Originally posted by howzzat Actually...under the given condition [of all the peripheral triangles ABC, BCD, CDE, DEA and EAB being equal in area], the pentagon has to be convex. This is because, equal area of each of the peripheral triangles will imply that the line joining AD is parallel to BC.
Similarly the line joining BE is parallel to CD; the line joining CA is parallel ...[text shortened]... t cannot but be convex. Thus stating the additional condition of convexity is redundant in fact.
Originally posted by AThousandYoung How does the area imply that?
That's coz both the triangles ABC and DBC can be viewed as having the common base BC. Hence their vertices A and D must be at equal heights from the base BC. This would mean that AD is parallel to BC. Similarly for the other sides CD, DE etc...
Originally posted by howzzat Actually...under the given condition [of all the peripheral triangles ABC, BCD, CDE, DEA and EAB being equal in area], the pentagon has to be convex. This is because, equal area of each of the peripheral triangles will imply that the line joining AD is parallel to BC.
Similarly the line joining BE is parallel to CD; the line joining CA is parallel ...[text shortened]... t cannot but be convex. Thus stating the additional condition of convexity is redundant in fact.
I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
Originally posted by preachingforjesus I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
That came from someone named preachingforjesus? Well, let's see if you can manage a proof. I certainly can't.
Originally posted by howzzat That's coz both the triangles ABC and DBC can be viewed as having the common base BC. Hence their vertices A and D must be at equal heights from the base BC. This would mean that AD is parallel to BC. Similarly for the other sides CD, DE etc...
This would mean that AD is parallel to BC.
I'm not sure I buy this part of your analysis. Can you prove it?
Originally posted by ranjan sinha " ABCDE is an irregular convex pentagon such that each of the five triangles ABC, BCD, CDE, DEA and EAB have equal area say 'x'.
Find the area of the pentagon in terms of 'x'."
Originally posted by rubberjaw30 Given: pentagon made up of five triangles
Given: all triangles have equal areas, x
fact: area of pentagon = sum of areas of inscribed triangles
therefore, area of pentagon equals x + x + x + x +x equals 5 x
No. Just sketch an irregular pentagon ABCDE and then draw the triangles specified in the problem, and you'll see why you are mistaken.
Originally posted by rubberjaw30 Given: pentagon made up of five triangles
Given: all triangles have equal areas, x
fact: area of pentagon = sum of areas of inscribed triangles
therefore, area of pentagon equals x + x + x + x +x equals 5 x
The triangles you're referring to are not the ones mentioned in the problem. You're referring to triangles with one vertex in the center.
Originally posted by AThousandYoung [b]This would mean that AD is parallel to BC.
I'm not sure I buy this part of your analysis. Can you prove it?[/b]
The triangles ABC and BCD have common base BC. Right?
Area of a triangle =(1/2)*Base*Height. Right?
If base is common, the heights of the two triangles must be equal. Hence if perpendiculars are drawn from A and D up to the line BC, these perpendiculars will have equal length. Under the conditions of the problem it seems clear that both the points A and D are on the same side of the line BC. Hence the line joining AD will have to be parallel to BC.
P.S. - Of course this argument will fail if A and D lye on opposite sides of the line BC.
Originally posted by preachingforjesus I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
I don't think you are right in believing so.
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.
I have constructed this example by working backwards using:-