The mystery of an irregular pentagon

Originally posted by AThousandYoung
I have yet to see a proof that any two segments are parallel.

howzzat explained this already on page 2.

It's easier to see if you just draw a part of the pentagon, say ABCD (A at the top, then continuing anti-clockwise). The side BC is included in two triangles, ABC and BCD. If you imagine these triangles rotated so that BC is a horizontal base for them then their area is equal to 1/2 * base * vertical height. Since ABC and BCD have the same base and area, their vertical height must also be equal, i.e. A is the same vertical distance above BC as D. Hence AD is horizontal and thus parallel to BC.

Originally posted by Fat Lady
howzzat explained this already on page 2.

It's easier to see if you just draw a part of the pentagon, say ABCD (A at the top, then continuing anti-clockwise). The side BC is included in two triangles, ABC and BCD. If you imagine these triangles rotated so that BC is a horizontal base for them then their area is equal to 1/2 * base * vertical height. Sinc ...[text shortened]... A is the same vertical distance above BC as D. Hence AD is horizontal and thus parallel to BC.

Ok, I understand now. Thanks.

The diagram for this proof involved a pentagon (ABCDE)
the internal diagonals were drawn and the inersections labled (abcde)such that: a is opposite A; B is opposite B; c is opposite C; etc.

Given: Pentagon ABCDE
Triangle ABC = Triangle BCD = Triangle CDE
=Triangle DEA = Triangle EAB
Therefore: height BY of Triangle ABE = height DZ of Triangle ADE
BD is parallel to AE
By corollary proofs:
AB is parallel to CE; BC is parallel to AD;
CD is parallel to BE; DE is parallel to AC

AeDE and ABaE are parlellograms
Ae = ED; aE = AB; Ba = AE = eD
Ba – ae = eD – ae and it follows that Be = aD
Triangle ABe is congruent to Triangle EaD because of "side side side"
angle ABD = angle BDE
therefore: ABDE is an isosceles trapezoid
angle BAE = angle AED and AB =ED
similar proofs establish that
AB = CD = AE = BC = DE
angle ABC = angle BCD= angle CDE= angle DEA= angle EAD

edits were to replace symbols that did not translate

Originally posted by preachingforjesus
The diagram for this proof involved a pentagon (ABCDE)
the internal diagonals were drawn and the inersections labled (abcde)such that: a is opposite A; B is opposite B; c is opposite C; etc.

Given: Pentagon ABCDE
Triangle ABC = Triangle BCD = Triangle CDE
=Triangle DEA = Triangle EAB
Therefore: height BY of Triangle ABE = height DZ of Tri ...[text shortened]... angle CDE= angle DEA= angle EAD

edits were to replace symbols that did not translate

angle ABC = angle BCD= angle CDE= angle DEA= angle EAD

If this is true, then it's not an irregular pentagon.

Originally posted by preachingforjesus
Triangle ABe is congruent to Triangle EaD because of "side side side"

I'm probably missing something, but where have you shown that AB=ED?

Edit: Sorry, I see now. AB=aE, Ae=ED and Be=aD.

Originally posted by preachingforjesus
Triangle ABe is congruent to Triangle EaD because of "side side side"
angle ABD = angle BDE

Angle ABD is opposite Ae, so surely the congruency of ABe and EaD means that angle ABD is equal to the angle opposite ED, namely EaD? (Which can be seen much more easily by considering the parallel lines AB and aE cutting the line BD).

Originally posted by AThousandYoung
[b] angle ABC = angle BCD= angle CDE= angle DEA= angle EAD

If this is true, then it's not an irregular pentagon.[/b]

In my previous post I stated my belief that this was a regular pentagon.

this is the proof

Originally posted by ranjan sinha
Take for example the irregular pentagon having angles
angle/_A=115 deg; angle/_B=125 deg; angle/_C=95 deg;
angle/_D=130deg & angle/_E=105 degrees..../_CAD=25deg; /_DBE=40deg; /_ECA=30deg; /_ADB=45deg; /_BEC=40deg.
This pentagon satisfies the conditions of the problem...

The supplements (180 - angle) of your interior angles should sum to 360 degrees, but they don't. They add up to 330 degrees. Furthermore, looking at a quick handsketch of the pentagon you described (rough since your angles don't add up), the interior triangles don't look they could possibly be close to the same area. CDE looks MUCH smaller than ABC, even considering the roughness of the sketch. How do you know that the pentagon you described has all of its interior triangles (ABE, ABC, BCD, CDE, ADE) the same area? With all the parallels in there (AB parallel to CE, BC parallel to AD, etc.) I'm strongly leaning toward that this must be a regular pentagon. Can you show us another counterexample? Thanks.

If the problem is valid as stated, then the total area (relative to area x) of an irregular pentagon that meets the given conditions is the same as that of a regular pentagon, which definitely meets the given conditions. I have found the area of the regular pentagon. It is 5x/4*[cos(54° )]^2. If there are any such irregular pentagons and the problem is valid, then this is also the answer for the general case. (Yes, this is cheating.)

Since the coefficient c of the answer where Area = cx is not something "neat" from geometry, like a square root, then I suspect the problem is not valid. It's difficult to imagine that the answer for any general irregular pentagon is based on the cosine of 54° squared.

Speaking of cheating, someone here is trying to find the answer by posting it in another forum: http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/648.shtml

Solution:
Assume ABCDE is regular. Let each side have length s. Let the center be point O. Divide the pentagon into 5 triangles ABO, BCO, CDO, DEO, AEO.
Area of entire pentagon = 5 * area of any one of these triangles.
Find the area of CDO. Area(CDO) = 1/2 * base * height.
Base = CD = s.
h / 0.5s = tangent 54°; h = 0.5s*tan 54°
Area(CDO) = 1/2 * s * 0.5s*tan 54° = 1/4 * s^2 * tan 54°.
Area(ABCDE) = 5/4 * s^2 * tan 54°

Find the area of any one of the interior triangles, such as ABE. The problem defines this to be x. Find x in terms of the side s:
Area(ABE) = x = 1/2 * base * h (a different h from the one above)
cos 54° = h / AB = h/s
h = s * cos 54°
base = BE
sin 54° = 0.5*BE / s
base = BE = 2s * sin 54°
x = 1/2 * (2s * sin 54° ) * (s * cos 54° )
x = s^2 * sin 54° * cos 54°

Now determine c in the equation Area(pentagon) = cx.
c = Area(pentagon)/x
c = 5/4 * s^2 * tan 54° / (s^2 * sin 54° * cos 54° )
c = 5/4 * (sin 54°/cos 54° ) / (sin 54° * cos 54° )
c = 5/[4 * (cos 54° )^2]

Area of the pentagon in terms of x:
Area = 5x/[4 * (cos 54° )^2]

Originally posted by HolyT
Solution:
Assume ABCDE is regular. Let each side have length s. Let the center be point O. Divide the pentagon into 5 triangles ABO, BCO, CDO, DEO, AEO.
Area of entire pentagon = 5 * area of any one of these triangles.
Find the area of CDO. Area(CDO) = 1/2 * base * height.
Base = CD = s.
h / 0.5s = tangent 54°; h = 0.5s*tan 54°
Area(CDO) = 1/2 * s * ...[text shortened]... )
c = 5/[4 * (cos 54° )^2]

Area of the pentagon in terms of x:
Area = 5x/[4 * (cos 54° )^2]

What's the "center" of an irregular pentagon? Or are you building on the Preacher's proof?

Originally posted by AThousandYoung
What's the "center" of an irregular pentagon? Or are you building on the Preacher's proof?

Read the previous post.

Originally posted by preachingforjesus


AeDE and ABaE are parlellograms
Ae = ED; aE = AB; Ba = AE = eD
Ba – ae = eD – ae and it follows that Be = aD
Triangle ABe is congruent to Triangle EaD because of "side side side"
angle ABD = angle BDE
therefore: ABDE is an isosceles trapezoid
angle BAE = angle AED and AB =ED
similar proofs establish that
AB = CD = AE = BC = D ...[text shortened]... angle CDE= angle DEA= angle EAD

edits were to replace symbols that did not translate

You are right only up to the point where you establish congruence of triangles ABe and EaD.

However your next inference(about /_ABD = /_BDE) is wrong. When the three sides of a triangle are equal to the three sides of another triagle respectively, the two triangles are congruent but the angles opposite the corresponding equal sides in the two triangles are equal. In fact
the /_ABe (=/_ABD) of triangle ABe is equal to /_EaD (and not /_EDa ,which of course is equal to /_BDE )of triangle Ead.

Your inference about /_ABD = /_BDE being a faux pas, is misleading and wrong.

Hence all further inferences drawn by you are erroneous.

Originally posted by HolyT
The supplements (180 - angle) of your interior angles should sum to 360 degrees, but they don't. They add up to 330 degrees. Furthermore, looking at a quick handsketch of the pentagon you described (rough since your angles don't add up), the interior triangles don't look they could possibly be close to the same area. CDE looks MUCH smaller than ABC, even co ward that this must be a regular pentagon. Can you show us another counterexample? Thanks.

In the counter-example there has been a minor typo error in the measure of /_A. Actually /_A=100 degrees. .

Thus the five angles of the pentagon are 100, 125, 95, 110 and 110 degrres respectively.

Apologies for the error. However, it is clear from the measures (given in the same post) of angles /_CAD=25 deg, /_DBE=40 deg, /_ECA=30 deg, /_ADB=45 deg and /_BEC=40 deg, that angle /_A of the pentagon should be 100 deg.

Plz check-up the sum of the exterior angles(80+55+85+70+70); it indeed adds up to 360 degrees.

Originally posted by HolyT
If the problem is valid as stated, then the total area (relative to area x) of an irregular pentagon that meets the given conditions is the same as that of a regular pentagon, which definitely meets the given conditions. I have found the area of the regular pentagon. It is 5x/4*[cos(54° )]^2. If there are any such irregular pentagons and the problem is vali posting it in another forum: http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/648.shtml

You are making an unfounded assumption without logically proving it. The so-called proof of 'preachingforjesus' has already been shown to be specious and wrong.

As mentioned in some earlier post in response to Fat Lady's similar conjecture, it has rightly been pointed out that even if the area of the pentagon be proportional to 'x' , even then the constant 'c' of proportionality may be different for different possible shapes of the pentagon with the same 'x'.

It has not even been established yet that the area of the pentagon is necessarily a linear function of x.

BTW.... I again reiterate that the solution of this problem involves no higher than Higher Secondary School level skill in math and geometry.

Originally posted by HolyT
Solution:
Assume ABCDE is regular. Let each side have length s. Let the center be point O. Divide the pentagon into 5 triangles ABO, BCO, CDO, DEO, AEO.
Area of entire pentagon = 5 * area of any one of these triangles.
Find the area of CDO. Area(CDO) = 1/2 * base * height.
Base = CD = s.
h / 0.5s = tangent 54°; h = 0.5s*tan 54°
Area(CDO) = 1/2 * s * ...[text shortened]... )
c = 5/[4 * (cos 54° )^2]

Area of the pentagon in terms of x:
Area = 5x/[4 * (cos 54° )^2]

If the pentagon were a regular pentagon; there is nothing left for proving. The area of the irregular pentagon is unnecessarily being assumed to be equal to the irregular pentagon with the same 'x'.