A Line in a Pentagon

 X A -Z- -Y- -- -- E B \ / \ / D-----C Consider a regular pentagon ABCDE with side length b. Draw isosceles triangle XCD, such that the area of XCD is the same as the area of the pentagon. XC crosses AB at point Y. XD crosses AE at point Z. Show that the length of YZ equals 2b*cos(72). [KD: This is the same as showing the distance YZ = BE - CD.] What is the area of XCD outside the pentagon in terms of b? For example, use b=4000.
Source: Reader Denis Borris.
Solutions were received from Radu Ionescu, Adrian Atanasiu, A.G. Clarke, Jozef Hanenberg, Paul Botham, John Hewson, Philippe Fondanaiche. Some representative solutions follow.
From Philippe Fondanaiche
As a reminder:
cos(2*pi/5) = [sqrt(5) - 1]/4
sin(2*pi/5) = sqrt(10 + 2*sqrt(5))/4
cos(pi/5) = [sqrt(5) + 1]/4
sin(pi/5) = sqrt(10 - 2*sqrt(5))/4
tan(pi/5) = sqrt(5-2*sqrt(5))

XA goes through the center O of the circle circumscribed to the pentagon ABCDE.
Moreover XA cuts CD at H and YZ cuts at K.
Let P the foot of the  perpendicular from Y on the line BC.

As area(XCD) = area (pentagon ABCDE) ,
XA = 5*OH and area(XAY) = area(BCY) or YK*XA=YP*BC
Let CD=b and YK=x

Then we have the following identities:b*(sqrt(5)+1
1) BE = 2*b*cos(pi/5) =b*(sqrt(5)+1)/2 = b*golden number
[see a recent Ken's POTW on the properties of the diagonal
of regular penatgon]
2) OH = b/(2*tan(pi/5)) = b/[2*sqrt(5-2*sqrt(5))]
3) OA=OC=b/(2*sin(pi/5)) = 2*b/sqrt(10 - 2*sqrt(5))
4) XA = 5*OH - OH - OA = 2*b/tan(pi/5) - b/(2*sin(pi/5)) = 2*b*sin(2*pi/5)
= b*sqrt(10 + 2*sqrt(5))/2
5) YP = YB*sin(2*pi/5) = (b-x/cos(pi/5))*sin(2*pi/5)

Therefore we have the relation:
x*2*b*sin(2*pi/5) = b*(b-x/cos(pi/5))*sin(2*pi/5) ==>
x = b*(sqrt(5)-1)/4 = b*cos(2*pi/5)
and we can check that
2*x = b*(sqrt(5)-1)/2 = BE - CD = b*(sqrt(5)+1)/2 - b

As a consequence area(XAYZ) = 2*area(XAY) = YK*XA
= 2*b*sin(2*pi/5) * b*cos(2*pi/5) = b^2*sin(pi/5)
= b^2*sqrt(10-2*sqrt(5))/4

From Radu Ionescu: 1) the area of XCD is the same as the area of the pentagon <==>the area of DEZ=the area of XAZ ==> EZ*ZD=XZ*ZA (1) ZA/ZE=ZD/ZX <==> EA/ZE=XD/ZX triangles XZY and XDC are similar ==> XD/ZX=DC/ZY (2) (1) and(2) ==> ZE=ZY Let be Z' on EC so that it is the simetric point of Z to EB and Y' on DB so that it is the simetric point of Y to EB ==>EZ=ZY=YB=BY'=Y'Z'=Z'E triangle Z'CB is isosceles ( angleBZ'C=18+36=angleZ'BC)==>Z'C=BC=b ==> YZ=EZ'=EC-b=EB-DC=2*b*cos(72) 2) the area of XCD outside the pentagon is 2*area(EDZ)=EZ*ED*sin108= 2*(b^2)*sin(108)*cos(72)=(b^2)*sin(36)= 94045648
From Jozef Hanenberg:
Let M be the centre of the pentagon and let T be the midpoint of CD.
Area(MCD) = Area(ABCDE)/5 = Area(XCD)/5.
It follows that XT = 5*MT, because triangles XCD and MCD share the
same base. Looking at triangle MCD we find
MT = b/(2tan(36)) and MA = MC = b/(2sin(36))
We find XA = XT - MT -  MA = 5b/(2tan(36)) - b/(2tan(36)) - b/(2sin(36))
= 4b/(2tan(36)) - b/(2sin(36)) = b*(4cos(36) -1)/(2sin(36))

XCD and ABCDE have the same area and they partially overlap.
If you look at the parts that don't overlap you find that
Area(XAY) = Area(YBC).
We use the formula for the area of a triangle: Area = 0.5 *a*b*sin(C),
where C is the angle formed by the sides a and b.

O.5*XA*AY*sin(XAY) = 0.5*BC*BY*sin(B). It follows that
b * (4cos(36) - 1)/(2sin(36)) * AY * sin(126) = b * (b -AY) * sin(108)).
Using sin(126) = cos(36) and sin(108) = sin (72) = 2sin(36)*cos(36)
and isolating for AY we get

AY = 4b*(sin(36))^2/(4cos(36) -1 + (4sin(36))^2)
= 4b*(1 - (cos(36)^2/(4cos(36) + 3 -  4(cos(36))^2)

Look in triangle AYZ and you find that
YZ = 2*AY*cos (36) = 8b*(1-(cos(36)^2)*cos(36)/(4cos(36) + 3 - 4(cos(36)^2).

We use cos 36 =(SQRT(5) + 1)/4. After substitution and a lot of
algebra you get YZ = b*(SQRT(5) - 1)/2 = 2*b*cos(72).
Here we use cos 72 = (SQRT(5) -1)/4.

The area outside the pentagon = 2*0.5*AY*XA*sin(126) = AY*XA*cos(36)
.....= b^2 * SQRT((5-sqrt(5))/8)

From John Hewson:
The following results are used later:

1/(2 cos 36) =  2cos 36 - 1    ......   (1)
1/(2 cos 36) = 2 cos 72   .....   (2)
2 cos 72 = 2 cos 36 - 1   .....   (3)

Produce XA to cut ZY, EB, and DC in Q, P, and O respectively.
Put ZY = w and AQ = q, and b = 1.
The internal angle of the pentagon is 108, and AEB = 36, and BED = 72.
The area of the pentagon is the sum of the areas of the triangles AEB, CBD,
and EBD.
That is sin 36 cos 36 + sin 36 cos 36 +(sin 36 + sin 72)/2 =(3 sin 72 +sin
36)/2, and that is equal to the area of DCX which is OX/2.
Thus OX = 3 sin 72 + sin 36
Now OA = sin 72 + sin 36 and XA = OX - OA = 2 sin 72.
Part 1.

There are two pairs of similar triangles: XDC and XZY; and AEB and AZY.
For each of these pairs we can equate the ratios of height to base:
AQ/AP = YZ/EB; and XQ/XO = YZ.
Thus q/sin 36 = w/(2 cos 36) or, using (2), q = 2w sin 36 cos 72; and
(2 sin 72 +q)/(3 sin 72 + sin 36) = w.
Eliminating q:
2 sin 72 = w(3 sin72 +sin 36 - 2 sin 36 cos 72)
That is:
4 sin 36 cos 36 = w(6 sin 36 cos 36 + sin 36 - 2 sin 36 cos 72)
Dividing through by sin 36 and using (3):
4 cos 36 = w(6 cos 36 + 1 - 2 cos 36 +1) = w(4 cos 36 + 2)
That is:  2 cos 36 = w(2 cos 36 + 1)
Dividing through by 2 cos 36 and using (1) we have:
1 = w(1 + 2 cos 36 -1);  w = 1/(2 cos 36), and using (2):
w = 2 cos 72.
For a pentagon of side b, YZ  = 2b cos 72 as required.

Part 2.
The area of XDC outside the pentagon is the difference of areas of the
triangles XZY and AZY.
Keeping b = 1 for the moment, the area is w XA/2.
Now w = 2 cos 72 and XA = 2 sin 72, and the required area is:
2 cos 72 sin 72 = sin 144 = 0.587785252292.
For a pentagon of side b the area of XCD outside the pentagon is
0.587785252292 b^2, and if b = 4000, it is 9404564.0.

My Solution (from Ken Duisenberg):
From the problem statement, we can deduce that the area of triangles XAY
and BYC are equal.  I'm going to assume the result and attempt to prove
the areas of the triangles are equal, which will in turn support the
problem statement, proving the result.  (Basically, the problem says
"Given Area XDC equals the pentagon, show YZ=BE-CD.", and I'm just proving
"Given YZ=BE-CD, show area XDC equals the pentagon."  Some may disagree
with this approach, but I believe the logic is sound.  I'm basically
showing you can't have one without the other.)

From C, draw a line perpendicular to DC, intersecting AB at point F.
BF=AY.  Triangles XAY and CFY are similar.

Now from our previous learning that the diagonal of a pentagon is the
golden ratio (see previous POTW), we know:

(P = golden ratio = [1 +sqrt(5)]/2)
(1) AB = P*AF = P*BY.       - Similar Triangles AB/2 to BE/2

(2)    AY = AB-BY
= P*BY - BY
BY/AY = 1/(P-1) = P
(2)    BY = P*AY

(3)    AF = P*BF            - Equivalence

(4)    FY = AF-AY
= P*AY - AY
AY/FY = 1/(P-1) = P
(4)    AY = P*FY

Through similar triangles, the area of XAY is P^2 times the area of CFY.
(Both length and width dimensions are larger by a factor of P, so the area
is larger by a factor of P^2.

Triangles CFY and CBY have the same height.  Triangle CBY is larger in
area by the ratio of BY/FY.  From 2 and 4, BF=(P^2)*FY, so the area of
triangle CBY is P^2 times the area of CFY.

So the area of triangles CBY and XAY are the same.  Since the rest of the
areas are common, the area of triangle XCD must equal the area of the
pentagon.

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