20. In $\triangle A B C$, points $P$, $Q$, and $R$ on sides $B C$, $C A$, and $A B$ respectively divide each side in the ratio $t:(1-t)$. The area of the triangle formed by the lengths of segments $A P$, $B Q$, and $C R$ is denoted as $K$, and the area of $\triangle A B C$ is denoted as $L$. Find $\frac{K}{L}$ (expressed in terms of $t$).
20. Draw lines through $A$ and $C$ parallel to $RC$ and $RA$, respectively, intersecting at point $D$. Then $ARCD$ is a parallelogram. Connect $DP$ and $DQ$. Then $\frac{CD}{CQ} = \frac{AR}{CQ} = \frac{tAB}{tAC} = \frac{AB}{AC}$, and $\angle DCQ = \angle BAC$. Therefore, $\triangle CDQ \sim \triangle ABC$. Hence, $DQ = \frac{BC}{AC} CQ = tBC = BP$, and $\angle CQD = \angle ACB$. Thus, $BPCQ$ is a parallelogram, so $PD = BQ$. $\triangle APD$ is a triangle with sides $AP$, $BQ$, and $CR$. $K = S_{\triangle APD} = S_{\triangle ABC} + S_{\triangle ACD} - S_{\triangle ABP} - S_{\triangle CPD}$,
and $S_{\triangle ACD} = S_{\triangle ACR} = \frac{AR}{AB} \cdot S_{\triangle ABC} = tL$, similarly $S_{\triangle APB} = tL$,
$$
S_{\triangle CPD} = \frac{\frac{1}{2} CD \cdot CP \sin \angle PCD}{\frac{1}{2} AB \cdot AC \sin \angle ABC} S_{\triangle ABC} = t(1-t) L \text{, so, } \frac{K}{L} = 1 - t + t^2 \text{. }
$$