From property of midpoints we have:
(X1Y1Z1) = (X1Y1Y2) - (Y1Y2Z1) = 1/2*(X1X2Y2) - (Y1Y2Z1) = 1/4*(X1X2X3) - (Y1Y2Z1)
Similarly we can show four equalities:
(X1Y1Z1) = 1/4*(X1X2X3) - (Y1Y2Z1)
(X2Y2Z2) = 1/4*(X2X3X4) - (Y2Y3Z2)
(X3Y3Z3) = 1/4*(X3X4X1) - (Y3Y4Z3)
(X4Y4Z4) = 1/4*(X4X1X2) - (Y4Y1Z4)
Sum these four equalities we have:
(X1Y1Z1) + (X2Y2Z2) + (X3Y3Z3) + (X4Y4Z4) = 1/2*(X1X2X3X4) - (Y1Y2Z1) - (Y2Y3Z2) - (Y3Y4Z3) - (Y4Y1Z4) (1)
Moreover Y1Y2Y3Y4 is Varignon parallelogram of X1X2X3X4 so:
(Z1Z2Z3Z4) = (Y1Y2Y3Y4) - (Y1Y2Z1) - (Y2Y3Z2) - (Y3Y4Z3) - (Y4Y1Z4)
= 1/2*(X1X2X3X4) - (Y1Y2Z1) - (Y2Y3Z2) - (Y3Y4Z3) - (Y4Y1Z4) (2)
From (1) and (2) we get the result:
(X1Y1Z1) + (X2Y2Z2) + (X3Y3Z3) + (X4Y4Z4) = (Z1Z2Z3Z4)

In fact, I want to evaluate the ratio R = (X1X2X3X4)/(Z1Z2Z3Z4) but it is not constant. By my observation, R >= 5 and very close to 5. When X1X2X3X4 is parallelogram then easy to show that R = 5 but I still not find the best evaluation in general case.

Best regards,
Bui Quang Tuan

Yes quite.

>In fact, I want to evaluate the ratio R =
>(X1X2X3X4)/(Z1Z2Z3Z4)

For the parallelogram, this is indeed simple. For the rest, I've no notion.

I am leaving in a couple of hours on a family trip. Shall be back by 11 Nov.

I don't know why, but the message I submitted a few moments ago is missing all the +(plus) signs !!
They were in the message I sent, as in the line above ?!

Regards,
Peter Scales

It's happened again.

My second message had the plus sign immediately before the (plus).

Regards,
Peter Scales

Dear Alex and Tuan,

Here and elsewhere Tuan has challenged us to find an expression for the area of the central quadrilateral (q) in relation to the area of a Vertex Middle Next Side convex Quadrilateral (Q). This is a nice problem, but it can become messy unless Q is defined in an appropriate manner. Here is an approach which simplifies the area calculation and makes it easy to understand the essence of the solution.

Forgive me if I change the notation, but there are many points to consider. But first some preliminary comments:

Intuitively Q/q attains an extreme value when it is most symmetrical (square) and an opposite extreme value when it is least symmetrical (triangular). In fact 5 <= Q/q <= 6 and this can be demonstrated graphically.

Let Quadrilateral be a square ABCD (with CD horizontal), with P,Q,R,S the respective midpoints of BC, CD, DA, AB.
PQRS is the Varignon parallelogram of ABCD.
Let AP/\DS = T, BQ/\AP = U, CR/\BQ = V, DS/\CR = W
Let BBp//AP, CCp//BQ, DDp//CR, AAp//DS and
let AP/\CCp = Cp, BQ/\DDp = Dp, CR/\AAp = Ap, DS/\BBp = Bp.
Denote by (ABCD) the area of polygon ABCD.

Then (AST)=(ARAp), (BPU)=(BSBp), (CQV)=(DQDp) and (DRW)= (RAAp).
So (ABCD) = 5*(TUVW) or Q = 5*q.
This could be called a proof by dissection.

And exactly the same proof applies to any parallelogram obtained from the square by stretching and/or shearing in the horizontal or vertical direction.
In math terms the square and parallelogram are affine equivalent.

Now transform the square, by allowing A to approach D (for example) while maintaining the above connectivities.
Then AP and BQ become medians, CR coincides with CD (and CA), while CS would be the third median, concurring with AP and BQ.

Q and q are both now triangles, and from the median properties of a triangle Q/q = 6.

Now for the proof:

Let ABCD be a convex Quadrilateral, with diagonals AC and BD intersecting at X, at angle theta.
Let P,Q,R,S and T,U,V,W be as defined above.
Let AX = a, BX = b, CX = c, DX = d.

Since we are dealing only with areas, there is no loss of generality in selecting theta = Pi/2, as Cavalieri's Principle or affine equivalence indicates.
For convenience we could use Cartesian Coordinates with Origin at X.
Then coords would be A(0,a), B(-b,0), C(0,-c), D(d,0).

Consider first the Quadrilateral ABCD with d=b and a not=c.
Let AP/\BD,= E, BQ/\AC = F, CR/\BD = G, DS/\AC =H.
Now find (EGWT) = (EDT) - (GDW):
Let BD/\SP = Bx, BD/\QR = Dx. Call BxE = t, EX = u, DxG = v, GX = w.
Triangles BxPE, PqAP and XAE are similar, yielding
t = b.c/(2.(c+2a)) and u = a.b/(c+2a).
Triangles DxRG, RsCR and XCG are similar yielding
v = a.b/(2.(a+2c)) and w = c.b/(a+2c).
Let SR/\AP = Sr, SR/\AC = Rs
Triangles STSr and DTE are similar.
Let perpendicular from T to ED be = k1.
Triangles SWR and DWG are similar.
Let perpendicular from W to GD be = k2.
Then k1 = a(u+b)/(2t+3u+2b) and k2 = a(2v+b)/(2(3v+2b))

Then (EGWT) = qa = k1(b+u)/2 - k2(v+b/2)/2.

Similarly find (GEUV) = (GBV) - (EBU):
Let perpendicular from V to GB be = j1.
Let perpendicular from U to EB be = j2.
Then j1 = c(w+b)/(2v+3w+2b) and j2 = c(2t+b)/(2(3b+2t))
Then (GEUV) = qc = j1(b+w)/2 - j2(t+b/2)/2.

Now Q = b(a+c) and q = qa + qc.

And after a deal of algebraic manipulation and simplification...

So Q/q = 4(3a+7c)(2c+3a)(3c+7a)(2a+3c)/(99.a^4+504.ca^3+794.c^2a^2+504.c^3a+99.c^4)

Note that Q/q is symmetrical in c and a.
Differential(Q/q) = 0 when c = a, and higher differentials confirm that Q/q is a minimum at c = a.

Then, at c = a, Q/q(min) = 4.10.5.10.5/2000 = 5

Consideration of b and d when c = a is identical.

So the conclusion is:
Q/q is a minimum of 5 when c = a and d = b. QED.

And these are exactly the conditions for ABCD to be a parallelogram.

Also note:
If any two adjacent values (a & b, b & c, c & d, or d & a) then the Quadrilateral degenerates into a Triangle and Q/q = 6

Another case:
If any one of a,b,c or d = 0 then Q/q = 56/11.

Other values can be explored for interest.

Regards,
Peter Scales

This ratio is always 5 no matter the shape of the quadrilateral, since any quadrilateral can be obtain by affine transformation from an square and if we construct the inner square it has 1/5 of the area of the exterior square, beside its area being equal to the sum of the four triangles.

And by the properties of the affine transformation see

https://www.cut-the-knot.org/triangle/pythpar/Geometries.shtml

the it is the obvious that the area ratio must be preserved

mpdlc

If I exactly understand you then your result is:
S(PQRS)/S(ABCD) = 1/5 for any convex quadrilateral ABCD.

But by my opinion it is may be not true. I take one limit case as in the image here. In this image, I choose D as near A and near segment AC as possible but keep triangle ABC fixed.

By this way, R can be as near midpoint Ro of AC as possible, S and P can be near A as possible and Q can be near Qo (intersection of BRo and AQ) as possible. It means that Area(PQRS) can be as near Area(AQoRo) as possible. Please note that triangle ABC still is not changed, therefore:

Area(PQRS)/Area(ABCD) can be near Area(AQoRo)/Area(ABC) and it can be near 1/6 as possible.

I think that the fact:
"Any convex quadrilateral can be obtained from a square by an affine transformation"
is not very clear and it is must be proved.

I am waiting for your opinion!
Thank you and best regards,
Bui Quang Tuan

If, a, b, c, d are the segments drawn from A, B, C, D and

P = a^b,
Q = b^c,
R = c^d,
S = d^a,

then, assuming A and D coincide, A, D, P, S are one and the same point, Q is the midpoint of AB and R is the center of ΔABC. The quadrilateral PQRS degenerates into ΔARQ whose area is indeed 1/6 of ΔABC.

So, what does that mean? It is not necessarily true that any quadrilateral is an affine image of a square. Which ones are?

It is not to hard to prove some following ratios:
ED/SD = 3
DE/PE = 4
AF/QF = 9
BG/RG = 7

From these, if BC = 2 then:
Area(RGC) = 1/7
Area(SHD) = 1/3
Area(PEA) = 1/2
Area(QFB) = 2/9

By our well known fact Area(PQRS) is sum of these four areas
Therefore Area(PQRS) = 151/126
Area(ABCD)/Area(PQRS) = 6/(151/126) = 5 1/151 > 5

So I think that the fact:
"Any convex quadrilateral can be obtained from a square by an affine transformation"
is not true.

Best regards,
Bui Quang Tuan

You are right an affine transformation does not permit get an arbitrary quadrilateral from a square, it must be a projective transformation but the metric properties of middle point of a segment is lost so it becames useless for our purpose to start with an square and use a mapping later.

Thank you

mpdlc