I tried to solve this chalenge, found a solution by guessing - which gives the answer 7*5=35, where the sub areas were 9*3=27, 6*8=48 and 8*4=32.

When trying to solve akgebricly, I found only 3 equations for 4 variables, which suggested infinite number of solutions.

So I built a geogebra file that proved it.

I will be happy to send it to any one who is intrested.

Amos, this work is wonderful and it's clear you spent a lot of time and energy thinking about this problem! I've enabled file sharing so that you can link your geogebra file! We'd love to see what you made.

Below is the original problem and Amos has answered the question: "

how many solutions exist?"I wonder if it's possible to use only one variable to describe every side length in this problem.

the geogebra file has extension ggb. It is very small (13kb) but the site cannot upload it.

In this file I used only one variable - the height of the rectangle whose area is 27.

I wonder if you can see it in the link

https://www.geogebra.org/m/napmqcvpThe link works! What's the smallest area for the unknown rectangle? What's the largest?

My answer depends on whether you require the unknown area and the yellowish area to have positive areas...

When changing parameter a by tenth of a unit, the maximum area is 46.56 (For a=3.9. For a=4 there is no solution. Maybe the area is a little bigger for 3.9<a<4) The minimum area is 3.65 (for a=2). (it probably goes to 0 for 1.9<a<2). (parameter a is the height of the rectangle whose area is 27).

My previous comment is not exact. It is based on my geogebra plan, which maybe is not complete. I will work on it and report again.

I have another area problem which I could not solve.

If some one succeeds to solve it, please ket me know.

Thank you for sharing Amos, this is an interesting problem. What steps did you take to solve it?

I found that the lower segments of the triangles with area 21 must be equal and the left segments of the triangles with areas 6 and 8 must be in relation 6:8. I tried to add segments and triangles but could not finalize the solution.