[Calculus] How do I find the smallest area of an isosceles triangle circumscribed on a circle with a radius a?

Question

Given a circle with a radius of "a", what is the smallest area of an isosceles triangle circumscribed on the circle (circle inside of triangle)? The area of the triangle will always be larger than that of the circle. I need to know HOW to do this. I can then figure the answer on my own. I tried using an angle in the open interval from 0 to pi/2. I figured the side lengths of the triangle in terms of the angle and came up with a function. The function didn't match my logic. Logically speaking, the area of the triangle will never approach 0 unless "a" approaches 0. The function arbitrarily approached 0 as the angle approached pi/2. Please help. I need this by Wednesday. Thanks for any and all the help!

Note: I am looking for a method and not the final answer. I need to know the function for all isosceles triangles with the given constraint. I can take it from there.

I figured it out. The area for the smallest isosceles triangle circumscribed on any circle with radius "a" is:

3a^2sqrt(3)

It took an entire page to simplify the equation to get one I could minimize. Thanks for all the help guys.

Answer 1

If O is the center of the circle, ABC the triangle, and D,E,F are the points of tangency on BC,CA and AB, respectively, then we can divide the triangle in three kites,

DOEC, EOFA and FODB

with two opposite angles right angles. The area of these kites depends on the arc distances DE, EF and FD (that is, on the angles DOE, EOF and FOD).

Split each of the kites into two congruent right triangles, conclude that DC = a/tan w where w = 1/2 of angle DOE, etc. The area of kite DOEC is easily seen to be a^2/tan w.

The total area of the triangle is therefore

a^2/tan u + a^2/tan v + a^2/tan w

where u, v, w are the angles EOF, FOD, DOE.

This shows that we must minimize the sum

A = 1/tan u + 1/tan v + 1/tan w

under the condition u + v + w = 180 degr. (We must also demand that u, v, w < 90 degr, otherwise the triangle is infinite.)

The derivative of 1/tan x is - 1/(sin x)^2. If we increase u by amount d and decrease v by amount d, the area changes by

dA = -d/(sin u)^2 + d/(sin v)^2

As long as v < u, this change will increase the area. The maximum is therefore found when v = u.

This shows that we must look for the situation where u = v = w = 60 degrees, and therefore the triangle is equilateral.

Answer 2

I will give a hint. The small area will occur when the the isosceles triangle were equilateral.

Answer 3

a divided by 2 isiocles could be equilater (at least 2 equal sides) so then take the reslt and multiply it by itself (not to itself as an exponent)

Answer 4

r = K/s = b sqrt[(2a - b)/(2a + b)]/2

K = rs
s = a + (b/2)

K = a(x + (y/2))
K = a((2x + y)/2)
K = (a/2)(2x + y)

Answer 5

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