Tricky circle problem?

Question

I have a circle of radius one, and draw a chord across it of length squareroot(3). In the minor segment I draw a rectangle - which has one side along the chord and has the opposite two points touching the edge of the circle. The question - what is the largest size of rectangle that can be fitted within the minor segment? Proof would be great!

When the question was set no calculators were allowed.

Answer 1

Let the center of the circle be (0,0).
Then
x² + y² = 1

If we let the chord with length sqrt3 be parallel to x-axis and above the x-axis (meaning y > 0), then the endpoints of the chord has x-coordinate x = ±sqrt3/2
If x = ±sqrt3/2 then
y = 1/2
(we will not choose -1/2 because y > 0)

The maximum length of the horizontal side of the circle is sqrt3. Notice that the horizontal side has a length twice the x-coordinate.
The maximum length of the vertical side is 1/2. Notice that the vertical side has a length 1/2 less than the y-coordinate.

Thus, You have the equations:
A = lw
l = 2x
w = y - 1/2
x² + y² = 1

solving for A in terms of y,
x = sqrt(1 - y²)
l = 2 sqrt(1 - y²)
A = lw = 2 sqrt(1 - y²)(y - 1/2)
A = sqrt(1 - y²)(2y - 1)

Getting the maximum y, solve for dA/dy = 0
dA/dy = (2y - 1)(1/2)[1/sqrt(1 - y²)](-2y) + sqrt(1 - y²)(2) = 0
-y(2y - 1)/sqrt(1 - y²) + 2 sqrt(1 - y²) = 0
y(2y - 1)/sqrt(1 - y²) - 2 sqrt(1 - y²) = 0
y(2y - 1) - 2(1 - y²) = 0
2y² - y - 2 + 2y² = 0
4y² - y - 2 = 0
y = [1 ± sqrt(1 + 32)]/8
y = (1 + sqrt33)/8
x = sqrt(1 - y²) = sqrt(30 - 2sqrt33)/8
w = y - 1/2
w = (sqrt33 - 3)/8
l = 2x = sqrt(30 - 2sqrt33)/4

^_^

Answer 2

Well, the first thing you want to do is shift the circle, to a logical place (in the xy-plane). By centering the circle on the y axis, you want to drop it down, so that the chord is at the x-axis. Since the chord is √3 in length, each side of the y-axis will have √3/2. So, you want to find the point on the unit circle that is (√3/2,y) with y>0. Solving for y is easy: y=√(1-(√3/2)^2)= √(1-3/4)= √(1/4)=1/2. So we drop the circle down 1/2.

The new Circle is given by x^2+(y+1/2)^2=1.

To maximize the rectangle, we can maximize the side of the rectangle on the right side of the y-axis.

Therefore, we can think of the rectangles made by choosing a point, call it (x,y) on the Circle, such that |x-0|•|y-0|=xy is maximized. We are given that x,y≥0 and x≤√3/2.

Since it is on the circle, we can solve for x in the circle. This gives:
x=±√(1-(y+1/2)^2), but since x≥0, we only need to consider the positive.

So we want to maximize xy=√(1-(y+1/2)^2)•y = A(y)

A'(y)=√(1-(y+1/2)^2)+y•1/2• (1-(y+1/2)^2))^(-1/2) •(-2(y+1/2)) =[(1-(y+1/2)^2)- y(y+1/2)]/√(1-(y+1/2)^2). This is 0 when the top is 0, so when 1-(y+1/2)^2-y(y+1/2) = 1-y^2-y-1/4-y^2-1/2y = 3/4-3/2y-2y^2= 0 or when 8y^2+6y-3=0

Solving for y we get y=(-6±√(36+96))/16 = (-6±2√(33))/16 = (-3±√33)/8 but -3-√33<0, thus the only y that can work is y=(-3+√33)/8. Solving for x, we get x=√(1-((-3+√33)/8+1/2)^2) = 1/8•√(30-2√33)

So the dimensions of half of the rectangle are 1/8•√(30-2√33) x (-3+√33)/8

And the dimensions of the whole rectangle are 1/4•√(30-2√33) x (-3+√33)/8

Answer 3

The maximum area = 0.369

Here's a slightly different solution:

Take the unit circle(r=1) to be centered at the origin.

For simplicity, taking the chord of length sqrt(3) to be parallel to the x-axis (and lying in the positive y quadrants), the chord's distance from the x-axis = sqrt(1- (sqrt(3)/2)^2) = sqrt(1- 3/4) = sqrt(1/4) = 1/2

Now let A (< pi/2) be the angle between the x-axis and a point on the circle having a y-co-ordinate greater than 1/2. When the radius is 1, the x co-ordinate is simply cosA. The area of the rectangles to be considered is simply given by (1) 2cosA(sinA -1/2). Taking the derivative of this function with respect to A and setting it equal to 0, we have:

2cosA(cosA)-2sinA(sinA-1/2)=0

cos^2(A) - sin^2(A) + sinA /2 =0

1- 2sin^2(A) + sinA /2 =0 or

2sin^2(A) - sinA /2 + 1 = 0. Solving this quadratic for sinA we get:
sinA = 0.84307 and cosA = 0.5378 (note: A = 57.47 degrees)

And the maximum area from(1) =

2 x 0.5378(0.84307 - 1/2) = 0.369

The dimensions of the rectangle are 1.0756 and 0.3431

Answer 4

The largest size (area) of rectangle that can be drawn is a square (since square is also a type of rectangle). This is a question of maxima and minima, but am sorry I dont have any proof for this.

Answer 5

Angles AXC and AYC are equivalent because they subtend an similar chord AC, so angles XAY and YCX are also equivalent. yet considering BX=XC we see that perspective ABC = perspective YCX = perspective XAY, and therefore AYB is an isosceles triangle with AY = by technique of = 12 cm. i'm hoping you've drawn an effective diagram so that you'll make experience of this. :) Edit: a good more convenient way of seeing it truly is to be conscious that perspective XAY and perspective YCX are equivalent as they both subtend an similar chord XY. So, as BX = XC means that perspective ABC = perspective YCX, we've that perspective XAY = perspective ABC and therefore AYB is isosceles with AY = by technique of = 12 cm.

Answer 6

about the size of a bourbon biscuit.

I have no proof, you'll just have to trust me.

Answer 7

u r extremely genius guy.
gr8 question buddy?