Re: Need help

On Thu, 24 May 2007 16:05:11 +0800, ALEX_001 <baowuji@xxxxxxx> wrote:

On Thu, 24 May 2007 01:02:34 -0500, quasi <quasi@xxxxxxxx> wrote:

On Wed, 23 May 2007 20:35:45 +0800, ALEX_001 <baowuji@xxxxxxx> wrote:

Can we improve a pentagon( all edges' length are the same) can be
completely covered by five circles(each circles 's center is a vertex
of the pentagon ,and the radius is the length of edge)?

Of course, I'm sure you meant "prove", not "improve".

Here's a sketch of a proof for the convex case ...

Assume there is a convex pentagon P which cannot be covered.

Since it's obvious that a regular pentagon can be covered, it follows
that P is not regular.

Without loss of generality, we can assume that the origin is not
covered, and the pentagon has sides of length 1.

Scanning counterclockwise, let the vertices v1,v2,v3,v4,v5 be at
distances r1,r2,r3,r4,r5 respectively from the origin.

Write the area A of P as the sum of 5 triangular areas ...

A=(1/2)*r1*r2*sin(t1) + ... + (1/2)*r5*r1*sin(t5)

where t_i is the angle between v_i and v_(i+1), with the convention
that v6=v1.

The angles t1,t2,t3,t4,t5 are all strictly between 0 and Pi and the
sum is 2*Pi.

By assumption, the origin is not covered, hence r1,r2,r3,r4,r5 are all
greater than 1, which gives

A > (1/2)*(sin(t1) + ... + sin(t5))

Since sin is strictly concave down on the interval (0,Pi), it follows
that

sin(t1) + ... + sin(t5) >= 5*sin((t1+t2+t3+t3+t4)/5) = 5*sin((2*Pi)/5)

Hence

A > (5/2)*sin((2*Pi)/5)

But the RHS is the area of a regular pentagon with sides of length 1.

Since the perimeters are equal, the regular pentagon must have area
greater area than P, contradiction.

The claimed result follows (for the convex case).

quasi

Thank you so much.

You're welcome.

Thanks for the problem, I enjoyed it.

Actually, I think my proof adapts easily to the following more general
statement ...

Let P be a convex pentagon and let r be the average side length.
If at each vertex of P a disk of radius r is placed, centered at that
vertex, then the 5 disks completely cover P.

The above statement applies to any convex pentagon. There's no
requirement for equal side lengths.

quasi
.

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