23 Sep 07
Originally posted by Fat LadyYou have rightly pointed out the error.
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.
[quote]
Originally posted by preachingforjesus
Triangle ABe is congruent to Triangle EaD ...[text shortened]... be seen much more easily by considering the parallel lines AB and aE cutting the line BD).
23 Sep 07
Originally posted by HolyTYour demonstration is invalid because you have not illustrated that any particular angle is 54 degrees.
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]
Furthere more: if all base angles of the central triangle is 54 degrees; then in order for all 5 of the given triangles to have equal area the pentagon must be regular.
wich I proved several posts ago.
24 Sep 07
1 edit
Originally posted by preachingforjesusMy demonstration is absolutely sound. The only invalid part might be that it is not proven to apply to the general case for irregular pentagons, but I clearly stated that multiple times in the course of introducing and giving my answer. Read my 3 posts in a row carefully. I stated at the outset that I was finding the area of a regular pentagon in terms of x as a "cheating" way of finding the answer for the general case.
Your demonstration is invalid because you have not illustrated that any particular angle is 54 degrees.
Furthere more: if all base angles of the central triangle is 54 degrees; then in order for all 5 of the given triangles to have equal area the pentagon must be regular.
wich I proved several posts ago.
I have no idea what you mean by "if all base angles of the central triangle is 54 degrees...." At any rate, I clearly described each step of what I was doing. The 5 interior angles of a regular pentagon are 108 deg. each. The 5 isosceles triangles that I was using to find the area all have interior angles of 54, 54, and 72 degrees.
24 Sep 07
Originally posted by ranjan sinhaNo, I made a conjecture or speculation and my tone and content were completely clear that I was not attempting to make a proof of my statement. I was using that conjecture to say that IF the problem was such that the case is the same for the regular pentagon and the irregular ones, then finding the result for the regular pentagon would also find the result for the general case.
You are making an unfounded assumption without logically proving it.
If the goal of this puzzle is NOT to find the area of the pentagon in terms of x, which is the only given value, then I would be extremely curious to know in terms of what else the area could be expressed that is general. What other values are out there which can be used to express the area for the general case? If the area is stated in terms of one or more of the angles of the irregular pentagon, then that's not really a general solution.
24 Sep 07
1 edit
Originally posted by ranjan sinhaDon't the the angles of a pentagon sum to 540 degrees?
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, /_ ...[text shortened]...
Plz check-up the sum of the exterior angles(80+55+85+70+70); it indeed adds up to 360 degrees.
29 Sep 07
3 edits
Originally posted by HolyTI am bored with this puzzle now. I am giving the solution. The solution for the general case runs as follows:-
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
Let ABCDE be the irregular pentagon satisfying the stated property. Join all the internal diagonals; a smaller inverted pentagon PQRST is created inside ABCDE. Let P be the tip opposite to A, Q opposite to B &c..&c.. A 5-pronged star is created inside the pentagon having its tips on the 5 corners of the pentagon. Let the area of the bigger pentagon ABCDE be w and that of the smaller inner pentagon PQRST be w'.
There are 5 triangles with their base on one of the sides of the internal pentagon and vertex on one of the tips of the outer pentagon (viz. triangles ASR, BTS, CPT, DQP and REP). Let their areas be a1, a2, a3, a4 and a5 respectively.These triangles form the five arms of the internal 'penta'-star.
Likewise 5 inwardly directed triangles are created which have their base on one of the sides of the outer pentagon and their vertices on the tips of the internal pentagon (viz. triangles PCD, QDE, REA, SAB and TBC). Let their areas be b1, b2, b3, b4 and b5 respectively.
According to the given condition,
a1+b3+b4 = a2+b4+b5 = a3+b5+b1 = a4+b1+b2 = a5+b2+b3 = x.
................................. .................eqns.(1)
It has been shown that the str. lines BE, CD are parallel; similarly str. lines CA, DE are parallel &c..&c..These results directly lead to triangles CBS and DRE being congruent; similarly triamgles DCT and ESA are congruent, Triangles EDP and ATB are congruent &c. &c..{Preacher had also gotten up to this result.}
This leads to:->
a2+b5 = a5+b2; a3+b1 = a1+b3; a4+b2 = a2+b4;
a5+b3 = a3+b5; a1+b4 = a4+b1.
...........................................eqns.(2)
Using eqns. (1) and (2) we get:-
a1=a2=a3=a4=a5=a(say).
b1=b2=b3=b4=b5=b(say). Let b/a = c (say).
Thus x = a+2b; w' = x-2a = 2b-a; w = 3x+a+b = 4a+7b....eqns.(3)
Hence w/x = (4+7c)/(1+2c)...............................................eqn(4).
The ratio c can be determined from simple geometry as follows:-
Join SD. Now, area of triangle SCD = area of triangle BCD = x;
[on account of having common base BC and equal height].
Hence area of triangle STD = x-a-b = b.
Now c = b/a = (Ar. tr.BTC)/(Ar. tr. BTS) = CT/TS...............eqn.(5)
Also, CT/TS = (Ar. tr.CTD)/(Ar. tr. STD) = (a+b)/b =1+1/c....eqn(6)
From eqns.(5) and (6), c = 1+1/c ; This leads to the quadratic eqn:-
c^2 - c - 1 = 0. As c > 0; hence c = [sqrt(5)+1]/2. Yes c happens to be none other than the golden ratio arising from simple geometrical considerations, even without the pentagon having to be regular.
Putting this value of c in eqn.(4) we get the required value of w/x.
29 Sep 07
Originally posted by HolyTAs shown in the post above, one need not resort to assuming the pentagon to be regular.
No, I made a conjecture or speculation and my tone and content were completely clear that I was not attempting to make a proof of my statement. I was using that conjecture to say that IF the problem was such that the case is the same for the regular pentagon and the irregular ones, then finding the result for the regular pentagon would also find the result fo ...[text shortened]... one or more of the angles of the irregular pentagon, then that's not really a general solution.
02 Oct 07
Originally posted by ranjan sinhaI have found a much simpler and shorter solution.
I am bored with this puzzle now. I am giving the solution. The solution for the general case runs as follows:-
Let ABCDE be the irregular pentagon satisfying the stated property. Join all the internal diagonals; a smaller inverted pentagon PQRST is created inside ABCDE. Let P be the tip opposite to A, Q opposite to B &c..&c.. A 5-pronged star ...[text shortened]... to be regular.
Putting this value of c in eqn.(4) we get the required value of w/x.
03 Oct 07
3 edits
Originally posted by Fat LadyNo..not at all..Here is my proof which is much much shorter than the one given in ranjan's post dated 29 th Sept. Borrowing the notations and symbols in that post, the area of the pentagon ABCDE is:-
Don't tell me, the margin is too small to include your proof?
w = Ar. of tr.ABC + Ar. of tr. ADE + Ar. of tr. ACD.
= x + x + (AC/SC)*x. =(2 + c)*x where c = AC/SC..............eqn.(1)
[AC/SC is the ratio of heights of triangles ACD and SCD {both triangles have the common base CD} and area of triangle SCD is x.]
Now the area of the pentagon ABCDE can also be written as:-
w = Ar. of tr. ABE + Ar. of parallelogram SCDE + Ar. of tr. BSC.
= x + 2x + (SC/AC)*(Ar. of tr. ABC) = 3x +(1/c)*x
= (3 +1/c)*x..............................................................eqn.(2)
From eqns. (1) and (2) we get 2+c = 3+1/c; hence c = {sqrt(5) +1}/2
Thus we get the required value of w/x.