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Given nine points connected as described by the black lines. G the middle of AC, I the middle of ED, and H middle of BF.

By adding segments connecting points in the drawing, create shapes with area 1/4 of the quadrilateral EFDB. As example - by adding segments EH and DH, two quadrilateral with area 1/2 of EFDB are created: EFDH and EHDB

At least 6 such shapes are known to me

enter image description here

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  • $\begingroup$ Are D and E essentially random or arbitrary points on BC and AB? $\endgroup$ – greenturtle3141 Jun 6 '19 at 2:17
  • $\begingroup$ They are arbitrary. EFBD is any quadrilateral. Let us keep it convex. $\endgroup$ – Moti Jun 6 '19 at 2:31
  • $\begingroup$ For the final answer, do you require one such shape or all six? $\endgroup$ – Brandon_J Jun 6 '19 at 2:35
  • $\begingroup$ @Brandon_J I'd imagine at least 6, since 2 are given and 2 more are relatively trivial. (Edit correction: the example technically isn't a given, whoops) $\endgroup$ – greenturtle3141 Jun 6 '19 at 2:37
  • $\begingroup$ @greenturtle3141 goodness, I'm tired - I didn't even see the given ones. Thanks. $\endgroup$ – Brandon_J Jun 6 '19 at 2:37
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Well, I have more...

HIFE
DFIH
DHIB
BIHE
HIA
HCI

So here's the thing... given your choice in actual point locations, either DHIB or BIHE is going to be a self-intersecting shape. As drawn in the question, BIHE is that shape. According to certain math, that's not a problem, but it doesn't seem acceptable to this puzzle. That means I'm really at 5 shapes total, and notably I never used point G. I can't think of how to use it either... it is collinear with HI which seems to hamper its usefulness.

My attempt as some kind of proof:

For any triangle, if you draw a line from one of the corners to the midpoint of the edge opposite that corner, you will always bisect the triangle... exactly in half. So triangle BFE is cut perfectly in half when we add line segment EH. That's why adding segments EH and HD divide BDFE in half, because it it really two separate triangles being bisected. Because point I is still a midpoint of HDFE, we are able to divide that quadrilateral in half using the same technique, thus making the required 1/4 area.

We can change our order to get the other two "trivial" solutions. We start with BDFE, but this time start by cutting it in half using segments BI and IF. The first half, BDFI, can be cut in half just like before, which you might expect would add two answers to our list... but it actually gives us a repeat. We cannot count HDFI twice, so we really just have 3.

So here's where it gets weird, you might notice that when we divide BDFE into BDFI and BIFE, we can divide BDFI into halves but we cannot divide BIFE... except we sort of can. If you follow the dividing rules, we can make a self-intersecting shape BIHE. The order of letters is important now, so note how the line segments cross themselves. According to this math, which I vaguely remember learning about over a decade ago, the area of a self-intersecting shape is positive OR negative depending on traveling CW versus CCW. I had to use CAD to check this, but indeed the area checks out if you subtract the small triangle from the big one.

enter image description here

So alternately, instead of that weirdness, you could imagine the points being situated a little different such that BIHE is normal looking and BDFI becomes the self-intersecting shape. I don't know why that makes it better, it's just worth noting.

To make the other two answers, I guess I got lucky. I gathered, intuitively, that segment HI was important and thought that maybe connecting it to point A or C might do something useful. It turned out to be the exact thing I needed and I didn't think about it too hard until now. I can't for the life of me figure out why it works. Oh well, it's not like I'm getting massive upvotes anyway.

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  • $\begingroup$ Your answer is correct! Do you have a proof for HIA and HCI? $\endgroup$ – Moti Jul 10 '19 at 19:50
  • $\begingroup$ @Moti I'll try to write up a proof for those tomorrow if I have time. I'm not sure I'm comfortable with subtractive area being part of the solution though... it kind of defies the meaning of "shape" in a casual setting like this. $\endgroup$ – Dark Thunder Jul 10 '19 at 19:59
  • $\begingroup$ I was just wondering, since there is an elegant short proof - about 4-6 lines. If you enjoy geometry as well as puzzling you might enjoy it. In any case you found 6. $\endgroup$ – Moti Jul 10 '19 at 20:22
  • $\begingroup$ @Moti well feel free to post it, because mine will not be elegant... I guarantee that. $\endgroup$ – Dark Thunder Jul 10 '19 at 20:25
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Partial answer. I assume it covers the trivial ones:

Quadrilaterals EHIF and HIFD are both 1/4 of the quadrilateral EFDB because they are 1/2 of quadrilateral EFDH.

Edited: Here is another one.

BIHD

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  • $\begingroup$ This is 1/3 of the solution. These are not the really challenging ones:) $\endgroup$ – Moti Jun 6 '19 at 3:41
  • $\begingroup$ I agree - these are the trivial ones. $\endgroup$ – ppgdev Jun 6 '19 at 3:49
  • $\begingroup$ There is another trivial and than some more DEMANDING $\endgroup$ – Moti Jun 7 '19 at 5:38

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