Problem:
A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by AB, has length 31. Find the sum of the lengths of the three diagonals that can be drawn from A.
Solution:
Label the remaining four vertices C,D,E, and F, in the natural order. Draw diagonals AC,AD, and AE. and let their lengths be x,y, and z respectively. Draw BD,BE, and DF and note that BD=z,BE=y, and DF=z, since chords of congruent arcs are congruent. Next we apply Ptolemy's Theorem: For a quadrilateral inscribed in a circle, the product of the lengths of the diagonals is equal to the sum of the products of the lengths of opposite sides. Using Ptolemy's Theorem on quadrilaterals ABCD,ABDE, and ADEF respectively, we obtain
xz=31β 81+81y(1)
y2=31β 81+z2
z2=812+81y(2)
Thus y2β31β 81=812+81y, implying
y2β81yβ81β 112=(y+63)(yβ144)=0
Since y cannot be β63, we must have y=144. Substituting in (2) and then in (1) we obtain z=135 and x=105. The sum of the three diagonals is then x+y+z=384β.
OR
Let R be the radius of the circle and 2x the measure of one of the central angles subtended by a side of length 81. Since there are five such sides in the hexagon, we must have x<36β. We now have
81=2Rsinx and 31=2Rsin(Οβ5x)=2Rsin5x(1)
and the sum of the lengths of the three diagonals from A is
2R(sin2x+sin3x+sin4x)(2)
We will use (1) to find the value of sinx and then use this to evaluate the sum in (2). By DeMoivre's formula,
from which sinx=Β±11β/6 or Β±34β/6. Since x<36β and sinx must be positive, we conclude that sinx=11β/6 and cosx=5/6. We can now evaluate the sum in (2):
Label the remaining vertices C,D,E,F. We shall find and make use of a few pairs of similar right triangles to calculate AD,AE, and AC, in this order.
Mark point Bβ² on AF so that AB=ABβ². Since β BAD and β FAD intercept congruent arcs, these angles are congruent, and it follows that β³BAD and β³Bβ²AD are congruent. Thus BD=Bβ²D. Since BD=FD is also true, we have Bβ²D=FD. If we now let M denote the midpoint of Bβ²F we see that β³AMD is a right triangle with leg AM of length 56. Now drop a perpendicular from D to the extension of FE and let N be the foot of this perpendicular. Since trapezoid ADEF is isosceles, we deduce that β FAD=β NED, and hence that β³AMDβΌβ³END. Thus
AD56β=ADAMβ=EDENβ=812ADβ81ββ
from which AD2β81ADβ2β 56β 81=0. The solutions to this quadratic are β63 and 144, hence AD=144.
The preceding argument has also shown that EN=63/2. The Pythagorean theorem now gives us DN=4511β/2, and a second application, to β³FND, gives FD=135.
This is also the length of AE.
To find AC, consider isosceles β³AEC. Let P be the midpoint of AC, so that β³PAE is a right triangle. This triangle is similar to β³MAD since β PAE=β MAD. We thus obtain
AMAPβ=ADAEβ
from which
AP=(AM)(AE)/AD=2105β
and AC=2AP=105. The required sum is therefore AC+AD+AE=384β.
