A brain teaser?

Question

show that maximum possible square cut from a circle is where the diagonal of square is the diameter of circle. Even show that the maximum possible circle from a square is where the diameter of circle is equal to side of square????

Answer 1

The second one's pretty obvious: pick any point as the centre of the circle, and the largest circle that can be built is one where the radius is the distance to the nearest side. To maximise this distance we need to use the centre of the square as the centre of the circle, when the radius will be half the length of a side of the square, and hence the diameter will be equal to the length of a side of the square.

For the first problem, pick any square inside the circle. If it has no points on the circumference, you can increase its size equally in all directions until it does. If it has one point on the circumference, you can increase its size in both directions moving away from that point (e.g. if the square is ABCD and B is on the circumference, you can move A and C further away from B and increase the size of the square that way).

Now we must have at least two adjacent points of the square on the circumference of the circle. We may or may not have the other points on the circumference. If not, we can move the square away from the side with both ends on the circumference, and have no points touching the circle, so we can expand the square again.

So we must have at least three corners of the square touching the circumference of the circle. This means that one diagonal of the square touches the circle at both ends. The ends of this diagonal subtend an angle of 90° at the other touching corner, so must subtend a central angle of 180°. So this diagonal forms a straight line through the centre of the circle, and must therefore be a diameter. (From which it follows immediately that the other diagonal must also be a diameter and the fourth corner must also touch the circle.)

Answer 2

Your first sentence is basically correct because the RADIUS of a CIRCLE can be a DIAGONAL of a SQUARE. You can actually draw several RADIUS line segments from the CENTER of the circle to the edge of the circle and make a square out of it. The number of squares cannot be quantitied because we do not know how many points are there in the circumference of a circle. But, we can show it in a formula. The circumference formula:

C = 2r * PI

Where C is the circumference, r is the radius and PI is the greek symbol p constant ratio of circles.

The second statement, is FALSE or INCORRECT. Either you missed something therefore it lacks supporting premise or it is altogether invalid. The diameter of a circle CANNOT be EQUAL to the side of any particular square because of the fact that you CANNOT SQUARE A CIRCLE.

Answer 3

"Show" at what level of rigor? Take the given solutions: try to expand the inner figure, and it leaks over the edges. shrink it anywhere, and you leave something smaller. In general, note that the three critical line segments are the circle's diameter and the square's diagonal and side.

For a little geometric rigor, try inscribing another circle inside the square (first part), or another square inside the circle (second part -- in this case, make the corners of the inner square lie on the side midpoints of the larger one).

Now, what happens to the innermost figure when you expand or shrink the middle one? Does this help your "proof" at all?

Are you allowed to use analytical methods? If so, try putting the outer figure centered at (0,0). Express the inner figure's area in terms of it's center and radius (distance to corner for the square). Find the maximum area for the given center. Express this in terms of the inner figure's center. Use this equation to determine the optimum placement for the figure's center. Reduce to common terms, and you'll get the same center (0,0).
Q.E.D.

Answer 4

Let b an h be the base and height of some inscribed rectangle of maximal area. Then b and h must maximize the quantity b*h subject to (b/2)^2+(h/2)^2=r^2, where r is the radius of the circle. Setting up Lagrange multipliers, we define
q=lambda*b*h+(b/2)^2+(h/2)^2.
0=dq/db=lambda*h+b/2
0=dq/dh=lambda*b+h/2
b=-2*lambda*h
0=lambda*(-2*lambda*h)+h/2
lambda^2=1/4
(-b/(2*h))^2=1/4
b/(2*h)=1/2
b=h
(b/2)^2+(h/2)^2=r^2
so
(b/2)^2+(b/2)^2=r^2
b=r*sqrt(2)
h=r*sqrt(2)

So the maximal-area rectangle is a square with side r*sqrt(2).

Answer 5

That gives me a headache.

Answer 6

where's that darn advil?!