thanks
royal crest
I don't understand T-T are you having an understanding problem or a practical problem?
vale bobcat
ree ok ill take a pic
royal crest
f(x)=2x+3 is a linear function, geometrically it's a line, you need to find the values for two xs, 0 and 1 for example, then draw the line passing through theml
vale bobcat
Its basic because im still in highschool but its my first time seeing this
royal crest
alright alright ^^
vale bobcat
royal crest
well, for the first function, since it's linear and you only need 2 points, take x = 0 and x = 10 then draw the line that passes through them
vale bobcat
the imaginary lines right
royal crest
for the rest, you'll need to try more values to get an intuition about the shape
vale bobcat
ok
I had algebra 2, 2 years ago so I forgot how to do f(x), im aiming to master this function
royal crest
good luck ♥
if you're not in a hurry, I advise you to have a look at khan academy
and patrickjmt
vale bobcat
thanks
but tomorrow is our quiz, and its gonna be our 3rd day in school XD
oh yeah, its just easier to use my common sense
dark sparrow
are you only given that HG || IJ and that HI = GJ
no text at all? none of it?
then what are you asked for?
frosty steppe
How much information do you need to define a triangle? How much do the two triangles labeled have in common?
If they both share enough information to determine the triangle then they’re the same. If not they could be different
silent plank
insufficient info to determine if they're equal
dark sparrow
idk there's a lot of possible piece of information that could be sufficient to decide either way
silent plank
was that line specifically in the answer?
think of a regular trapezium
it would have the same properties as the figure above
frosty steppe
There are pairs of triangles that satisfy the pictured conditions but still aren’t congruent
silent plank
parallel top/base
legs of equal length
pic doesn't have to be to scale so you can only use that info provided
void nymph
ABCD is a rhombus that rotates, point E is its center, how could i give the x and y i would expect C to be at from E at any given rotation of ABCD
dark sparrow
ABCD is not a rhombus by any stretch...
it's a parallelogram at best
anyway @void nymph are you familiar with matrices
void nymph
not at all, what are those?
oh should i've said quadrilateral instead
@dark sparrow
dark sparrow
ok nevermind no matrix talk then ok fair we can do this w/o that
alright how about polar coordinates
void nymph
no
dark sparrow
shit
void nymph
ive only done up to tenth grade trigonometry
and since im going to be plugging this into code, the simpler the better really
dark sparrow
i mean i could just give you the formula to rotate a point (x,y) by an angle θ around the origin
if that's what you want
void nymph
that would be perfect thanks
dark sparrow
ok
rotating $(x,y)$ by $\theta$ around the origin sends it to $$(x \cos(\theta) - y \sin(\theta), x \sin(\theta) + y \cos(\theta) )$$
somber coyoteBOT
Ann:
dark sparrow
i can explain how i obtained this formula, if you want. though it might be a bit tough given your unfamiliarity with polar coords, which i'm kinda bummed out by!
dark sparrow
this should work
void nymph
yeah but knowing me im probably going to stuff something up and have to revise my code over and over until it works
rose monolith
why is inverse cos ( - sqrt3 / 2) = 150
and not -30?
pls @rose monolith tag me on response
rose monolith
yes
let me have a think
OHHH TYSM
I SEEE
cos(-30) = (cos30)
i see now tysm ❤
upper karma
TFW INVERSION ACTUALLY WORKS
dark sparrow
what
eager pendant
@upper karma problem?
upper karma
@eager pendant number 5 on page 5 here https://cms.math.ca/Reports/cmo/2007-cmo-report-en.pdf (caution: contains solutions)
eager pendant
ooh i've done #3 and #4 from this before
eager pendant
wait nvm i've done this problem before lol
upper karma
lol
eager pendant
||radical centrer + show stuff is cyclic||
dark sparrow
yeah because your AB and CD don't meet
your diagram is fundamentally wrong
now THAT's more like it
you deleted the pic containing the original problem statement
okay so you're interested in triangles APD and CPB
so yknow
maybe make those more visible
and also make it more visible which pairs of segments are known to be equal
i mean
it'd probably be better to do it on paper
so that you don't have precise control over what you're not supposed to
still
you've deleted the original problem statement
there were two pairs of segments that you were told are equal
which ones were they
yes that's what the bot's for
somber coyoteBOT
euler/fermat:
dark sparrow
...
color
why did you even need to color them
but ok
ok
AP = CP and BP = DP
ok so you wanted diagram-making advice
my advice is
don't even touch any computerized diagram-making tool
and don't use coordinates like EVER unless the problem involves them somehow
make your diagram such that no two lines appear parallel unless they're explicitly KNOWN to appear parallel
make no angle appear right unless it is KNOWN to be right
everything that is KNOWN should be MARKED, such as right angles, or other angles that happen to be known too
segments that are equal should be marked with notch patterns
like this
segments that aren't KNOWN to be equal should not be made to seem equal
point is
you need to make your diagrams a bit sloppy
neatness is bad because at best it locks you into "ok this is true in this one case but what about the general case"
and at worst it makes you come to incorrect conclusions that can be busted with a counterexample
make your diagram such that no two lines appear parallel unless they're explicitly KNOWN to appear parallel
make no angle appear right unless it is KNOWN to be right
segments that aren't KNOWN to be equal should not be made to seem equal
of course
that doesn't mean that you should draw a segment of length 2 appearing longer than one of length 7
point is
a diagram is, first and foremost, a means of displaying the information that you know, AND NOTHING ELSE.
somber coyoteBOT
euler/fermat:
dark sparrow
oh, my apologies
i connected the wrong segments
there
easy fix
you mean APD and CPB?
i can add another piece of known information to my diagram, like so:
this isn't given directly but it is derived from what is
do you mean "you need to argue that angles APD and CPB are the same"
if so: they're vertical angles
well i'm not sloppy about making straight lines straight
and also because i don't take the diagram at face value
i mean ok
this is in microsoft paint and i used the line tool
were this on a piece of paper, i would probably have made a line that looks straight
but i wouldn't give much of a shit about making it PERFECTLY straight
what matters is that i know that the line i made represents an idealized straight line in the idealized setting of euclidean geometry
and that i know that AB and CD are both supposed to be straight
and that i know that two intersecting straight lines make two pairs of vertical angles
not at the moment no
i mean exactly what i said, i don't have anything to give you right now
¯_(ツ)_/¯
i said nothing about whether or not i'll encounter any such resources later
i'd say that the relative order of the vertices matters, but not necessarily the absolute order
but that's gonna get lost on you
so i'm not even gonna try
i'd say it does but i'd rather have a diagram
sounds about right then
no, because in other settings you may not even have any points named T, S or Q
oh my god
why do you keep deleting your previous messages that contain images
this is really fucking annoying and you should stop doing that
somber coyoteBOT
euler/fermat:
dark sparrow
yeah but WHY DID YOU HAVE TO GO AND FUCKING DELETE IT
OH DOES IT REALLY.
it takes as much space as a message that's 5 lines long, at best.
yikes.
i don't know! maybe the authors of some low-quality books don't follow that convention!
do you expect me to be OMNISCIENT and be able to CLEANLY DELINEATE EVERYTHING FOR YOU in EVERY SINGLE FUCKING BOOK IN EXISTENCE
because whether you want it or not THIS IS WHAT YOU'RE MAKING IT SOUND LIKE
and if you haven't taken any of the at least 10 hints that i've dropped now i am ACTUALLY REALLY FUCKING UPSET RIGHT NOW and your incessant questions are actually ONLY MAKING IT WORSE
oh and now you're trying to placate me with a half-assed attempt to redeem yourself
lusty quest
dawg you can just google that
@upper karma its to learn how to google, just make sure to ask questions and explain why they stump you
dark sparrow
wow
@upper karma what the hell
why did you just
why did you just up and delete your own half of the fucking dialogue
you make me sound like a weirdo talking to herself
why
seriously i don't understand
why would you leave the answer hanging as if it was screamed into the void
quiet mason
i didnt merely paste that i did it to show sloth something he was asking for
trail minnow
it was no mere copy paste
lmfao
@dark sparrow OH MY GOD READING THIS IS FUCKING HILARIOUS OUT OF CONTEXT HAHAHA
dark sparrow
see euler
trail minnow
upper karma
👀
upper karma
lmao
lusty quest
ok so something like i/4-i is a complex equation, what does it mean by absolute value and arguments?
lusty quest
so i have to multiply the absolute value and add the argument
digital stirrup
Hello. First of all, i/4 - i is a complex number, not a complex equation. They're kind of weird at first, but what's cool about complex numbers is that they are 2D numbers. Check out this page, I loved this article so much when I first read it: https://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/
dark sparrow
not when dividing
digital stirrup
Oops, I didn't see someone was already answering
dark sparrow
and it's very likely they meant parentheses again
and wanted to write $\frac{i}{4-i}$
somber coyoteBOT
Ann:
dark sparrow
or i/(4-i)
digital stirrup
So Backyard Babushka, the argument of a complex number a + bi is its angle (which is easily visible in the trigonometric form of a complex number, and can be found with tan^-1 (b/a)) and the absolute value is found by sqrt(a^2 + b^2). The link I posted helps connect this stuff visually, and there's other really good pages on that site for understanding the basics of complex numbers. The trigonometric form of a complex number is R * (cos x + i sin x), where x is the angle or argument of that complex number, and R is the absolute value of that complex number. Think about what happens when we take two complex numbers in this format: A * (cos x + i sin x) / B * ( cos y + i sin y), where A is the absolute value of the first complex number and x is the argument or angle of the first complex number, and B is the absolute value of the second complex number and y is the argument or angle of second complex number. What is the argument of the new number? What is the angle of the new number? (remember how to rationalize denominators when doing complex number division: you multiply numerator and denominator by the conjugate of the denominator) Also keep in mind your trig identities for sine (x + y) and sine (x - y), and cosine (x + y) and cosine (x - y) (Sum and difference identities)
upper karma
Hi guys, need help with solid geometry
Isn't the solution the volume of the complete sphere divided by 8 ?
hard gale
well they want the surface area not the volume :/
upper karma
thanks man ! I'm not so good at English and I'm kinda shallow haha
upper karma
So, is the solution pi*4*3/4 + pi ? I've used all my attempts on edx
It's supposed to be the area of the circle with radius 2 times 3/4 (the 3 flat sides of the octant) plus the full surface area of the sphere divided by 8, that is 4*pi*r^2 / 8 (and simplified gives me just pi)
hard gale
4/8 = 1
that's what you just did @upper karma
and don't hesitate to ping back if i don't answer in this type of situations, pretty busy atm
upper karma
found the error, thanks guys
upper karma
Hi
upper karma
Hey
silent plank
have you drawn a rough sketch?
and from your diagram what would be the length and width of the big rectangle
what is the width of the path on the left/top side?
and that would make the total lengths/widths?
well you have 'w' on the left, 'w' on the right and 11 in the middle
stiff forge
Hello, could I ask a simple question here?
digital stirrup
Sure
stiff forge
it is correct to spell that the domain of Cotangent and Cosecant are all real numbers except {nπ} ?
digital stirrup
Yes, that is the domain of the cotangent and cosecant functions.
cosmic flower
sinx + acosx = a
sly marlin
Divide by $\frac{1}{\sqrt{1+a^2}}$, I guess
somber coyoteBOT
Element118:
quiet mason
uhm
mild vale
well uh
mild vale
no
mild vale
my knowledge of trig is very fragmented- this I did not know
quiet mason
koh o
mild vale
for reference, was solving this question: The polar curve r(theta)=1-2cos(theta) for 0<=theta<=theta
let R be the region in the third quadrant enclosed by the curve and x-axis
which integral represents area R
I got the 1/2r(theta)^2 correct
:P
pliant matrix
upper karma
lmao
quiet mason
lol
dark sparrow
you approached this problem in a way that lends itself too easily to fuckups
it's kinda hard to tell what you even did
your work looks like a string of random equations with no rhyme or reason to them
i'm not going to force any one method on you, and i'm also going to set aside the fact that you pinged me despite me being right here
but what i'd do is make this line a number line
with J at 0 and K at 1
it then becomes almost painfully obvious that U and V are 7/8 and 7/6
so their midpoint is 49/48
and from there, WJ/WK is quite clearly 49
this seems to be the most headache-free solution
your method is probably fine if executed correctly but it's just hard to keep track of everything like that when you've got all those letters piling up
i mean ofc you weren't told to make it a number line lol
it's basically coordinates but in one dimension
idk 🤷
like uh
no really idk if the setting of a problem is a straight line i find it a rather obvious idea to at least try to make it a number line
abstract snow
is it ok if i ask a question?
dark sparrow
abstract snow
lol
need halp with a geometry proof
its Prove <ABC + <BCD + <CDA + <DAB always = 0 (mod 180)
and the angles are all directed angles (clockwise) so <ABC = 360 - <CBA
and the < is an angle sign, not a greater than
frosty steppe
Is this in a quadrilateral? @abstract snow
frosty steppe
Alternatively, think of a car driving across the quadrilateral until it reaches its starting point again. After going all the way around the shape, it will have made one full turn
Perhaps that's a more intuitive way of thinking about it
abstract snow
nah cuz it could be something like
A-----B
\ /
\ /
/
/
C-----D
so it doesn't need to be a quadrilateral
abstract snow
well ig you could just figure it out by vizualizing it
but is there a way to prove it?
idk if that makes sense?
frosty steppe
Of course, I'm having trouble of thinking of a way to formalize that idea
That's half the journey
abstract snow
yee lol
frosty steppe
Do you know how exact your proof needs to be? I can think of a few things but they're not super rigorous
abstract snow
thats ok
it doesn't need to be super exact i think
i did this, but its not a really good proof cuz i used the beginning assumption in the proof
honestly i think i just proved 720 = 720
frosty steppe
When you start with an assumed statement, then you'll get to something like that
Yeah this is tough. Do you see what I'm saying about the car making a full turn along this path?
abstract snow
yee, i get that part
frosty steppe
Then your guess is as good as mine about how to put that into numbers
I'll think about it
abstract snow
ok thank you!
frosty steppe
You think too
abstract snow
lol i def will
upper karma
If the point (k, 5) is 3 units from the point B, find the coordinates of F. B(-3, 4)
Any help would be appreciated I just dont understand this questio
spark stag
what's F?
upper karma
No Idea?
spark stag
do you have any more information? a diagram?
upper karma
nope I was just given that question and the point B
spark stag
oook, i guess i'm just gonna assume F is the set of points that satisfy the relation given
upper karma
Ill keep trying to work it out
might need to ask my teacher because this seems like a bad question
spark stag
this is waht i'm assuming the question means
it wants you to find the coordinates of F_1 and F_2
i think?
and F_1 and F_2 are the points (k,5)
such that they're a distance 3 away from B
upper karma
yes
spark stag
we can find these by drawing a right triangle
and jamming pythagorean theorem
and then using those findings to find the coordinate of the point
upper karma
wait I think I have answer, (-3 + 2 sqrt(2), 5)
spark stag
that corresponds to point F_2 in my diagram
spark stag
we know the side of length 3
that's given by the question
the side of length 1 can be found by considering
F has y value 5
B has y value 4
so their difference in y value is 1
upper karma
yes
spark stag
and then pythagorean theorem lets us find the 2sqrt2
upper karma
nice
spark stag
sqrt(9^2 - 1^2) = sqrt(8) = 2sqrt2
so now, since the coordinate of x for point B was -3
we add 2sqrt2 to that
-3 + 2sqrt2
the only weird bit is that
-3 - 2sqrt2
is also correct
so maybe its meant to be $(-3 \pm 2\sqrt{2}, 5)$?
somber coyoteBOT
Namington:
upper karma
yep I agree
spark stag
¯_(ツ)_/¯
upper karma
Thanks for you help
olive solar
,rotate
somber coyoteBOT
olive solar
honestly i just went up to where it returns an integer at all. once it returns an integer, you can just square it if it's negative and keep it if it's positive
spark stag
^
dark sparrow
to make sure i understand your handwriting
you're asked to find the smallest positive integer $n$ such that $\left( \frac{2i}{1+i} \right)^n$ is a positive integer?
somber coyoteBOT
Ann:
quiet mason
4
spark stag
uhhh
check your work there
@quiet mason
wait misread
fk
although youre still wrong
dark sparrow
k so 2i/(1+i) = 1+i just to get that out of the way i guess
spark stag
yeah, as noted
so now we just want to consider
when that i "disappears"
since as colorado says above
dark sparrow
the argument of 1+i is π/4 aka 45° aka 1/8 of a full turn
spark stag
ah, thats a good intuition too
upper karma
$(i+1)^n = \br{\sqrt 2 e^{i \frac \pi 4}}^n$
somber coyoteBOT
dark sparrow
raise the thing to the 8th power and you get 16, a positive integer. raise it to any power below 8 and you get something that either isn't real or is negative
spark stag
yep
you can verify this by cases, if youre patient
but ann's visualization is the best way to look at this
dark sparrow
honestly if you don't have a geometric picture of complex multiplication in mind
then you should work on that
bc like
uh
yeah
upper karma
lol
spark stag
pfft, just consider the group (C \ {0}, *) and consider group operations
truly a much better intuition
dark sparrow
namington i'm serious
i'm trying to impart some actually useful intuition that one can use to do shit here
hrghgrnggrhngngh
spark stag
alright, i'll back off the joking; I didn't intend to distract from your intuitive explanation
worthy root
I am so active on this server contribute valuable question and jokes but I am still not honourable
I am so underrated
upper karma
What is two isoceles right triangles, both placed opposite to each other's hypotenuse?🤔
spark stag
i... dont understand your question
upper karma
It's a square @spark stag
spark stag
what is the "opposite of [a] hypotenuse"
do you mean "placed so they share a hypotenuse"?
upper karma
Indeed
spark stag
...why are we sharing trivial facts
upper karma
Hm no specific reason, we just do
worldly scroll
can someone help me with this
dark sparrow
what have you tried and where are you stuck
dark sparrow
and how exactly did you "jump into using cosine"?
worldly scroll
|AB|.Cos90 = 4
dark sparrow
no, |BD|/|AB| is cos(ABD)
worldly scroll
but I'm not sure how to get it
dark sparrow
consider that some of these triangles in the picture might be similar
worldly scroll
I can see something do with the ratio of 4 and 5 maybe
dark sparrow
no
consider that some of these triangles in the picture might be similar
forget about the 4 and 5 for now
worldly scroll
ok
worldly scroll
I really have no idea
strong lion
Help!! 😦 sitting here for hours trying to proof this equation :((( can somebody help me? 😃
eager pendant
@strong lion how are points defined
strong lion
AT = 1/2c ; L is the point where the height over AB is ; M is defined by half the angle of y; E is the point where Incircle touches AB
Angle b isn’t defined, it’s about a general triangle
astral mirage
Can any of you fine ass gentlemen help me with a problemo
Desperate and really don't know what to do
It's on conditional probability
it'd be much appreciated
spark stag
we're selecting a 7th grade student, so our "possible choices" only include 7th graders
of which there are 40
so we have 40 "possible" choices
out of those 40, 11 want to join Medical Science
so our probability is...
spark stag
yep.
astral mirage
Good sir I'd suck the dick clean off your body if I could, thank you
spark stag
...that's a new one
astral mirage
I have one more if you're still available for the help?
not gonna lie, this one's been getting me
spark stag
probability (when everything has an even chance) can be thought of as
$P(\text{something}) = \frac{\text{possible ways for that "something" to happen}}{\text{possible ways for anything to happen}}$
somber coyoteBOT
Namington:
spark stag
in the picture, the denominator is 52 choose 3 for all of them, which makes sense
theres 52 cards in a deck, and we choose 3
so theres 52 choose 3 possible ways to pick 3 cards
for each option, we want to check whether the numerator "counts" the number of ways to meet the criteria
for example
let's look at #1
P(2 aces and 1 ten)
how many ways are there to select 2 aces out of 4?
(since there's 4 aces in a deck)
astral mirage
wouldn't there be 2?
spark stag
no:
Ace of Spades + Ace of Hearts
Ace of Spades + Ace of Clubs
Ace of Spades + Ace of Diamonds
that's already 3 different ways
and, of course, there's more
this is why "choose" (nCr) exist
nCr counts the ways to choose r objects out of n
we're choosing 2 aces
out of 4 aces
so nCr = 4C2
so the 4C2 part works
but what about the "tens"?
how many ways are there to choose 1 ten out of a deck?
leaden nova
Yo namington
spark stag
?
leaden nova
Priest is shy to tell you this but he doesn't really know how to answer the questions you are asking
spark stag
well, I'm afraid that I cant give a full ground-up introduction of combinations/permutations
if thats the case, you might want to review that part of a textbook or look it up on khan academy or something
leaden nova
He knows what it means
astral mirage
Truthfully I am here to ask help for cece
now that she is here, just gonna casually boop out
leaden nova
Sorry we are confusing... I just need the help.
somber coyoteBOT
euler/fermat:
limpid basin
So basically there’s a formula that states that the area of ADE is (AD)/(AB) * (AE)/(AC) * [ABC]
So you want to replace those ratios with their respective values and then find the complement of that area
You can prove it pretty straight forward
Let D’ be a point on AC such that DD’ is parallel to BC. Then [ADD’] = (AD/AB)^2 * [ABC]. [AED] = [ADD’] * (AE/AD’) (ratio of the bases). And since DD’ and BC are parallel, AD/BC = AD’/AC, so AD’ * AB = AD * AC. We can then multiply by 1 = (AD’*AB)/(AD*AC) to get [AED] = (AD/AB)^2 * (AE/AD’) * [ABC] * (AD’*AB)/(AD*AC) = (AD/AB) * (AE/AC) * [ABC]
Dunno
Maybe intermediate will
Wtf?
Oh you mean WOOT?
Why do they only have one intermediate book :I
Anyways I’m pretty sure they cover it in AoPS vol 2
Lol idk
I mean...
This formula is just handy I don’t think it’s that widely used lol
Btw what was your AMC score this year?
oh >.>
No bueno
Wait so did you take BMO or something?
It’s the British MO
Idk what your country does
Lol trash
we never talk about my amc score
AMC 10 :(
?
Also we might want to move to #competition-math this is offtopic
silent plank
yes
dark sparrow
your question seems to suggest you know some proof of the similarity of a pair of triangles but for whatever reason it doesn't qualify as rigorous to you
do you mind sharing that
dark sparrow
you know some proof of the similarity of a pair of triangles but for whatever reason it doesn't qualify as rigorous to you
do you mind sharing that
you have a proof you feel uneasy about
share that proof
well ok before we do that, we need to have a definition for congruence of triangles
and at least from what i've seen, SSS is taken as the definition, and then to prove SAS and ASA one reduces them both to SSS
you've yet to answer my request, though.
you have a proof you feel uneasy about
share that proof
uh
native grove
Hey guys, I asked a question in beta regarding a circle that circumscribes three other circles. Could anyone here give me a hand over there? :)
strong lion
Was somebody able to solve my problem?
dark sparrow
what problem
strong lion
Proofing that equation
AT = 1/2c ; L is the point where the height over AB is ; M is defined by half the angle of y; E is the point where Incircle touches AB
in a general triangle
It’s part of a different problem, however I managed to break it down to this point
But know I’m stuck
Now*
eager pendant
@strong lion what's the entire problem
oh M is the intersection of the angle bisector and AB
buoyant idol
ok this is gna seem real ez
but fuck it idfk how to do this
i need to find the ratio of the grey area
for some reason japanese maths r strangely hard
the syllabus is extremely different and frankly kinda embarrasing how i cant solve this questiton
anyone?
the answer is 1:15
dark sparrow
you need to find the ratio of the areas of the grey triangles?
buoyant idol
okok
also i dont think ur allowed calculators so ye
no crazyz stuff like idk......
buoyant idol
ikr
its in japanese but the most important parts r in numbers
it says PR:RB is 1:6 which i have no fckin idea how they got
dark sparrow
sorry, this is making my brain hurt a bit
buoyant idol
smae
trail minnow
@buoyant idol can you translate if its not too much trouble?
buoyant idol
sure i can try
trail minnow
try to do so as literally as possible, otherwise we can't really understand the hypotheses
dark sparrow
are the circled katakana a, i and u like equation numbers
buoyant idol
(1) draw a line through point Q, parallel to AD, as so in the right picture/image and draw point S, T in the intersections
triangle BAP is the equivalent (but different size) of triangle BTS, the ratio of this is 1:2
If we write TK as k AP is 2k. This means TQ = AD = 2k*=6k
Therefore QS=6k-k = 5k
triangle QSR is the equivalent (but different size) of triangle APR, the ratio of this is AP:QS = 2:5. This means that AR:RQ = 2:5, PR:RS = 2:5
from PR:RS is 2:5, PR:RB= 2:(5+2+5) = 1:6
from AR:RQ=2:5 and PR:RB= 2:(5+2+5) = 1:6, triangle ARP: triangle RBQ = AR * PR : RQRB = 21:5*6 = 1:15
thats the best i think i can do
does that make sense?
silent plank
i found it
buoyant idol
u did!?
silent plank
at the start,
tri BAP similar to tri BTS
buoyant idol
ye
silent plank
PR:RS is in a ratio of 2:5
silent plank
and
PR+RS = BS
buoyant idol
ahhh
silent plank
combining that info will get that 1:6
buoyant idol
ye
zenith copper
Hey,
I have a question regarding the locus definitions of an ellipse.
I am familiar with the definition "the set of points for which the sum of the distance between two points is a constant value",
however my 11th grade math book here gives the definiton of "the set of points with equal distance to a circle and a point within the circle"
I can't find anything online about this second definition. Is anybody familiar with it? do there exist any other alternative locus definitions? how do they relate, and how can you prove they are equivalent?
thanks
silent plank
understand it fully now?
buoyant idol
yeye i got it
u see how easily hard japanese math is?
its ez but hard well at least for me
oh ye i still need ur help cuz therres 6 more of those questions
steady forum
Er need help, i used 2sinxcosx divided bu cosx to get 1/2sinx then inversed to get 30 degrees and 150 degrees, im not sure how to get the 3rd and 4th answer which is 90 degrees and 270 degrees
Pm me if you know thank you
silent plank
factor instead of eliminating it completely
steady forum
Ok let me tey
Tey
Try
Got it
Inversed cos x-1 (0) is 90 or -90 which is 0 or 270 degrees
Thank you
silent plank
same for any future trig questions otherwise you'll lose solutions
steady forum
Ok, i will always consider factorising!
buoyant idol
can someone help me with a similiar question as befire
steady forum
When does sin45= 1/sqaure rooted 2 and when oes it equal sqaure rooted 2 over 2
Because i thought sin45 degree is like this
hard gale
1/sqrt(2) = sqrt(2)/2
buoyant idol
wait whot
hard gale
same thing
steady forum
buoyant idol
oh ye lmao
hard gale
yeah
steady forum
Er
buoyant idol
what onion said
steady forum
Ok wait let me try the question with both solutions
hard gale
$$\frac{1}{\sqrt{2}} = \frac{\sqrt 2 }{\sqrt 2 \sqrt 2} = \frac{\sqrt 2}{2}$$
somber coyoteBOT
emeric75:
steady forum
Im in grade 8 lol
steady forum
I think?
buoyant idol
its so much harder than u think
steady forum
The smaller triangle ontop of the orange one is sam right
steady forum
Zzz
buoyant idol
it doesnt say it is
steady forum
U need the bottom small triangle to minus the small trinagle at the right side
Its werd
Its a rhombus
So uh
Wait
Wtf
buoyant idol
whut
steady forum
The orange tri area is 4?
steady forum
Ok no bye this is damn hard lol
buoyant idol
lmao
steady forum
I go revise for my test forst later i come help u if i got lah
buoyant idol
u from asia?
steady forum
Sg
buoyant idol
lmao
trail minnow
kek
upper karma
Hello lah
tropic turret
As a Singaporean, I can confirm we use "lah" compulsively.
silent plank
@buoyant idol was there a solution provided?
unless i'm mistaken you can use ratio of areas several times which will get ||76||
scarlet flower
@buoyant idol what answer are you searching for?
silent plank
are you completely stuck?
buoyant idol
sorta
silent plank
mmk, i'll give partial hints.
do you know how to find the ratio area of similar (2D) shapes?
buoyant idol
i think so
but how do u know theyre similar
ik the bigger triangle on top of the small one is around 64
buoyant idol
idk abt the small triangle next to that one is tho
silent plank
now focus on the triangle at the base
buoyant idol
but how do u know theyre equal
buoyant idol
sure
hold on
I GOT IT
thx
i swear to god my heart is gna pop lmao
i got 5 more question due tmrw night
AND ITS 2 AM
and im having an existential crisis abt whats real and how to define reality
silent plank
do you need your solution dbl checked?
buoyant idol
dbl?
silent plank
double.
silent plank
not enough info to say that yet. all you have is 1 angle
you'll need to go through a few more steps
buoyant idol
yeye hold on
here
the pink shows theyre the same tri
its a bit confusing cuz most of them r pink
but ik that if 2 anglles r the same and 1 line is the same, theyre the same tri
lmao
uhhhh
i still got a lot so im j gna continue posting here for those that want to solve it
AMND and MBCN r the same area and u have to find x
small raptor
What does the parallel tell you regarding 74 and 3z-19?
agile tree
all it says is In the figure below, h is parallel to L and j is parallel k. Find the values of x and z.
small raptor
Right, and that is key. There are properties regarding the angles.
Do you know them?
agile tree
no i don't thats what i got stuck on i tried looking them up but was confused what i thought was the corresponding angle theorem
small raptor
Well, if the lines are crossing parallels, then you have congruent angles so, in this case, what you see above is the same for below. Now, because line k is equal to line j, as they are also parallel to one another, then their angels also are congruent to each other. In other words, 3z-19 is equal to 74 degrees.
If you need a visual aid on this
agile tree
so should i solve the equation 3z-19=74?
small raptor
Right, for z.
small raptor
no
agile tree
because they are congruent?
small raptor
Remember the rule of Supplementary Angles in that they equal to 180 degrees (due to being a line)
small raptor
No problem.
buoyant mirage
does trig ever stop being a giant pain?
mental wharf
yes
buoyant mirage
i just feel like in all the math classes i've had, nobody's ever really explained trig well, maybe that's the issue
frosty steppe
@buoyant mirage it’s definitely a lot less intuitive. It’s only useful and not pretty for its own sake which is kind of frustrating
At least the trig curriculum taught in hs
It’s the gateway to cool stuff though
small raptor
What is it that trip you up thr most? I find practicing graphing the basics in both unit circle and wave form to help a lot
buoyant mirage
all of it really. i took an entire class on trig and the only thing i got out of it was "sohcahtoa"
the more i lurk here, the more i realize my math education was... not good
small raptor
Happens. Pick up a trig book and slowely work through it
buoyant mirage
wish i could, currently grinding calc though :P
zenith copper
you are gonna need it once you reach trig 2ish kinda level
buoyant mirage
yeah, just hit some calc problems that required some trig knowledge
so im grinding trig so i can grind calc lmao
classes start in 2 weeks and im so screwed
zenith copper
you can do surprisingly much in two weeks :)
agile tree
Hey guys i was wondering if anyone can help me solve this
dark sparrow
what's giving you trouble here?
agile tree
im not really sure on how to solve it
im doing some geo review and forgot but im pretty sure i can use trig
spark stag
you "can use trig", yes, but thats pretty vague
lets start with the left triangle, the one with h
buoyant mirage
rusty on trig, but i would find the length of the hypotenuse and then use that to find the length of h
spark stag
can you figure out which trigonometric ratio should we use there?
...why
you can just find h directly
agile tree
yea i use tangent namington on the first angle
buoyant idol
so can anyone help e w the problem from yesterday.... :/?
small raptor
Hum, sounds like a difficult problem indeed. I wouldn't know how either as I never done that, but I would probably start with the area of the whole triangle as 1/2*bh - 17, the 17 being the combined total area of the bottom. this would give you bh = 34, so then all you need to do is find the base. At least, that's my approach. I have no idea if that would be correct
upper karma
@small raptor wat're u responding to
small raptor
There was a image and question, but for whatever reason, they decided to delete it.
queen python
gleaming yoke
need help with this q :" )
silent plank
what have you done so far?
dark sparrow
you got to $QR = \frac{8}{\frac{\sqrt{2}+\sqrt{6}}{4}} \cdot \frac{\sqrt{3}}{2}$
somber coyoteBOT
Ann:
dark sparrow
why did you stop?
gleaming yoke
that gives -4√6 + 12√2 cm
dark sparrow
,w simplify 32/(sqrt(2)+sqrt(6)) * sqrt(3)/2
somber coyoteBOT
Requested by Ann#0413
dark sparrow
ok yes you did the simplification correctly
...where'd you get it from that the answer is $4\sqrt{6}$ though?
somber coyoteBOT
Ann:
dark sparrow
no, your answer is definitely correct. i think the book has an error in its answer key
does it only provide an answer or a worked solution as well?
gleaming yoke
nope does not include any worked solutions
but look here (mind the messy working haha)
when using sin45 instead of sin75 i get 4√6
i think theres been a typo too
dark sparrow
okay yeah maybe the 75 was meant to be a 45
as written your answer is correct though
gleaming yoke
agreed. 180-75-60(which are the angles provided) makes the other unknown angle =45. so yeah prolly an error in the textbook.
anyway thank you so much for the help, much appreciated. have a great day @dark sparrow < 3
peak mesa
Anyone knows how i rotate a 3d cube by x or y axis by local space? I'm trying but the result isnt correct...
Any sugest mention me
dark sparrow
what have you tried and where are you stuck?
peak mesa
https://gyazo.com/2419bcb448748063e72e02d16eb65ae2
With that code
'-'
I can see that there is a rotation but all wrong
with this
I don't know what else to do, I've tried a little of everything
@dark sparrow
dark sparrow
okay uh
my message was in response to a question that has since gotten moved by the person asking it
digital stirrup
Maybe there's something to do with matrix multiplication? I don't know much about rotating around an axis, but that might have something to do with it
small raptor
I thought a matrix is involved myself, I know it's the case with Quaternion rotation, but that said, it's not really my area. Probably better going to a programming or CS server
digital stirrup
@peak mesa check out rotation matrices
I think they're exactly what you're looking for.
Check this out:
Apply those matrices as transformations will rotate all points by theta about one of the axes
peak mesa
But the location have 3 values and witch thing have 9 values
wise quartz
it looks like the original formulas you gave are correct tho (for rotating around the x axis in a right-handed system)... can you try it on a smaller / more static example? the problem may be elsewhere in the code
digital stirrup
Then multiply by one of the rotation matrices above
wise quartz
what are the coordinate axes in the animations?
matrices are a useful tool but (unless i'm missing a sign error?) they will just reduce to the same formula you posted at first
digital stirrup
Just looking at what I wrote above, you would actually multiply [x, y, z] by these matrices:
It depends on how you want to set it up
wise quartz
sendor: they already are, those reduce to the original lines of code
digital stirrup
Oh okay
I suck at programmin ghaha
I was just pointing out a slight difference (where I was wrong), in that if you want to rotate [x, y, z,] you multiply it by that second image I posted it, not the first one
wise quartz
basically they're computing [x y z] = Q_x(theta) [this.x this.y this.z]
digital stirrup
Oh, alright. So that's already in his code
wise quartz
i mean maybe there's a sign error, i miss those constantly, but it's the right idea :-)
digital stirrup
Ah, cool, thanks. Just looking at the first and second images I posted, they are different in that you might get a sign error depending on how you set up the multiplication. Since the columns of the first matrix have to equal the rows of the second matrix, if you want to multiply a vector by a rotation matrix you use the second image, and if you want to multiply the rotation matrix by a vector you use the first image. Otherwise there would be an error in the direction of rotation
So that could be where the problem is, idk
wise quartz
it's hard to tell for sure but the last animation looks like there's an intermediate state where the cube maps to a plane... none of these matrices shd do that for any parameters, even under transpose etc, which is why i'm suspicious about a problem elsewhere in the code
digital stirrup
Ah ok, yeah that could be
agile tree
Which properties are true for these quads?
digital stirrup
@upper karma I love that picture too. That site is absolutely fantastic, it's helped me so much
Vaultier, picture a trapezoid and parallelogram. See which ones have right angles, parallel sides, and notice their side lengths
mighty gull
its a nice picture, with my one problem with it being it assuming tan(theta)<1, but its still a nice pic
digital stirrup
Ah yeah, that's true. It's an interesting way to memorize the identities, and as long as you keep in mind what tangent and cotangent and all the other ratios really are, you'll be fine
buoyant mirage
i'm trying to solve part of this problem: http://tutorial.math.lamar.edu/Solutions/CalcI/TrigEquations_CalcII/Prob4.aspx
specifically, trying to solve the equation cos(3y/7)=0. solving gives two solutions, 3y/7 = (π/2) + 2πn and 3y/7 = (3π/2) + 2πn However, the solution on the page says "Notice that we can further reduce this down to 3y/7 = (π/2) + πn".
How do the two solutions reduce down to one? I'm kinda lost
wise quartz
each of the two solutions is an infinite pattern repeating every 2pi, and the two are offset from each other by pi, so they merge into a single infinite pattern repeating every pi
buoyant mirage
ah, that makes sense. thank you for the explanation!
upper karma
on the euler line we have equally spaced: the circumcenter O, centroid G, then some point X, then the orthocenter H
is there any name or characteristic property for X?
pastel swift
