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Serious Daily math problem thread

Here's a solution for problem #0017
Screenshot 39

Also see @mNFwTJ3wz9 's solution. I like his better than mine, because it's more elegant and requires less laborious calculations.

...cont.
Now we make a rectangle around the whole thing.

putting c into the equation

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
y^2 + (1 - 5/2) *y + (25/16 - 1) = 0
y^2 - 3/2 * y + 9/16 = 0
(y - 3/4)^2 = 0
y = 3/4

View attachment 414017
View attachment 414020
View attachment 414019
Area(purple) = tri (green) + rect(red) - curve(blue) - circle(yellow)

Area =
+ 1/2 * 1/2 * sqrt(3/4)
+ sqrt(3/4) * 3/4
- (sqrt(3/4)^3) / 3
- pi/3 * 1/2 * 1^2

what integral?

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448
 
Problem #0018
F23CBFF4 225F 433E 9006 D26A713E56AD

in the diagram, there is a semi-circle, with a circle inscribed in it such that it is tangent to the midpoint of the diameter of the semi-circle as well as the top of it. if AB is perpendicular to the diameter of the semi-circle and tangent to the circle and has length 1, what is the area of the orange circle?

i will post my solution tomorrow. have fun

@mNFwTJ3wz9
@Divergent_Integral
@SelfCrucified
@grondilu
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt
 
As in how do you integrate the big square root
i used a trig sub.
let x=sin(u), then dx=cos(u)du

then we are integrating
sqrt(1-(sinu)^2)*cos(u)=(cosu)^2

to integrate (cosu)^2, you can use integration by parts or the cosine double angle identity: cos(2u)=2(cosu)^2-1.

then, you can rewrite everything back in terms of x.
 
is there a particular reason for subbing in trig? Trig integrals have never been my strong suit
when dealing with roots, we like to use trig-subs to take advantage of identities like sin^2+cos^2=1. It may be easier to integrate trig functions rather than having to deal with roots.
 
Any thoughts about a Daily Statistics/ Econometrics/ Financial Mathematics/ Actuarial Mathematics etc problem thread? As well as Candle Stick Charting/ Game Theory etc question thread?
 
Any thoughts about a Daily Statistics/ Econometrics/ Financial Mathematics/ Actuarial Mathematics etc problem thread? As well as Candle Stick Charting/ Game Theory etc question thread?
Shut the fuck up cracker that is cracker nerdy white boi shit on the G
 
For your edification and intellectual stimulation I will post a math problem on here daily (or at least a few times a week). I will try to make the problems as accessible as possible, in that I will assume no really advanced mathematical knowledge. Just a bit of cleverness will do.

You may post your solutions in the thread. If none are received, I will give the solution myself the next day. Good luck, and happy problem solving!

Problem #0001

Let a, b, c be positive numbers such that a + b + c = 1. Prove that (1 - a)(1 - b)(1 - c) >= 8abc. Under what conditions on a, b, c does the equal sign hold?
 
Any thoughts about a Daily Statistics/ Econometrics/ Financial Mathematics/ Actuarial Mathematics etc problem thread? As well as Candle Stick Charting/ Game Theory etc question thread?
Financial math threads would be useful for stockcels.
 
Problem #0018
View attachment 414552

in the diagram, there is a semi-circle, with a circle inscribed in it such that it is tangent to the midpoint of the diameter of the semi-circle as well as the top of it. if AB is perpendicular to the diameter of the semi-circle and tangent to the circle and has length 1, what is the area of the orange circle?

i will post my solution tomorrow. have fun

@mNFwTJ3wz9
@Divergent_Integral
@SelfCrucified
@grondilu
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt
r = diameter of inscribed circle
(1)^2 + (r/2)^2 = r^2, r -> sqrt(3/4)
area = pi(sqrt(3/4)/2)^2 = pi/3
 
r = diameter of inscribed circle
(1)^2 + (r/2)^2 = r^2, r -> sqrt(3/4)
area = pi(sqrt(3/4)/2)^2 = pi/3
Nice, you got it right

good work:feelsthink:
 
New problem:

For N=16, find the minimal collection {x1, x2, ... } of unique numbers 1 <= xn <= N

Such that every number k from 1 to N can be written as:

k = x + y OR k = x - y OR k = x

Where x and y are some numbers from this collection

Also, give a general way to derive this minimal collection
 
Last edited:
post the next problem @WorldYamataizer
 
Problem #0019

if x>0 is a real number satisfying the equation
(x^2)+12sqrt(x)=5​
Then determine the exact value of the expression
x+2sqrt(x).​

@mNFwTJ3wz9
@Divergent_Integral
@SelfCrucified
@grondilu
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt
@copeful
 
I turned on my TI-84 to use the equation solver and found the batteries are dead. I don't have any AAA batteries...
Copy paste and wolf ram alpha gives the latter expression approximately equal to 1. It would be more interesting to work it out though.
 
Last edited:
I turned on my TI-84 to use the equation solver and found the batteries are dead. I don't have any AAA batteries...
Copy paste and wolf ram alpha gives the latter expression approximately equal to 1. It would be more interesting to work it out though.
This problem is very challenging. But yes, the expression is exactly 1.

Here’s a hint: Factor the given equation, somehow. Do you see any squares?

it took me personally a few hours to come up with a solution. its one of those “rabbit-from-a-hat” problems where u have to make a specific move that isn’t too obvious
 
This problem is very challenging. But yes, the expression is exactly 1.

Here’s a hint: Factor the given equation, somehow. Do you see any squares?

it took me personally a few hours to come up with a solution. its one of those “rabbit-from-a-hat” problems where u have to make a specific move that isn’t too obvious
We can rewrite (x^2)+12sqrt(x)=5 as
x^4-10x^2-144x+25=0

When I was in high school years ago I would surely be able to solve this.
 
we don't necessarily need to solve for x, then plug that value into x+2sqrt(x).

i checked Wolfram and it gives that the exact value of x would be 3-2sqrt(2).

Then x+2sqrt(x)=3-2sqrt(2)+2sqrt(1-2sqrt(2)+2)
=3-2sqrt(2)+2sqrt((1-sqrt2)^2) <-----(Perfect square)

which does evaluate to 1.
 
whats the identity for the square root?
As in how do you integrate the big square root
i used a trig sub.
let x=sin(u), then dx=cos(u)du

then we are integrating
sqrt(1-(sinu)^2)*cos(u)=(cosu)^2

to integrate (cosu)^2, you can use integration by parts or the cosine double angle identity: cos(2u)=2(cosu)^2-1.

then, you can rewrite everything back in terms of x.
Whoa niggas. I thought this was gonna be high school math only. We're throwing calc 2 up in this bitch? How far are we taking this?
is there a particular reason for subbing in trig? Trig integrals have never been my strong suit
The real reason is boring. It was trial and error, and mathematicians noticed that certain substitutions make integration less painstaking and tedious.
 
Here's my solution to the last problem
Problem #0019

if x>0 is a real number satisfying the equation
(x^2)+12sqrt(x)=5​
Then determine the exact value of the expression
x+2sqrt(x).​
Screenshot 42
 
Bumping for math
 
I haven’t done maths like this since first year uni, rusty af
 
Last seen Jan 30, 2021


:feelscry:
 
[UWSL]Let A be a regular cube, where the centers of its faces are the vertices of an octahedron. In turn, the centers of the faces of this formed octahedron are vertices of another cube. Obtaining consecutively octahedrons and cubes infinitely, determine the ratio of the sum of the volume of all inscribed polyhedra to the volume of the initial cube[/UWSL]

[UWSL][/UWSL]
 
[UWSL]Let A be a regular cube, where the centers of its faces are the vertices of an octahedron. In turn, the centers of the faces of this formed octahedron are vertices of another cube. Obtaining consecutively octahedrons and cubes infinitely, determine the ratio of the sum of the volume of all inscribed polyhedra to the volume of the initial cube[/UWSL]
You can make seperate threads for problems you know. Nobody gonna read sixth page of a math thread
 

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