- 01 Sep '07 00:26

I'm with Mephisto... where is point F in a pentagon, irregular or not?*Originally posted by ranjan sinha***ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.**

Any takers for this puzzle? - 01 Sep '07 01:25 / 1 edit

5x, assuming "DEF" is really "DEA" and "EFA" is really "EAB".*Originally posted by ranjan sinha***ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.**

Any takers for this puzzle?

Oh, no, I'm wrong.

Is F the center? - 09 Sep '07 13:39

Both of you -you and Mephisto are right....*Originally posted by jebrydzagin***I'm with Mephisto... where is point F in a pentagon, irregular or not?**

Apologies for the inadvertant error....

The puzzle should actually read as follows:-

" 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'."

However, in spite of the inadvertant error, the spirit and the substance of the puzzle was quite clear...

I stand corrected...Plz solve the puzzle now... - 14 Sep '07 14:28 / 1 edit

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.*Originally posted by ranjan sinha***Both of you -you and Mephisto are right....**

Apologies for the inadvertant error....

The puzzle should actually read as follows:-

" 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'."

However, in spite of the inadvertan ...[text shortened]... nce of the puzzle was quite clear...

I stand corrected...Plz solve the puzzle now...

Similarly the line joining BE is parallel to CD; the line joining CA is parallel to DE. And so on... for each of the five sides...

Hence the pentagon ABCDE , under such conditions, cannot be concave. It cannot but be convex. Thus stating the additional condition of convexity is redundant in fact. - 15 Sep '07 10:47

How does the area imply that?*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.