The octagon ABCDEFGH is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon ACEG is a square of area 5 and the polygon BDFH is a rectangle of area 4, find the maximum possible area of the octagon.
The answer is A = 6.7082
Ok, yet again I don’t have a proof, but a hunch. We have no control over the size of the BDFH rectangle, but we do have control over the orientation of it relative to the square. My hunch is that the maximum possible area of the octagon occurs when it is symmetric over the diagonals of the square (when the sides of the square and the sides of the rectangle are at 45 degree angles). I’ll explain how I got there at the end. For now…
I will call the sides of the square S, the diagonal of the square D, the base of the rectangle B, and the height of the rectangle H. Starting with the square, the area is S^2 = 5. So S = sqrt(5). The diagonal then is D^2 = S^2 + S^2 = 2*S^2 = 2*5 = 10. So D = sqrt(10).
Next, the square is inscribed in the circle, so the diameter of the circle is the diagonal of the square, which is D. This means that the rectangle’s diagonal must also be D. So D^2 = B^2 + H^2. Well, the area of the rectangle is B*H = 4. So H = 4/B. Substituting, we get D^2 = B^2 + (4/B)^2 = B^2 + 16/B^2. Well, multiply both sides by B^2 and we get D^2*B^2 = B^4 + 16, or 0 = B^4 – 10B^2 + 16. This is basically 0 = x^2 – 10x + 16, it’s just that x=B^2. So I can use the quadratic function to find x, and then B = sqrt(x). Well, x = (10 +/ sqrt(10^24*1*16))/(2*1) = (10 +/ sqrt(10064))/2 = (10 +/ sqrt(36))/2 = (10 +/ 6)/2 = 5 +/ 3. So x = 8 or 2. So B = sqrt(8) or sqrt(2). Let’s say B = sqrt(8) = 2sqrt(2) since it’s bigger. Then H = 4/B = sqrt(2).
Ok, so to recap we have S = sqrt(5), D = sqrt(10), B = 2sqrt(2), H = sqrt(2).
Now I need to find the Perpendicular distance, P, between a side of the square and a vertex of the rectangle. The easiest way for me to calculate it (keeping radicals for as long as possible, not just using decimals) is using a coordinate system. I will have the origin at the center of the circle, with the diagonals of the square along the axes, and the short sides of the rectangle parallel to the xaxis). So two corners of the square are at (0, D/2) and (D/2, 0), and a corner of the rectangle is at (H/2, B/2).
The line equation for the side of the square is y = (1)x + D/2. Next, we want an equation for a line that is perpendicular to the side of the square, and passes through (H/2, B/2). Well, this line’s slope will be the negative reciprocal of the squares slope. So m = (1/(1)) = 1. So far we have y = x + b. Well, we have one pair of x and y, so that will help us find b. B/2 = y = x + b = H/2 + b. So b = (BH)/2.
So our two equations are
y = (1)x + D/2
y = x + (BH)/2
Next, we’ll find where these two lines intersect, then find the distance between that intersection point and the vertex of the rectangle.
(1)x + D/2 = y = x + (BH)/2
D/2 = 2x + (BH)/2
2x = D/2 – (BH)/2
2x = (DB+H)/2
x = (DB+H)/4
So y = (2DD+BH)/4 = (D+BH)/4
Alrighty, so the distance, P, between (x, y) and (H/2, B/2) is just
P^2 = (H/2 – x)^2 + (B/2 – y)^2 = [(2H – D + B – H)/4]^2 + [(2B – D – B + H)/4]^2
P^2 = [(H – D + B)/4]^2 + [(B – D + H)/4]^2 = 2[(B – D + H)/4]^2 = (B+H–D)^2/8
P^2 = (2sqrt(2) + sqrt(2) – sqrt(10))^2/8 = (3sqrt(2) – sqrt(10))^2/8
P^2 = [(3sqrt(2))^2 –2*3sqrt(2)sqrt(10) + (sqrt(10))^2]/8 = (9*2 – 6sqrt(20) + 10)/8
P^2 = (28 – 12sqrt(5))/8 = (7 – 3sqrt(5))/2
P = sqrt[(7 – 3sqrt(5))/2]
So the area of the octagon is
A = S^2 + 4*(S*P/2) = S^2 + 2(S*P) = 5 + 2[sqrt(5)*sqrt[(7 – 3sqrt(5))/2]]
A = 5 + 2sqrt[5(73sqrt(5))/2] = 5 + 2sqrt[10(73sqrt(5))/4]
A = 5 + sqrt(70 – 30sqrt(5))
So A = 6.7082
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Ok, here’s my explanation of my hunch.
Forgetting about dimensions for a second, draw a square inscribed in a circle. Then draw a rectangle inscribed in the circle. The centers of the two should line up. The area of the octagon is simply the area of the square, plus the 4 triangles that are defined by one edge of the square and one vertex of the rectangle. Two triangles on opposite sides of the square will be identical. We’ll say the height of a triangle is the Perpendicular distance between the vertex of the rectangle and the side of the square. Since we have two pairs of triangles, we’ll have P1 and P2. Calling the length of the sides of the square S, the equation for the area of our octagon is simply:
A = S^2 + 2*(S*P1/2) + 2*(S*P2/2) = S^2 + S*P1 + S*P2 = S^2 * S(P1+P2)
We have no control over S, as that is basically given. So the area of the octagon is maximized when P1+P2 is maximized. Well, I can make P1 really big (the distance from the midpoint of the side of the square to the circle), but then P2 is really small. I can rotate the rectangle from side to side and see the areas of those triangles change. As one grows the other shrinks. This leads me to believe that (P1+P2) is maximized when P1=P2. This occurs when the octagon is symmetric over a diagonal of the square. So the sides of the rectangle are perpendicular to the diagonals of the square (or at 45 degree angles to the sides of the square).

Posted by nikki
on 20041025 18:09:40 