Show square inscribed in circle has maximum area

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 7K views
t_n_p
Messages
593
Reaction score
0

Homework Statement


Show that the square, when inscribed in a circle, has the largest area of all the 4-sided polygons. Try to show that all sides of a quadrilateral of maximal area have to be of
equal length.

The Attempt at a Solution


How do you start?
I don't get how showing the sides are equal will help prove that the square has maximal area.
Any hints?
 
Physics news on Phys.org
Try to express surface of the inscribed 4-sided polygon as a function of the length of its sides. General case will be more difficult, but for a specific case of rectangle it should be relatively easy to express lengths as a function of one variable.

Once you have that, find maximum.
 
Borek said:
Try to express surface of the inscribed 4-sided polygon as a function of the length of its sides. General case will be more difficult, but for a specific case of rectangle it should be relatively easy to express lengths as a function of one variable.

Once you have that, find maximum.

I don't quite understand.
Say I have a square, let the side length = a.

area = a^2.
Then maximize by taking derivative?
 
Real easy to prove with coordinate geometry, but you need to know the parametric equation of a circle and the area of a quadrilateral.

Any point on a circle (with origin as centre and radius r) is (r cosx, r sinx)
and area of a quadrilateral with vertices (x1, y1), (x2, y2), (x3, y3), (x4, y4),
is #8 here http://www.mathisfunforum.com/viewtopic.php?id=3301

take four points, put them in the formula, the result becomes obvious
 
that gives me the area, but it doesn't prove that it is the maximum area for a 4 sided shape.
 
it does! the result for area of a general quadrilateral in a circle comes to be (assuming radius r and angles A, B, C, D
r2/2[sin(A-B) + sin(B-C) + sin(C-D) + sin(D-A)]
the maximum value of which (2r2) is attained only when each of the angular differences is 90 degrees which results in a square

did u try solving it? you need basic trignometric formulae