Atrisinājums

Apzīmējam nogriežņu garumus \(OF = x_1\), \(OG = x_2\), \(BF = CG = y_1\), \(AF = DG = y_2\) (skat. 12. att.).

Tā kā \(\sphericalangle BFO = \sphericalangle DGO = 90^{\circ}\) un \(\sphericalangle BOF = \sphericalangle DOG\), tad \(\triangle OFB \sim \triangle OGD\) pēc pazīmes \(\ell\ell\). Tātad \(\frac{x_1}{x_2} = \frac{y_1}{y_2} = k\) un \({\displaystyle \frac{S_{OFB}}{S_{OGD}} = k^2}\), no kā iegūstam, ka \(S_{OGD} = \frac{4}{k^2}\).

Apskatām trijstūru laukumu attiecību:

\[\frac{S_{OFB}}{S_{OFA}} = \frac{\frac{1}{2}OF \cdot BF}{\frac{1}{2}OF \cdot AF} = \frac{BF}{AF} = \frac{y_1}{y_2} = k\;\;\Rightarrow{}S_{OFA}=\frac{4}{k}.\]

Apskatām trijstūri \(AOD\) (sk. attēlu, kur \(OM\) ir trijstūra \(AOD\) augstums):

\[S_{ODA} = S_{OMA} + S_{OMD} = S_{OFA} + S_{OGD} = \frac{4}{k} + \frac{4}{k^2} = 63.\]

Atrisinot iegūto kvadrātvienādojumu \(63k^2 - 4k - 4 = 0\), iegūstam:

\[ k = \frac{4 \pm \sqrt{16 + 4 \cdot 4 \cdot 63}}{126} = \frac{4 \pm 32}{126};\]

\[k_1 = \frac{36}{126} = \frac{2}{7}\;\;\text{un}\;\;k_2 = -\frac{28}{126}\;\text{(neder)}.\]

Līdz ar to esam ieguvuši, ka \(S_{OFA} = \frac{4}{k} = 4 : \frac{2}{7} = 14\) un \(S_{OGD} = \frac{4}{k^2} = 4 : \left( \frac{2}{7} \right)^2 = 49\). Tā kā \(\triangle BOE \sim \triangle DOA\) pēc pazīmes \(\ell\ell\), tad \(\frac{S_{BOE}}{S_{DOA}} = \left( \frac{OB}{OD} \right)^2 = k^2\) un

\[S_{BOE} = S_{DOA} \cdot k^2 = 63 \cdot \frac{4}{49} = \frac{36}{7}.\]

Izmantojot laukuma īpašības, iegūstam * \(S_{ABD} = S_{BOF} + S_{OFA} + S_{DOA} = 4 + 14 + 63 = 81\); * \(S_{ABD} = S_{BCD} = \frac{1}{2}S_{ABCD}\) * \(S_{OECG} = S_{BCD} - S_{BOE} - S_{OGD} = 81 - \frac{36}{7} - 49 = 32 - \frac{36}{7} = \frac{188}{7} = 26\frac{6}{7}\).