16 Sep 07
Originally posted by howzzatYou are quite right.
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.
16 Sep 07
3 edits
Originally posted by AThousandYoungThat'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...
How does the area imply that?
16 Sep 07
Originally posted by howzzatI believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
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.
16 Sep 07
Originally posted by howzzatThis would mean that AD is parallel to BC.
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...
I'm not sure I buy this part of your analysis. Can you prove it?
16 Sep 07
Originally posted by ranjan sinhaIs the answer simply (2 + phi)x?
" 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'."
[where phi = golden ratio]
16 Sep 07
Originally posted by LemonJelloGiven: pentagon made up of five triangles
Is the answer simply (2 + phi)x?
[where phi = golden ratio]
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
16 Sep 07
1 edit
Originally posted by rubberjaw30No. Just sketch an irregular pentagon ABCDE and then draw the triangles specified in the problem, and you'll see why you are mistaken.
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
16 Sep 07
Originally posted by rubberjaw30The triangles you're referring to are not the ones mentioned in the problem. You're referring to triangles with one vertex in the center.
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
17 Sep 07
2 edits
Originally posted by AThousandYoungThe triangles ABC and BCD have common base BC. Right?
[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]
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.
17 Sep 07
1 edit
Originally posted by preachingforjesusI don't think you are right in believing so.
I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
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:-
/_CAD=25deg; /_DBE=40deg; /_ECA=30deg; /_ADB=45deg; /_BEC=40deg.
This pentagon satisfies the conditions of the problem. But obviously it is not a regular pentagon.