Let $AB = CD = a$ and $BC = AD = b$. Then we have:

$$\frac{AX}{XB} = k \implies \frac{a}{XB} = k+1 \implies XB = \frac{a}{k+1}$$

From this we obtain that:

$$AX = \frac{ka}{k+1}$$

We get simular results for all other segments. Now check that $XYZV$ is inscribed in $ABCD$ and we have:

$$P_{ABCD} - P_{XBY} - P_{YCZ} - P_{ZDV} - P_{VAX} = P_{XYZV}$$

Now using the condition and the formula for area of a triangle we have:

$$P_{ABCD} - \frac{XB\cdot YB \cdot \sin \angle XBY}{2} - \frac{YC\cdot ZC \cdot \sin \angle YCZ}{2} - \frac{ZD\cdot VD \cdot \sin \angle ZDV}{2} - \frac{VA\cdot XA \cdot \sin \angle VAX}{2} = \frac{86P_{ABCD}}{100}$$

Substitute for all the segments and the recall the formula for area of a paralelogram: $P = ab \sin \alpha$ and you'll get:

$$P_{ABCD} -\frac{2kP_{ABCD}}{(k+1)^2} = \frac{86P_{ABCD}}{100}$$

Divide by $P_{ABCD} \not = 0$ and you'll get:

$$\frac{7}{50} = \frac{2k}{(k+1)^2}$$
$$7(k+1)^2 = 100k$$
$$7k^2-86k + 7 = 0$$

Solbing it you'll get: $k_1 = \frac 17 \left(43 - 30\sqrt{2}\right)$ and $k_2 = \frac 17 \left(43 + 30\sqrt{2}\right)$. Since $k_2 &gt;2$ the only solution is $k_1$

The $ABCD$ paralelograms sides are $AB,BC,CD,DA$. On these line segments there are points in the same order: $X,Y,Z,V$.
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