Question 471487: show that the maximum area of a rectangle inscribed in a circle is a square
Found 3 solutions by MathLover1, karaoz, richard1234:
Answer by MathLover1(20849)   (Show Source):
Answer by karaoz(32)  (Show Source):
You can put this solution on YOUR website! Show that a rectangle inscribed in a circle will have the maximum possible area when it is a square.
We will do this in two steps. First, we will need to prove this:
(1) Diagonal of any rectangle inscribed in a circle is a diameter of the circle. (This is essentially the converse of Thales' theorem).
Let A, B, C and D be the vertices of a rectangle inscribed in a circle and let AC and BD be the diagonals of this rectangle.
We can now focus on any one of the four triangles: ABC, BCD, CDA and DAB.
Let's take ABC. AC side of this triangle is the diagonal of the rectangle.
AB and BC are two sides of the rectangle, which potentially may be of different length
(but we will prove they must be the same if the area of the inscribed rectangle is maximised).
The angle at B is the right angle since it is one of the angles of the rectangle, which by definition has four right angles.
Hence, looking at the triangle ABC, we can see that this is a right-angled triangle inscribed in the circle.
Therefore to prove (1), we need to show that side AC of any such triangle must be a circle's diameter.
Proof: Choose any three points A, B and C on the circle and connect these points to make a triangle ABC.
Let's suppose that the claim is that the angle at B is right angle.
We will show that if this is true then it must follow that AC is a diameter of the circle.
Connect the centre of the circle (O) with each of the vertices of the triangle creating the segments OA, OB and OC.
Let's call the angle defined by the path OAB as α and the angle defined by the path OCB as β.
Since OA, OC and OB are all of equal length (equal to the length of the circle's radius) then all three inner triangles (OAB, OAC and OBC) are isosceles.
We will next want to find the angle between OA and OC (the angle at O made out by segments OA and OC) using only angles α and β as given.
The angle between OA and OB is equal to 180° - 2α (due to the fact that OAB is isosceles and the fact that the sum of all three internal angles in a triangle sum to 180°).
Similarly, the angle between OB and OC is equal to 180° - 2β.
Finally, since all the three angles at O add up to 360° (full circle), it follows that the angle between OA and OC is equal to 360° - (180° - α) – (180° - β) = 2(α+β).
However, the angle at B of the original triangle ABC is equal to α+β (follows from the fact that OAB and OBC triangles are isosceles).
This angle was claimed to be 90° and therefore the angle between OA and OC is equal to 180°.
This means that the points A and C and the centre of the circle (O) are collinear.
In other words, the centre of the circle is lying on the straight line segment AC, which in turn means that AC must be a diameter of the circle.
So, we can conclude that each diagonal of the rectangle inscribed in a circle must be a circle's diameter.
Now, we can return to the original question and prove that the area of a rectangle inscribed in a circle will be maximised when its sides are equal, i.e. when the rectangle is in fact square.
(2) A rectangle inscribed in a circle will have the maximum possible area when it is a square.
This can be done either using trigonometry or using Lagrangean for optimising a function with equality constraint.
Use of trigonometry is less complex, so let’s see that one first.
USING TRIGONOMETRY:
First, visualize the rectangle as made out of two large right-angled triangles (for example, ACB and ACD).
These two triangles are equivalent.
Next, add another such a triangle to the rectangle ABCD so that one of the rectangle’s sides is also one leg of the added triangle.
We now have three equivalent triangles.
Observe only the two of these three triangles that share the same leg.
These two triangles make an isosceles triangle, which has the same area as the rectangle ABCD.
The length of the two equal sides of the isosceles triangle is equal to the length of the rectangle's diagonal.
Due to what was proven in (1), this is the same as the length of the circle’s diameter, which we will denote by d.
With some basic understanding of trigonometric functions (sin and cos), it should not be too difficult to show that area of any triangle with sides a and b, and the angle between them γ, can be expressed as:
A = ˝ ab sin γ
In our particular case, the area of the triangle (which is equivalent to the area of the original rectangle) will be equal to
A = ˝ d2 sin γ,
where γ is really the same angle as the angle formed by two diagonals of the rectangle.
Since d is fixed for any given circle, the question now boils down to:
For what value of γ the function A = ˝ d2 sin γ is going to take the maximum possible value?
Maximum value for sin function is 1 and therefore the question further boils down to:
For what value of γ we have sin γ = 1?
The answer is 90° meaning that our rectangle, when stretched to cover the maximum possible area while still being inscribed in a circle, will have to become square.
USING LAGRANGEAN
Area of the rectangle is A = ab, which we want to maximise.
However, considering what was proven in (1), it is clear that a and b will have to satisfy the following condition:
a2 + b2 = d2, where d is the length of the circle’s diameter.
Since we are not concerned about the actual length of the diameter (it will have to be fixed for any given circle), we can think of d2 as being any constant C.
So, the problem can be described as:
Maximise ab while satisfying the condition a2 + b2 = C
The corresponding Lagrangean is:
L = ab – λ( a2 + b2 – C)
The solution is obtained by letting each of the partial derivatives to be equal to 0 and then solving the obtained system of equations.
δL/δa = b – 2λa = 0
δL/δb = a – 2λb = 0
δL/δλ = a2 + b2 – C = 0.
From the first and the second equation, it follows that a2 – b2 = 0, which is equivalent to:
(a + b)(a – b) = 0,
which can be true only if a = b or if a = -b.
Since we only care about positive lengths of rectangles’ sides, then we disregard the second solution.
We are left with a = b, which proves that for the area of the rectangle inscribed in a circle to be maximised, its sides must be equal.
In other words, the rectangle will have to be a square.
Answer by richard1234(7193)  (Show Source):
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