Just an interesting problem.

[SLD]

assuming that the box is an actual square, what’s the area of the circle?

 

[Jarhyn]

View attachment 46284

assuming that the box is an actual square, what’s the area of the circle?

So I'm just going to talk through this rather than doing the solution: find where x^2=x. So, 1. This gives the upper right vertex of the square.

Then, the center of the circle is somewhere on the line y=-x+1

The circle itself is defined by (y-m)^2+(x-n)^2=r where m and n are coordinates on the above line.

I played around on desmos for a minute but I didn't recognize any of the numbers that approximated the solution. I'll probably play more with it later, but n is somewhere around 1/1.56 and m is somewhere around 1/2.76, with radius around 1/7.76, however I wasn't constraining these to the line, either, and doing all of it by eyeball.

 

[lpetrich]

I used Mathematica for the solution, and you might want to try it with your favorite computer-algebra software.

For the coordinates being {horizontal, vertical} and the circle having radius r, it has center {r,1-r}.

To find the intersection with the parabola with the tangents having the same direction, one can use a rational function for points on the circle:

\( \displaystyle{ x = r + r \frac{2t}{1+t^2} ,\ y = 1 - r + r \frac{1-t^2}{1+t^2} } \)

One must find the point where the circle intersects the parabola, and the tangents of the two curves must be parallel.

\( \displaystyle{ X_c = X_p ,\ X'_c \cdot \epsilon \cdot X'_p = 0 } \)

where (epsilon) is the antisymmetric symbol, in 2D, {{0,1},{-1,0}}.

One can find x and r as functions of t:

\( \displaystyle{ x \to \frac{t}{t^2-1} ,\ r \to \frac{\left(t^2+1\right) \left(t^2-t-1\right) \left(t^2+t-1\right)}{2 (t-1)^2 t^2 (t+1)^2} } \)

and an equation for t itself:

\( \displaystyle{ -t^5+t^4+t^3+3 t^2-t-1 = 0 } \)

Its solutions (numerical): {-0.627345 - 1.13355 i, -0.627345 + 1.13355 i, -0.442894, 0.660266, 2.03732}

The fifth one gave reasonable results: point {0.646632, 0.418133} and radius 0.361027.

 

[SLD]

View attachment 46284

assuming that the box is an actual square, what’s the area of the circle?

So I'm just going to talk through this rather than doing the solution: find where x^2=x. So, 1. This gives the upper right vertex of the square.

Then, the center of the circle is somewhere on the line y=-x+1

The circle itself is defined by (y-m)^2+(x-n)^2=r where m and n are coordinates on the above line.

I played around on desmos for a minute but I didn't recognize any of the numbers that approximated the solution. I'll probably play more with it later, but n is somewhere around 1/1.56 and m is somewhere around 1/2.76, with radius around 1/7.76, however I wasn't constraining these to the line, either, and doing all of it by eyeball.

Why do you say that the center lies on -x+1? Is that just from eyeballing it?

 

[Jarhyn]

View attachment 46284

assuming that the box is an actual square, what’s the area of the circle?

So I'm just going to talk through this rather than doing the solution: find where x^2=x. So, 1. This gives the upper right vertex of the square.

Then, the center of the circle is somewhere on the line y=-x+1

The circle itself is defined by (y-m)^2+(x-n)^2=r where m and n are coordinates on the above line.

I played around on desmos for a minute but I didn't recognize any of the numbers that approximated the solution. I'll probably play more with it later, but n is somewhere around 1/1.56 and m is somewhere around 1/2.76, with radius around 1/7.76, however I wasn't constraining these to the line, either, and doing all of it by eyeball.

Why do you say that the center lies on -x+1? Is that just from eyeballing it?

It's a square and is making contact with two sides that come to a corner with the circle. The center of such a circle that is tangent to two adjacent sides of a square will always be on the diagonal that bisects that angle.

The only time this is not going to be true is if the circle only touches a single edge or the rectangle is not a square.

This is because the distance from each side is going to be equal to the radius, this it is equally "down" from the corner as it is "left" from the corner. As the square is size 1, because y=x=x^2 defines the upper right corner, the x intercept of this line must be 1, and because the slope from that intercept must place the Y intercept at 1, this defines an inverse relationship between x and y.

The center of the circle therefore must be somewhere on the line y=-x+1.

Anyway I would then try and find line perpendicular to the parabola which, where it intersects y=-x+1, is equidistant to y=0.

Then I have it's radius. Solve for area.

 

[steve_bank]

I think Jaryn is on the right track.


x = x^2 at x = 1, so the box is 1x1.

With tangents of the circle on the x and y axis in the upper left corner of the box the center of the circle is (x,1-x), a line with a slope -1.

y = mx +b
m = -1
x intercept is (1,0)
0 = -1 + b
b = 1
y = -x + 1

Solve for he intercept of the line and curve..
-x + 1 = x^2
x^2 + x - 1 = 0
Quadratic formula
ax^2 + bx + c = 0
x = (-b +-sqrt(b^2-4ac))/2a
xp = (-1 + sqrt(5)/2 = 0.618034
yp = xp^2 = 0.381966

Radius r = distance from center of circle (x,1-x) to (xp,yp) intercept point.

Distance formula.
r = sqrt( (x - xp)^2 + ((x-1) - yp)^2)
Plug in the numbers for xp and yp and you should get another quadratic,

 

[SLD]

After looking at it, I do see what Jarhyn is saying. It must be on the line -x+1.

Any circle shoved into that corner has to have that property. Duh.

Hey, while I did major in math I graduated 42 years ago. Now I’m stupid.

I sometimes fantasize about retiring and going back to study math all over again and take it further than before. But I fear I’d have to start over at Algebra I.

 

[steve_bank]

Thanks for a problem to cogitate on SLD.

 

[steve_bank]

The only data point to nail; it down is the intersection of curves. The radius of the circle is x, so when the x matches the distance from center of the circle to the curves along the diagonal line that us the radius,

One way is an iterative solution. Iterate x vales and when x is close to equaling the distance from center to the curve it is done. A common method when all you need is a number.

import math as ma
import numpy as np
import array as ar
import matplotlib.pyplot as plt


n = 100000
y1 = ar.array('d',n*[0])
y2 = ar.array('d',n*[0])
cc = ar.array('d',n*[0])
x = np.linspace(0,1,n)

#find intercept of line and curve
#x^2 = -x + 1
#x^2 + x -1 = 0
a =1
b = 1
c = -1
xp =(-b + np.sqrt(b**2 - 4 *a*c))/(2*a)  #x intercept
yp = xp**2 #y intercept
print("Intercept (x,y)  ",xp,"  ",yp)

#line and cirve
for i in range(n):   
        y1[i] = -x[i] + 1
        y2[i] = x[i]**2
        
#iterate until x the radius of the circle equals
 #the distence from the cener of the circke to the curve
for i in range(n):
        yc = -x[i] + 1
        r = np. sqrt( (x[i]-xp)**2  + (yc - yp)**2 ) #distance formula
        if abs(r-x[i]) < .00001:
               print("Circle Center(x,y)  ",x[i],"  ",yc)
               cc[i] = yc #plot point
               area = np.pi*r**2
               print("Area  ",area)
               break
                                  

[fig, p] = plt.subplots(1)
p.plot(x,y1,linewidth=2.0,color="k")
p.plot(x,y2,linewidth=2.0,color="k")
p.plot(x,cc,linewidth=2.0,color="k")
p.grid(color='k', linestyle='-', linewidth=1)
p.grid(which='major', color='k',linestyle='-', linewidth=0.8)
p.grid(which='minor', color='k', linestyle='-', linewidth=0.4)
p.minorticks_on()
plt.show()

 

[Jarhyn]

The only data point to nail; it down is the intersection of curves. The radius of the circle is x, so when the x matches the distance from center of the circle to the curves along the diagonal line that us the radius,

One way is an iterative solution. Iterate x vales and when x is close to equaling the distance from center to the curve it is done. A common method when all you need is a number.

import math as ma
import numpy as np
import array as ar
import matplotlib.pyplot as plt


n = 100000
y1 = ar.array('d',n*[0])
y2 = ar.array('d',n*[0])
cc = ar.array('d',n*[0])
x = np.linspace(0,1,n)

#find intercept of line and curve
#x^2 = -x + 1
#x^2 + x -1 = 0
a =1
b = 1
c = -1
xp =(-b + np.sqrt(b**2 - 4 *a*c))/(2*a)  #x intercept
yp = xp**2 #y intercept
print("Intercept (x,y)  ",xp,"  ",yp)

#line and cirve
for i in range(n):  
        y1[i] = -x[i] + 1
        y2[i] = x[i]**2
       
#iterate until x the radius of the circle equals
 #the distence from the cener of the circke to the curve
for i in range(n):
        yc = -x[i] + 1
        r = np. sqrt( (x[i]-xp)**2  + (yc - yp)**2 ) #distance formula
        if abs(r-x[i]) < .00001:
               print("Circle Center(x,y)  ",x[i],"  ",yc)
               cc[i] = yc #plot point
               area = np.pi*r**2
               print("Area  ",area)
               break
                                 

[fig, p] = plt.subplots(1)
p.plot(x,y1,linewidth=2.0,color="k")
p.plot(x,y2,linewidth=2.0,color="k")
p.plot(x,cc,linewidth=2.0,color="k")
p.grid(color='k', linestyle='-', linewidth=1)
p.grid(which='major', color='k',linestyle='-', linewidth=0.8)
p.grid(which='minor', color='k', linestyle='-', linewidth=0.4)
p.minorticks_on()
plt.show()

Well, as I said, that point on the line whose distance, perpendicular to Y=1 will be equal to the distance to of the line perpendicular from the parabola to that same point, and this will naturally be the point where the tangent curves meet because it will be a radius of the circle that touches the parabola there.