4. Zadatak

Neka su $\alpha$, $\beta$ i $\gamma$ redom mjere unutranjih kutova trokuta $ABC$ pri vrhovima $A$, $B$ i $C$, te neka je

$$\alpha'=|\angle NAE|,\qquad \beta'=|\angle FBP| \qquad\text{i}\qquad \gamma'=|\angle QCM|.$$

Vrijedi

$$\alpha'=360^\circ-90^\circ-90^\circ-\alpha=180^\circ-\alpha$$

i analogno

$$\beta'=180^\circ-\beta,\qquad \gamma'=180^\circ-\gamma.$$

Površina trokuta $ABC$ se može izraziti na sljedeće načine:

$$P=\frac{ab\sin\gamma}{2} =\frac{bc\sin\alpha}{2} =\frac{ac\sin\beta}{2}.$$

Slijedi

$$P_{BFP} =\frac{ac\sin\beta'}2 =\frac{ac\sin(180^\circ-\beta)}2 =\frac{ac\sin\beta}2 =P,$$

i analogno

$$P_{MCQ}=\frac{ab\sin\gamma'}2=P, \qquad P_{EAN}=\frac{bc\sin\alpha'}2=P.$$

Vrijedi

$$P=\frac{ab\sin\gamma}{2} =\frac{5\cdot6}{2}\cdot\frac{\sqrt2}{2} =\frac{15\sqrt2}{2}.$$

Prema poučku o kosinusu vrijedi

$$c^2 =a^2+b^2-2ab\cos\gamma =6^2+5^2+2\cdot6\cdot5\cdot\frac{\sqrt2}{2} =61+30\sqrt2.$$

Površina šesterokuta $EFPQMN$ iznosi

$$\begin{aligned} P_s &=4P+a^2+b^2+c^2\\ &=4\cdot\frac{15\sqrt2}{2}+36+25+61+30\sqrt2\\ &=122+60\sqrt2. \end{aligned}$$