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Wizard Joined: Oct 14, 2009
• Posts: 25739
March 10th, 2020 at 3:48:07 PM permalink A circle of radius 1 is tangent to a parabola of equation y=x2. What is the area of the region in red, between the circle and parabola?

Usual rules:

1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
3. Beer to the first satisfactory answer and solution, subject to rule 2.
�Extraordinary claims require extraordinary evidence.� -- Carl Sagan
ssho88 Joined: Oct 16, 2011
• Posts: 655
March 10th, 2020 at 5:27:21 PM permalink
Area = 3*(3)^0.5/4 - PI/3 = 0.25184, I will post the calculations if the answer is correct

https://ibb.co/1GKFM66

Q(0,c) is center of the circle, P(a,b) is the intersection point.

Gradient of Line PQ = m1 = (b-c)/a

Gradient of tangent to the parabolic at P(a, b) = m2 = dy/dx = 2x = 2a

So, m1 * m2 = -1

(b-c)/a * 2a = -1

c= 0.5 + b ------------- Eq1

Parabolic EQuation,

y = x^2, So, b = a^2 ---------------Eq2

Circle Equation,

x^2 + (y-c)^2 = 1

So, a^2 (b-c)^2 = 1 ---------------Eq3

From Eq1, Eq2 and Eq3,

b + (b- 0.5 - b)^2 = 1

b = 0.75 = 3/4

c = 1.25 = 5/4

a = (0.75)^0.5 = (3)^0.5/2

AREA A = PI/4 -(2*PI + 3(3)^0.5)/24

AREA B = 3(3)^0.5)/24

AREA A + AREA B = PI/4 -2*PI/24 = PI/6

RED Area = 2 * [a*b - (AREA A + AREA B)]

= 2 * [(3)^2/2 * 3/4 - PI/6]

= (3)^0.5 * 3/4 - PI/3

= 0.25184

EDITED
Last edited by: ssho88 on Mar 10, 2020
Wizard Joined: Oct 14, 2009
• Posts: 25739
March 10th, 2020 at 6:08:19 PM permalink
Quote: ssho88

Area = 3*(3)^0.5/4 - PI/3 = 0.25184, I will post the calculations if the answer is correct

I agree this is the correct answer. For the beer, you must show your work. There is still opportunity for anyone else to swoop in before to claim it, except those already on the "beer list."
�Extraordinary claims require extraordinary evidence.� -- Carl Sagan
ssho88 Joined: Oct 16, 2011
• Posts: 655
March 10th, 2020 at 7:25:55 PM permalink
Quote: Wizard

Quote: ssho88

Area = 3*(3)^0.5/4 - PI/3 = 0.25184, I will post the calculations if the answer is correct

I agree this is the correct answer. For the beer, you must show your work. There is still opportunity for anyone else to swoop in before to claim it, except those already on the "beer list."

https://ibb.co/1GKFM66

Q(0,c) is center of the circle, P(a,b) is the intersection point.

Gradient of Line PQ = m1 = (b-c)/a

Gradient of tangent to the parabolic at P(a, b) = m2 = dy/dx = 2x = 2a

So, m1 * m2 = -1

(b-c)/a * 2a = -1

c= 0.5 + b ------------- Eq1

Parabolic EQuation,

y = x^2, So, b = a^2 ---------------Eq2

Circle Equation,

x^2 + (y-c)^2 = 1

So, a^2 (b-c)^2 = 1 ---------------Eq3

From Eq1, Eq2 and Eq3,

b + (b- 0.5 - b)^2 = 1

b = 0.75 = 3/4

c = 1.25 = 5/4

a = (0.75)^0.5 = (3)^0.5/2

See attached image,
AREA A = PI/4 -(2*PI + 3(3)^0.5)/24

AREA B = 3(3)^0.5)/24

AREA A + AREA B = PI/4 -2*PI/24 = PI/6

RED Area = 2 * [a*b - (AREA A + AREA B)]

= 2 * [(3)^2/2 * 3/4 - PI/6]

= (3)^0.5 * 3/4 - PI/3

= 0.25184

Wizard Joined: Oct 14, 2009
• Posts: 25739
March 10th, 2020 at 7:37:56 PM permalink
Quote: ssho88

A bit cryptic, but I'll accept that. Well done -- I owe you a beer. Please consider yourself on the "beer list" from now on, which means you're on a 24-hour delay for future problems. �Extraordinary claims require extraordinary evidence.� -- Carl Sagan
Dalex64 Joined: Feb 10, 2013
• Posts: 1067
March 11th, 2020 at 12:31:18 PM permalink
I was thinking about solving this in a different way. Please let me know how wrong I am.

given the formula for the circle that you can find from ssho88's work, what I was thinking of would be to calculate the area under the curve of the circle between 0 and b,
subtract the area under the curve of the parabola between a and b,
and double it.
Wizard Joined: Oct 14, 2009
• Posts: 25739
March 11th, 2020 at 2:01:50 PM permalink
Quote: Dalex64

I was thinking about solving this in a different way. Please let me know how wrong I am.

given the formula for the circle that you can find from ssho88's work, what I was thinking of would be to calculate the area under the curve of the circle between 0 and b,
subtract the area under the curve of the parabola between a and b,
and double it.

That would work. It still leaves the hard part of the problem, finding the point where the circle and parabola are tangent and finding the area under the circle for that portion.
�Extraordinary claims require extraordinary evidence.� -- Carl Sagan
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5863
March 11th, 2020 at 4:37:24 PM permalink
I get the same answer, but in a slightly different method from everybody else, although the first part appears to be the same as ssho88's method (and, to be honest, I had to look up one of the integrals, so even had I entered on time, it would have been with an asterisk):

Let (0,c) be the center of the circle, and (t, t2) the point of tangency with x > 0
At (t, t2), the tangent to the parabola has slope 2t, so the perpendicular to the tangent has slope -1/(2t)
This means the line containing the center of the circle and the tangent point is y = -1/2t x + c
Since (t, t2) is on this line, t2 = -1/(2t) * t + c = c - 1/2 -> c = t2 + 1/2
Since the circle has radius 1, (t - 0)2 + (t2 - c)2 = 12 = 1
Substitute t2 + 1/2 for c: t2 + (t2 - (t2 + 1/2))2 = 1
t2 + (1/2)2 = 1 -> t2 = 3/4 -> c = 5/4 and t = sqrt(3)/2

The red area = the area underneath the circle between the tangent points - the area underneath the parabola between the tangent points
Since the graphs are symmetrical with respect to the y-axis, this = 2 * (the area underneath the circle - the area underneath the parabola) for 0 <= x <= sqrt(3)/2
The graph of the bottom semicircle of the circle is y = 5/4 - sqrt(1 - x2)
This = 2 INTEGRAL(0, sqrt(3) / 2) {5/4 - sqrt(1 - x2) - x2) dx}
= 2 (INTEGRAL(0, sqrt(3) / 2) {5/4 dx} - INTEGRAL(0, sqrt(3) / 2) {sqrt(1 - x2) dx} - INTEGRAL(0, sqrt(3) / 2) {x2 dx})
The one in the middle is the one I had to look up; go down to #8
Area = 2 (5/4 * sqrt(3) / 2 - sqrt(3)/4 sqrt(1 - 3/4) + 1/2 arcsin (sqrt(3)/2) - (sqrt(3)/2)3 / 3)
= 2 (5/8 sqrt(3) - 1/8 sqrt(3) - 1/2 * PI/3 - 1/8 sqrt(3))
= 2 (3/8 sqrt(3) - PI/6)
= 3/4 sqrt(3) - PI/3

Wizard Joined: Oct 14, 2009
• Posts: 25739
March 12th, 2020 at 7:34:48 PM permalink
Here is my solution in PDF form.

Yay math! (version 2)
Last edited by: Wizard on Mar 13, 2020
�Extraordinary claims require extraordinary evidence.� -- Carl Sagan
Doc Joined: Feb 27, 2010
• Posts: 7274
March 13th, 2020 at 11:56:33 AM permalink
Quote: Wizard

Here is my solution in PDF form.

Just curious -- why do you have the coordinates of some points in your figures shown as (x,y) (as for the point of tangency) and others shown as (y,x) (as for the center of the circle and for the point on the y axis even with the point of tangency)?
Wizard Joined: Oct 14, 2009
• Posts: 25739
March 13th, 2020 at 2:36:32 PM permalink
Quote: Doc

Just curious -- why do you have the coordinates of some points in your figures shown as (x,y) (as for the point of tangency) and others shown as (y,x) (as for the center of the circle and for the point on the y axis even with the point of tangency)?

You're absolutely right. I just fixed it. Please check the previous link again.
�Extraordinary claims require extraordinary evidence.� -- Carl Sagan
Doc Joined: Feb 27, 2010