Serious Daily math problem thread

[WorldYamataizer]

Problem #0017

In the diagram, a circle of radius 1 is internally tangent to y=x^2.
Determine the black area between the circle and parabola.

I will share a written solution tomorrow, same time. Happy problem-solving

@mNFwTJ3wz9
@Divergent_Integral
@SelfCrucified
@grondilu
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

 

[mNFwTJ3wz9]

Problem #0017
View attachment 413879
In the diagram, a circle of radius 1 is internally tangent to y=x^2.
Determine the black area between the circle and parabola.

I will share a written solution tomorrow, same time. Happy problem-solving

@mNFwTJ3wz9
@Divergent_Integral
@SelfCrucified
@grondilu
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

Brute forcing this

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 2

contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4

 

[WorldYamataizer]

Brute forcing this

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 2

contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4

i got “h” (i used “h” instead of “c”) the same way.

i am just evaluating an integral rn and im done.

this problem is sort of just algebraic bashing. next time, i’ll post a problem that requires more of an insightful construction or idea

 

[mNFwTJ3wz9]

Brute forcing this

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 2

contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4

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

putting c into the equation

Spoiler: 3

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

Spoiler: 4

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

i am just evaluating an integral rn and im done.

what integral?

Spoiler: Final ans

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448

 

[WorldYamataizer]

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

putting c into the equation

Spoiler: 3

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

Spoiler: 4

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?

Spoiler: Final ans

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448

im impressed, you did the question the “elegant” way. i just brainlessly did the area under the circle minus the area under the parabola.

this integral.

 

[WorldYamataizer]

Here's a solution for problem #0017

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

Spoiler: 3

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

Spoiler: 4

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?

Spoiler: Final ans

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448

 

[WorldYamataizer]

Problem #0018

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

 

[mental_out]

im impressed, you did the question the “elegant” way. i just brainlessly did the area under the circle minus the area under the parabola.
View attachment 414068
this integral.

whats the identity for the square root?

 

[WorldYamataizer]

whats the identity for the square root?

what do u mean?

 

[mental_out]

what do u mean?

As in how do you integrate the big square root

 

[WorldYamataizer]

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.

 

[mental_out]

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

 

[WorldYamataizer]

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.

 

[LittleBoy]

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?

 

[BummerDrummer]

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

 

[LittleBoy]

Shut the fuck up cracker that is cracker nerdy white boi shit on the G

It's money making dough...

Computer science and engineering mathematics come next.

 

[SocialzERo]

Shut the fuck up cracker that is cracker nerdy white boi shit on the G

 

[Michinomiya Hirohit]

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?

 

[dogcel256]

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.

 

[LittleBoy]

Financial math threads would be useful for stockcels.

Yup. Bullseye.

 

[copeful]

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

Spoiler

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

 

[WorldYamataizer]

Spoiler

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

 

[ScornedStoic]

me right now