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Here is the question, it is no 3

 

https://web2.0calc.com/questions/a-few-geometry-problems-please-help-and-explain-thx

 

Let ABCDEF be a convex hexagon. Let A', B', C', D', E', F' be the centroids of triangles FAB, ABC, BCD, CDE, DEF, EFA, respectively. (a) Show that every pair of opposite sides in hexagon A'B'C'D'E'F' (namely A'B' and D'E', B'C' and E'F', and C'D' and F'A') are parallel and equal in length. (b) Show that triangles A'C'E' and B'D'F' have equal areas.

 

 

 Aug 23, 2020
edited by gwenspooner85  Aug 23, 2020
edited by gwenspooner85  Aug 23, 2020
edited by gwenspooner85  Sep 4, 2020
edited by gwenspooner85  Sep 4, 2020
edited by Melody  Sep 4, 2020
edited by Melody  Sep 4, 2020
edited by gwenspooner85  Sep 10, 2020
edited by Melody  Sep 11, 2020
 #1
avatar+118687 
+1

Do you have, or can you find the address of where this was asked last time?

It really is polite to continue it on the original thread, and it is also better for future people who might find it in a search.

 Aug 23, 2020
 #2
avatar+118687 
+2

Here is one answer

https://web2.0calc.com/questions/a-few-geometry-problems-please-help-and-explain-thx#r3

 

Here is another:

https://web2.0calc.com/questions/please-help_7172

 

 

Have you seen it at a different address again?

 Aug 23, 2020
 #3
avatar+781 
+2

It was the latter link, and I added it to my question. But does anyone have a hint or something to help me out here?

gwenspooner85  Aug 23, 2020
 #4
avatar+118687 
+3

Let ABCDEF be a convex hexagon. Let A', B', C', D', E', F' be the centroids of triangles FAB, ABC, BCD, CDE, DEF, EFA, respectively. (a) Show that every pair of opposite sides in hexagon A'B'C'D'E'F' (namely A'B' and D'E', B'C' and E'F', and C'D' and F'A') are parallel and equal in length. (b) Show that triangles A'C'E' and B'D'F' have equal areas.

 

 

 

Hi Gwen,

I have spent time playing with this question. 

Graphically I have convinced myself that it is true.

I tried to follow Gavin's logic in the other answer but he seemed to jump a big step. Maybe what he was saying would have been more obvious if his pic had still been visible, but I doubt it.

 

However I did simply the problem to look only at one half of hexagon, which is a quadrilateral.  The properties of each quadrilateral will be the same so if you can prove 

1)     B'C' is parallel to AD   and that     (that is what it always appears to be from my diagram.)

2)       B'C' = one third of AD                  (also what it always appears to be from my diagram.)

 

then, by extension, you can prove what you are being asked.

 

Here is the interactive diagram that you can play with.

Only point A is fixed, the rest of the diagram can be dragged wherever you want. The stated requirements will stay correct.

https://www.geogebra.org/classic/hpyqwy5r

 Aug 25, 2020
edited by Melody  Sep 11, 2020
 #5
avatar+118687 
+1

Maybe someone else would like to continue from where I have left off ?

 Aug 25, 2020
 #6
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0

Attn: Gwenspooner85

You said on your other post that you have now figured out this question.

Can you please put your answer here so other people (namely me) can learn as well.

Thank you.

 Aug 26, 2020
 #7
avatar+781 
+2

I redacted my solution.

gwenspooner85  Aug 26, 2020
edited by gwenspooner85  Sep 10, 2020
 #8
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0

Thanks Gwen,

It is such a pity that your pics have been blocked.

It is not your fault, I know this.    sad

Melody  Aug 27, 2020
 #9
avatar+781 
+1

Oh...why are they blocked though? I could still type up the solution here though, but there won't be any diagrams.

gwenspooner85  Aug 27, 2020
 #10
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I do not know why it is blocked, there are a lot of bugs lately.

I have sent you a private message.

Melody  Aug 27, 2020
 #12
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0

Thanks very much guest.   cool   laugh  cool

 Aug 27, 2020
 #13
avatar+2511 
+3

Tá céad fáilte romhat You are very welcome

GingerAle  Aug 27, 2020
edited by Guest  Aug 27, 2020
 #14
avatar+781 
+2

How was the guest able to post them though?

gwenspooner85  Aug 27, 2020
 #15
avatar+1094 
+2

Just telling you, guest did not go into your computer and steal all those files.

 

What he/she did was long press (if on mobile) the image, or left click (if on PC or MAC) and select "copy image". This will give the link to the image, which gives the picture.

 

I think the reason why guest's pictures were not blocked (or sent for moderation) was because they were the web2.0calc links, which the system probably recognizes as 'innocent'. Maybe.

 

Have a good day Gwen!

 

:)

ilorty  Aug 27, 2020
 #16
avatar+118687 
0

Thanks Ginger :)     And thanks ilorty too   wink

Melody  Aug 27, 2020
edited by Melody  Aug 27, 2020

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