Finding the area of the 4th triangle, given the areas of the other 3, and all the 4 form a rectangle

Let BC = 1 unit. Then, $4 = y = [\triangle CBQ] = \frac {1.QC}{2}$ yields QC = 8 units

Let DQ = s units. Then AB = s + 8 units

$5 = z = [\triangle ABP] = \frac {(s+8).AP}{2}$ yields $AP = \frac {10}{s + 8}$ units

$DP = 1 – AP = … = \frac {s – 2}{s + 8}$

Similarly, $3 = (\frac {1}{2}) {s}{\frac{ s – 2}{s + 8}}$.

The above result is a solvable quadratic with roots $s = 12$ or $–4$ (rejected)

Therefore, the rectangle occupies 20 square units and x, y, z occupy …. Giving f = ...

Let BC = h and AB = k

QC = 8/h (since area BCQ = 4)

AP = 10/k (since area ABP = 5)

(h - 10/k)(k - 8/h) = 6 (since area DPQ = 3)

Now: k = 20/h or k = 4/h (h <> 0)

Since the area of the rectangle is k*h and larger than 12 we have to disregard the second solution.

Using the first solution we get 20 as the area of the rectangle and f = 8.

Since there only is one solution, I'd call the problem complete.

Fantastic problem! Thanks for sharing. I solved it with Euler Math Toolbox (see below). However, I'd like to see a more geometric solution. Probably, I am just blind right now.

The method of the solution is straightforward. $xh$ is $x$ depending on $a$ and $b$ (computing the area as side times side divided by 2). $fh$ is $f$ depending on $a$, $b$, $z$ and $y$. Then, surprisingly, we can get rid of the total area $a*b=x+y+z+f$

>xh &= (b-2*z/a)*(a-2*y/b)/2

$${ (a - {{2y} \over b} ) (b - {{2z} \over a} )} \over 2$$

>fh &= expand(a*b-(xh+y+z))

$$ {{ab} \over 2} - {{2yz} \over {ab}}$$

>fhh &= subst(x+y+z+f,a*b,fh)

$$ {{z + y + x + f} \over 2} - {{ 2yz} \over {z + y + x + f}} $$

>&solve(f=fhh,f), function f(x,y,z) &= f with %[2]

$$ [ {f = -\sqrt {z^2 - 2yz + 2xz + y^2 + 2xy + x^2}, f = \sqrt {z^2 - 2yz + 2xz + y^2 + 2xy + x^2}} ] $$

$$ \sqrt {x^2 - 2yz +2xz + y^2 + 2xy + x^2} $$

>f(3,4,5)

$$ 8 $$