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 square has a side equal to sqrt(5) and therefore a diagonal of length sqrt(10), which is also the diameter of the circle, whose radius is therefore sqrt(10)/2.
If rectangle BDFH has shorter and longer sides x and y respectively, then xy = 4 and y=sqrt(10-x^2). This can be solved numerically and the numbers come out to sqrt(2) and 2*sqrt(2), which can be verified by plugging into the two equations. These subtend arcs of 53.13... and 126.86... degrees along the circle.
The only variable is the orientation of rectangle BDFH relative to square ACEG. The short sides of the rectangle could either bracket opposite corners of the square, or be bracketed by two opposite pairs of corners of the square. In the former case, triangles are formed on all four sides of the square to complete the octagon and Heron's formula can be used to compute their areas, once the sides have been computed as chords of the circle based on the central angle (arc length) separating them from the vertices of the square. In the latter case Brahmgupta's formula can be used as two quadrilaterals are appended to the square to make the octagon, and these quadrilaterals are inscibed in the circle.
The following program tabulates the total area (four triangles plus 5 or two quadrilaterals plus 5) at 2-degree intervals and at two local maxima for the arc separating a corner of the rectangle from a corner of the square:
DECLARE FUNCTION asin# (x#)
DEFDBL A-Z
CLS
pi = ATN(1#) * 4#
r = SQR(10) / 2
x = 2
DO
xP = x
x = 4 / SQR(10 - x * x)
LOOP UNTIL x = xP
PRINT x, 4 / x
sep1 = 2 * asin(x / 2 / r) * 180 / pi
sep2 = 180 - sep1
PRINT sep1, sep2
FOR a1 = 1 TO sep1 STEP 2
b1 = sep1 - a1
a2 = 90 - a1
b2 = 90 - b1
chord1 = 2 * r * SIN(a1 / 2 * pi / 180)
chord2 = 2 * r * SIN((a2 / 2) * pi / 180)
s = .5 * (chord1 + chord2 + SQR(5))
area = SQR(s * (s - chord1) * (s - chord2) * (s - SQR(5)))
chord1 = 2 * r * SIN(b1 / 2 * pi / 180)
chord2 = 2 * r * SIN(b2 / 2 * pi / 180)
s = .5 * (chord1 + chord2 + SQR(5))
area = 2 * (area + SQR(s * (s - chord1) * (s - chord2) * (s - SQR(5))))
area = area + 5
PRINT USING "### ##.#########"; a1; area
NEXT a1
a1 = sep1 / 2
b1 = sep1 - a1
a2 = 90 - a1
b2 = 90 - b1
chord1 = 2 * r * SIN(a1 / 2 * pi / 180)
chord2 = 2 * r * SIN((a2 / 2) * pi / 180)
s = .5 * (chord1 + chord2 + SQR(5))
area = SQR(s * (s - chord1) * (s - chord2) * (s - SQR(5)))
chord1 = 2 * r * SIN(b1 / 2 * pi / 180)
chord2 = 2 * r * SIN(b2 / 2 * pi / 180)
s = .5 * (chord1 + chord2 + SQR(5))
area = 2 * (area + SQR(s * (s - chord1) * (s - chord2) * (s - SQR(5))))
area = area + 5
PRINT a1; area
FOR a1 = 1 TO 45 - sep1 / 2 STEP 2
chord1 = 2 * r * SIN(a1 / 2 * pi / 180)
chord2 = 2 * r * SIN(sep1 / 2 * pi / 180)
chord3 = 2 * r * SIN((90 - a1 - sep1) / 2 * pi / 180)
s = .5 * (chord1 + chord2 + chord3 + SQR(5))
area = 2 * SQR((s - chord1) * (s - chord2) * (s - chord3) * (s - SQR(5))) + 5
PRINT USING "### ##.#########"; a1; area
NEXT a1
a1 = 45 - sep1 / 2
chord1 = 2 * r * SIN(a1 / 2 * pi / 180)
chord2 = 2 * r * SIN(sep1 / 2 * pi / 180)
chord3 = 2 * r * SIN((90 - a1 - sep1) / 2 * pi / 180)
s = .5 * (chord1 + chord2 + chord3 + SQR(5))
area = 2 * SQR((s - chord1) * (s - chord2) * (s - chord3) * (s - SQR(5))) + 5
PRINT a1; area
FUNCTION asin (x)
asin = ATN(x / SQR(1 - x * x))
END FUNCTION
The output is:
1.414213562373095 2.82842712474619
53.13010235415598 126.869897645844
1 6.051443390
3 6.148785077
5 6.238635417
7 6.320884940
9 6.395433439
11 6.462190087
13 6.521073552
15 6.572012093
17 6.614943650
19 6.649815917
21 6.676586408
23 6.695222506
25 6.705701507
27 6.708010644
29 6.702147104
31 6.688118029
33 6.665940513
35 6.635641575
37 6.597258130
39 6.550836942
41 6.496434568
43 6.434117290
45 6.363961031
47 6.286051265
49 6.200482915
51 6.107360231
53 6.006796669
26.56505117707799 6.70820393249937
1 6.008497746
3 6.024112280
5 6.037869919
7 6.049753899
9 6.059749743
11 6.067845273
13 6.074030624
15 6.078298262
17 6.080642986
18.43494882292201 6.081138830084191
where the first line shows the lengths of the sides of the rectangle, and the second shows the total arc cut by these sides as chords.
The maximum area shows up as 6.081138830084191, when the rectangle's short edges do not bracket a corner of the square, and are centered on one of the sides of the square, at which point the corner of each rectangle is 18.43... degrees of arc from a corner of the square.
P.S., of course 6.708... > 6.081..., and the answer with the short side of the rectangle straddling a corner of the square is the largest, as per later posts.
Edited on October 25, 2004, 8:54 pm
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Posted by Charlie
on 2004-10-25 16:00:08 |
solution
