I've been working on this. Inside the pentagon are five line segments that each make up one side of one of the five triangles; AC, BD, CE, BE and AD. These line segments intersect with one another at five points, which we can name F, G, H, I and J.
Now each of the five triangles with area = x are subdivided into three triangles, and share one of the small triangles with the large triangle clockwise, and one small triangle is shared with the large triangle counterclockwise. Thus ABC might be split into ABF, AFG and ACG. BCD will also contain the small triangle ACG.
Thus we will get five equations of the form
ABF + BFG + BCG = x = ABC.
Another such equation would be
BCG + CGH + CDH = x = BCD
Both have BCG, so we can subtract the two to get
ABF + BFG - CGH - CDH = x
That's as far as I've gotten. This sort of thing can be done several more times with different pairs of triangles. This may lead to the answer.
Originally posted by Fat LadyAw man, your reply with the answer immediately made me feel kind of stupid. Glad you didn't work it out thoroughly after all.
To be honest I assumed that you were correct in saying that any pentagon with that property would have the same area and then worked out the area for a regular pentagon, which is easy. However even armed with the correct answer I couldn't prove it for the general case.
Originally posted by Fat LadyI have never said so....Where has it been said that any pentagon with that property will have the same area?
To be honest I assumed that you were correct in saying that any pentagon with that property would have the same area and then worked out the area for a regular pentagon, which is easy. However even armed with the correct answer I couldn't prove it for the general case.
Originally posted by ranjan sinhaWhen I say "the same area" I mean cx, where x is the area of each of the five internal triangles defined by three consecutive vertices of the pentagon. If it is true that any such irregular pentagon will have area cx then it is also true for a regular pentagon, which is just a special case.
I have never said so....Where has it been said that any pentagon with that property will have the same area?
Originally posted by Fat LadyYou seem to be right in assuming the area of the pentagon being some constant times the area of each peripheral triangle. I had expressed the same view when I said that the area of the pentagon should be a linear function of x. It seems reasonable to assume that the area (of the pentagon) should be proportional to x from dimensional considerations too. But the constant of proportionality may vary with the shape. thus your constant "c" may be different for different possible irregular shapes with same x.
When I say "the same area" I mean cx, where x is the area of each of the five internal triangles defined by three consecutive vertices of the pentagon. If it is true that any such irregular pentagon will have area cx then it is also true for a regular pentagon, which is just a special case.
Originally posted by howzzatRight... you seem to be right.
You seem to be right in assuming the area of the pentagon being some constant times the area of each peripheral triangle. I had expressed the same view when I said that the area of the pentagon should be a linear function of x. It seems reasonable to assume that the area (of the pentagon) should be proportional to x from dimensional considerations t ...[text shortened]... thus your constant "c" may be different for different possible irregular shapes with same x.
Originally posted by AThousandYoungMy math is faulty.
I've been working on this. Inside the pentagon are five line segments that each make up one side of one of the five triangles; AC, BD, CE, BE and AD. These line segments intersect with one another at five points, which we can name F, G, H, I and J.
Now each of the five triangles with area = x are subdivided into three triangles, and share one of ...[text shortened]... n be done several more times with different pairs of triangles. This may lead to the answer.
ABF + BFG + BCG = x
BCG + CGH + CDH = x
ABF + BFG - CGH - CDH = 0
Not = x.
Originally posted by Fat LadyI have yet to see a proof that any two segments are parallel.
So if AB is parallel to EC and BC is parallel to AD, then we could call the intersection of AD and EC F, say, and the parallelogram ABCF would have area 2x. Then we just need to find the area of that tricky blighter FCD.