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I am sophomore in pure mathematics and at my university they are teaching kids who will participate in the local olympiad. One day I saw this question written in one of the classrooms:
Let [math] \square ABCD[/math] be a quadrilateral. Let [math] E [/math] be the midpoint of [math] \overline{\rm BC} [/math] and [math] F [/math] be the midpoint of [math] \overline{\rm DC} [/math]. If the lines connecting [math]A[/math] and these midpoints trisect the diagonal [math] \overline{\rm DB} [/math] then prove the quadrilateral is a parallelogram.
I wrote it down thinking it was a trivial kids problem and then started trying it to prove it by the usual tricks. Construct some parallel lines. some similar triangles and boom. But no. I HAVE BEEN DOING THIS FOR DAYS. Until now. This is my proof:
Consider the arbitrary points [math]B,C,D[/math] in space such that they make a triangle. Let the midpoint of [math] \overline{\rm BC} [/math] be [math] E [/math] and the midpoint of [math] \overline{\rm CD} [/math] be [math] F [/math]
By construction there exists a point A such that [math] \square ABCD[/math] is parallelogram and if it is a parallelogram then it can be proven that the lines [math] \overline{AE} [/math] and [math] \overline{AF} [/math] trisect the line [math] \overline{BD} [/math]
(this fact can be proven almost directly).
Lets also name [math]X[/math] as the point where [math] \overline{AE} [/math] intersects [math]\overline{BD}[/math] and [math]Y[/math] as the point where [math] \overline{AF} [/math] intersects [math]\overline{BD}[/math]
As [math]B,C,D[/math] are arbitrary, the original statement is equivalent to saying:
The only point [math]A[/math] such that [math]\overline{AE}[/math] and [math]\overline{AF}[/math] trisect the diagonal is the point [math]A[/math] such that the quadrilateral is a parallelogram.
(cont)
Fuckng word limit.
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>>8632410
So now lets assume the contrary to this new equivalent statement. Suppose there exists another point [math]A' \neq A[/math] such that this statement is true for it.
That means that [math]\overline{A'E}[/math] passes through X or Y and that [math]\overline{A'F}[/math] passes through X or Y.
Lets assume that [math]\overline{A'E}[/math] passes through X and the other through Y. Then we get the two equalities:
[math]\overline{A'E} = \overline{XE} =\overline{AE} [/math] and [math]\overline{A'F} = \overline{YF} =\overline{AF} [/math] but this implies that [math]A' = A[/math], contradicting our hypothesis. If we consider the reverse case then we reach the same contradiction. Therefore this negation is contradictory and thus the proposition is true.
Holy fuck. Now that I am not completely a brainlet... can someone prove the original statement using only normal geometry? No fancy logic. Just by connecting some lines, proving some congruences, etc. Please.
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>>8632411
going from your picture (and not what you wrote)
if DF = FE = EB and DF+FE+EB=DB then necessarily they must all be trisections of DB
if you place a point X on BD such that AX is perpendicular to BD, then the triangles DXA, AFE and AXB will all have the same area
maybe that points you in the right direction; I'd try to be of more help but I'm too drunk
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>>8632530
>going from your picture (and not what you wrote)
The labelings in my picture and from a similar stack overflow problem I found when googling a proof for my problem. It describes the same situation.
>then necessarily they must all be trisections of DB
Yeah. For my theorem this is assumed. What needs to be proven is that this implies that quadrilateral is a parallelogram.
>if you place a point X on BD such that AX is perpendicular to BD, then the triangles DXA, AFE and AXB will all have the same area
Why?
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>>8632554
because the area of a triangle is the length of any side multiplied by the altitude from that side
and the altitude from a side is just the line perpendicular to that side through the third point
also i made a mistake; i should have said that the area of ADF = AFE = AEB
basically they're all triangles with the same base and the same height
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>>8632573
>basically they're all triangles with the same base and the same height
I don't think this is well justified.
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i might be drunk but i still know that these three shaded regions have the same area
its because the red line and the green line are perpendicular and their products are the same
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>>8632618
>i might be drunk but i still know that these three shaded regions have the same area
Yeah but that is not the point. This is why having diagrams in geometry is really deceptive. Of course those triangles have equal area but that is because the drawing is already a parallelogram.
You can't assume anything about the shape of the quadrilateral to then prove what the shape of the quadrilateral is. That is bad logic.
The idea is that you are given a quadrilateral with some properties and you have to prove that those properties actually imply that it is a parallelogram.
A more appropiate diagram would be one with the points B, C, D and the point A being a question mark, as what you want to prove is similar to where is that point A at?
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>>8632635
you know the more i look at this problem the more im feeling you should just resort to brute-forcing the problem with linear algebra
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>>8632673
>you should just resort to brute-forcing the problem with linear algebra
I know you can probably do this with linear algebra by turning all segments into vectors and then doing some magic but I think that is also problematic.
I mean, this is a kids problem. Meant for high school kids. You are not supposed to try the more enlightened methods. I was able to prove it using only geometry but also doing a lot of logical trickery. I am interested in a pure kiddy friendly geometry proof that Euclid would be proud of.
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>>8632687
in non-American high schools the students would just do the linear algebra, but whatever
I think your logical proof by contradiction is the closest you're going to get
Euclid isn't rolling in his grave over it
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