Another way would be to split the region outside the octagon into 8 triangles: 4 with area 1/12 of ABCD and 4 with area 1/8 of ABCD.

The first 4 are

MBX₁, NCX₃, PDX₅, QAX₇.

The other 4 are

BNX₂, CPX₄, DQX₆, AMX₈.

1 - 4(1/8 + 1/12) = 1 - 4·5/24 = 1 - 5/6 = 1/6.

Consider the region J formed by intersecting triangles APB and CMD. Tile the plane with the entire parallelorgram figure, and observe that J tiles across ABCD 3 times, as does region K, formed similarly to J but sideways. The region H formed by intersecting triangles PAN and QCM tiles 4 times as does the similarly formed region I.

It is obvious that triangle ABQ and the 7 others of this form each have 1/4 the area of ABCD, as do regions Now consider the area formed of these 8 triangles, together with H, I, J, and K. These areas together cover every part of ABCD 3 times, but the octagon, and only the octagon, 4 times. Hence their area is 3a + o. But considering their sizes, it is also 8(a/4) + 2(a/4) + 2(a/3) = 19a/6. Therefore, o = a/6.

This is perhaps not the shortest way to solve the problem, but the tiling is very pretty. I seem to recognize in it a standard Arabic decorative motif.

--Stuart Anderson

These regions are hexagons. Did you mean here the parallelograms AX₃CX₇ and BX₁DX₅?

I had meant to say "intersecting angles" rather than "intersecting triangles." Perhaps clearer would have been "intersecting regions embraced by angles." In any event, I did mean the parallelograms you mentioned.

Clearly.

I got an applet here

https://www.cut-the-knot.org/Curriculum/Geometry/OctagonByOverlap.shtml

A very nice solution.

Thank you! The page is very nice looking and clearly shows the grids and overlapping areas. This is pretty much exactly as I visualized it, but prettier.

A couple of small notes: I spell my last name with "son" not "sen" (on both pages of solutions). Also, in the applet, when you click on "shape 3" you get the shape described in the text below as shape 4, and vice versa.

Looking back at my post, I see also that you have corrected an error I made there: I erroneously switched which parallelograms tiled 3 times and which ones tiled 4 times. Of course you understood my intention despite this error; thank you for correcting it. I mention it here for the benefit of anyone who is reading this thread and may not have seen the applet pages yet.

--Stuart Anderson

I apologize. One was a misprint, the other the result of economizing with copy&paste. I am sorry.

>Also, in the
>applet, when you click on "shape 3" you get the shape
>described in the text below as shape 4, and vice versa.

Thank you. Everything is fixed now.

Sorry for typo mistakes here: all "rectangle" should be corrected truly as "parallelogram"

Best regards,
Bui Quang Tuan

I have found two another proofs:

DIVIDE AND CONQUER PROOF
Because symmetries of this configuration so some facts can be proved similarly and we can use tactic "divide and conquer" to prove this problem as following:
Step 1: divide ABCD into four area equal parts by MP, NQ then
Area(X1X2X3X4X5X6X7X8)/Area(ABCD) = 1/6 <=> Area(X1X2OX8)/Area(AMOQ) = 1/6
Step 2: divide AMOQ into two area equal parts by AO then
Area(X1X2OX8)/Area(AMOQ) = 1/6 <=> Area(X1X2O)/Area(AMO) = 1/6
Step 3: divide AMO into two area equal parts by AX2 then
Area(X1X2O)/Area(AMO) = 1/6 <=> Area(X1X2O)/Area(AX2O) = 1/3
It is true because X1 is centroid of ABD and X1O/AO = 1/3

GENERALIZED FOR CONVEX QUADRILATERAL PROOF
Another proof can be used also in generalized fact for any convex quadrilateral as following:
ABCD is covex quadrilateral. M, N, P, Q are midpoints of sides AB, BC, CD, DA respectively.
O is intersection of MP, NQ
X2, X4, X6, X8 are midpoints of segments OM, ON, OP, OQ respectively.
X1 = QX2 /\ MX8
X3 = MX4 /\ NX2
X5 = NX6 /\ PX4
X7 = PX8 /\ QX6
(Or X1, X3, X5, X7 are centroids of QOM, MON, NOP, POQ respectively)
then Area(X1X2X3X4X5X6X7X8) = 1/6*Area(ABCD)

The proof is based on two very simple and wellknown facts:
1. In a triangle two medians bound with two their sides one quadrilateral with area = 1/3 area of the triangle. (or: three medians divide the triangle into six equal area triangles).
2. In a convex quadrilateral midpoints of four sides bound one parallelogram with area = 1/2 area of the quadrilateral.
From these facts:
Area(X1X2X3X4X5X6X7X8) = 1/3*Area(MNPQ)
Area(MNPQ) = 1/2*Area(ABCD)
therefore:
Area(X1X2X3X4X5X6X7X8) = 1/3*1/2*Area(ABCD) = 1/6* Area(ABCD)

Best regards,
Bui Quang Tuan

I have upgraded my page.

I found that problem in a 2001 Mathematics Teacher article.

It can be found here

https://www.cut-the-knot.org/temp/Octagon.pdf

>ratio of two perimeters:
>rp = Perimeter(ABCD)/Perimeter(X1X2X3X4X5X6X7X8)

In the article there are tables upon tables of coordinates of the vertices of the octagon and the equations of the side lines. Too horrible to contemplate but perhaps useful in computing the perimeter.

>Of course this ratio is not constant but by my observation
>(still not proved):
>2 < rp < 3

P(X1X3X5X7) < P(X1X2X3X4X5X6X7X8) < P(X1X3X5X7)*3/2

P(X1X3X5X7) = P(ABCD)/3

1/3 < P(X1X2X3X4X5X6X7X8)/P(ABCD) < 1/2

3 > P(ABCD) / P(X1X2X3X4X5X6X7X8) > 2

>If this fact is true then we can see some nice expressions
>such as:
>6 = 2*3
>6 = (3+2)+(3-2)

How did you do that?