Greater

[cinelli]

.                  ..... 
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A....................................D 
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B....................................C 
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The figure shows a square ABCD. AB is tangent to a circle, diameter 4 inches, and the size of the square is such that vertices C and D lie on the circle. The question is, which is greater - the circumference of the circle or the perimeter of the square?

Cinelli

[UncleEbenezer]

We're given the diameter as 4. So the circumference is 4π, and the problem reduces to whether the side of the square is greater or less than π.

Consider a line through points P, Q and R, where P is the midpoint of AB, Q of CD, and R where the line intercepts the circle. P to R is the diameter.

Then consider the angles of triangle CPR at each vertex. Let S be the distance CQ; i.e. half the side of the square. The angles are then
at C, atan(2) + atan((4-2S) / S)
at P, atan(1/2)
at R, atan(S / (4-2S))

and of course the sum of the angles is π.

Now we could solve for S, but we're not asked to do that; merely whether S is greater or less than π/2. Plugging in π/2 as S in the above produces a sum that falls just short of π, from which we can infer that the perimeter is shorter than the circumference.

p.s. If the next problem is their respective areas, that now follows trivially

[Gengulphus]

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A....................................D 
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B....................................C 
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The figure shows a square ABCD. AB is tangent to a circle, diameter 4 inches, and the size of the square is such that vertices C and D lie on the circle. The question is, which is greater - the circumference of the circle or the perimeter of the square?

Spoiler...

Using a co-ordinate system in which the centre of the circle is at (0,0), the scale is inches, and half the side of the square is S inches, A is at (-2,S), B at (-2,-S), C at (2S-2,-S) and D at (2S-2,S). By Pythagoras' theorem, C and D lie on the circle if and only if (2S-2)^2 + S^2 = 2^2, which multiplies out to 5S^2 - 8S + 4 = 4, or (5S-8)*S = 0. So either 5S-8 = 0, i.e. S=1.6, or S=0. The circumference of the circle is 4π = 12.56637..., and the perimeter of the square is 8S = 12.8 or 0.0. So either could be greater on the wording of the question - but given that a square of side 0, while mathematically entirely consistent with that wording, is a rather degenerate case and not exactly consistent with your diagram, I suspect the intended answer is that the perimeter of the square is greater!

An alternative technique is simply to spot the applicability of 3-4-5 Pythagorean triangles:

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A....................................D 
.   .                               /| . 
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..                           5/      |4   . 
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.                         /          |      . 
.                       /            |      . 
+----------------------+-------------+      .
.           5           \     3      |      . 
.                         \          |      . 
.                           \        |      . 
..                           5\      |4    . 
. .                             \    |    . 
.  .                              \  |   . 
.   .                               \|  . 
B....................................C 
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            ...              ... 
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with distances measured in units of 2/5ths of an inch to make the radius of the circle 2 inches. But that technique is liable to miss spotting the alternative degenerate solution!

Gengulphus

[GrahamPlatt]

Had a think, can’t do a clever proof.
However I can say that the circle is greater... (diameter = d)
Minimum case
If C & D are moved diametrically opposite the point of contact of the tangent AB, then the quadrangle collapses into two lengths which are the diameter of the circle, (and two “heights” of zero magnitude). Minimum perimeter of the quad = 2d
Maximum case
C and D swing up and down to the apex and base of the circle. So AB & CD are now the diameter and AD and BC are the radius.
So the max perimeter = 3d

Circumference of circle = 3.142d. Greater than max of quad.

[Gengulphus]

Had a think, can’t do a clever proof.
However I can say that the circle is greater... (diameter = d)
Minimum case
If C & D are moved diametrically opposite the point of contact of the tangent AB, then the quadrangle collapses into two lengths which are the diameter of the circle, (and two “heights” of zero magnitude). Minimum perimeter of the quad = 2d
Maximum case
C and D swing up and down to the apex and base of the circle. So AB & CD are now the diameter and AD and BC are the radius.
So the max perimeter = 3d

I'm afraid those only show that the minimum case is at most 2d and the maximum case at least 3d. If you try other positions for C and D, you can get both smaller and larger quadrangle perimeters.

E.g. swing C and D up and down to the "northeast" and "southeast" extremes of the circle. Then AB and CD are each twice the radius divided by the square root of 2 = ~0.7071d, and AD and BC are each the radius plus the radius divided by the square root of 2 = ~0.8536d, and the perimeter is twice the sum of those = ~3.1213d. And again, that only shows that the maximum is at least ~3.1213d.

Gengulphus

[cinelli]

Gengulphus has the right answer. It is interesting that UncleEbenezer came to the wrong conclusion in saying that the circumference is greater. Also his angle at C, which he gives by the fearsome expression atan(2) + atan((4-2S) / S), is a right angle as it is the internal angle in a semi-circle.

I am bound to point out that GrahamPlatt is solving a different problem from that set – by finding the perimeter of a rectangle rather that a square. This is a legitimate question in its own right but as Gengulphus has pointed out, you can’t just take two points on the circumference and say that, since the perimeters at these points are below the circle's circumference, this must be true for the infinity of rectangles obtained as D slides round the circle. If you are asking for the maximum perimeter of a rectangle ABCD, I calculate the value is

2*(1+sqrt(5))*r .=. 6.4721*r

where r is the radius of the circle. This is just a little more than the square’s perimeter, 6.4*r. The circle’s circumference is 2*pi*r .=. 6.2832*r.

For the optimum rectangle the sides are in the ratio of (1+sqrt(5))/2, the golden ratio.

Cinelli

[Gengulphus]

I am bound to point out that GrahamPlatt is solving a different problem from that set – by finding the perimeter of a rectangle rather that a square. ...

To be fair to GrahamPlatt, the square concerned is one of the rectangles he considered, so if the maximum for the rectangles had been under pi times the diameter of the circle, it would have been a solution. So it was a reasonable approach to solving the problem - just not one that ends up working.

Gengulphus

[GrahamPlatt]

Yes, that was my thinking. That if points C&D were both situated at the Easternmost point of the circle, that represented the minimum “rectangle”. Placing C & D to the North and South of the circle gives an easy calculation of the perimeter, which is 3d. I had then (wrongly) surmised that the value for the perimeter would in all cases lie somewhere between these two values.