《2026年通城学典全国中考试题分类精粹中考数学》

15. (2025·天津)在平面直角坐标系中，O为原点，等边三角形ABC的顶点A，B的坐标分别为$(0,2)$，$(0,-1)$，点C在第一象限，等边三角形EOF的顶点E的坐标为$(-\sqrt{3},0)$，顶点F在第二象限。
(1) 填空：如图①，点F的坐标为$\left( -\frac{\sqrt{3}}{2},\frac{3}{2} \right)$，点C的坐标为$\left( \frac{3\sqrt{3}}{2},\frac{1}{2} \right)$。
(2) 将等边三角形EOF沿水平方向向右平移，得到等边三角形$E'O'F'$，点E，O，F的对应点分别为$E'$，$O'$，$F'$。设$OO' = t$。
① 如图②，若边$E'F'$与边AB相交于点G，当$\triangle E'O'F'$与$\triangle ABC$重叠部分为四边形$OO'F'G$时，试用含有t的式子表示线段GA的长，并直接写出t的取值范围；
② 设平移后重叠部分的面积为S，当$\frac{3\sqrt{3}}{4} \leq t \leq \frac{3\sqrt{3}}{2}$时，求S的取值范围。

答案：

(1) $\left( -\frac{\sqrt{3}}{2},\frac{3}{2} \right)$ $\left( \frac{3\sqrt{3}}{2},\frac{1}{2} \right)$
(2) ① 由题意，知$\angle F'E'O' = \angle OEF = 60°$，$O'E' = OE = \sqrt{3}$。$\because OO' = t$，$\therefore OE' = O'E' - OO' = \sqrt{3} - t$。$\therefore OG = \tan 60° · OE' = \sqrt{3}(\sqrt{3} - t) = 3 - \sqrt{3}t$。$\therefore AG = OA - OG = 2 - 3 + \sqrt{3}t = \sqrt{3}t - 1$。当点$F'$落在$y$轴上时，$O$为$O'E'$的中点，$\therefore t = \frac{\sqrt{3}}{2}$。当点$E'$与点$O$重合时，$t = \sqrt{3}$。当$\triangle E'O'F'$与$\triangle ABC$重叠部分为四边形$OO'F'G$时，$\frac{\sqrt{3}}{2} ＜ t ＜ \sqrt{3}$ ② 当$\frac{3\sqrt{3}}{4} \leq t ＜ \sqrt{3}$时，重叠的部分为四边形$OO'F'G$。如图①，过点$F'$作$F'M \perp x$轴于点$M$。由
(1)和
(2)①，可知$F'M = \frac{3}{2}$，$OG = 3 - \sqrt{3}t$，$OE' = \sqrt{3} - t$，$\therefore S = S_{\triangle O'E'F'} - S_{\triangle OE'G} = \frac{1}{2} × \sqrt{3} × \frac{3}{2} - \frac{1}{2}(\sqrt{3} - t)(3 - \sqrt{3}t) = -\frac{\sqrt{3}}{2}(t - \sqrt{3})^2 + \frac{3\sqrt{3}}{4}$。$\because - \frac{\sqrt{3}}{2} ＜ 0$，$\therefore$当$t = \frac{3\sqrt{3}}{4}$时，$S$取得最小值，为$-\frac{\sqrt{3}}{2} × \left( \frac{3\sqrt{3}}{4} - \sqrt{3} \right)^2 + \frac{3\sqrt{3}}{4} = \frac{21\sqrt{3}}{32}$。易知$\frac{21\sqrt{3}}{32} \leq S ＜ \frac{3\sqrt{3}}{4}$。记$BC$与$x$轴的交点为$N$，当$t = \sqrt{3}$时，易知此时点$E'$与点$O$重合，点$O'$与点$N$重合，重叠的部分恰为$\triangle O'E'F'$，如图②。$\therefore S = \frac{1}{2} × \sqrt{3} × \frac{3}{2} = \frac{3\sqrt{3}}{4}$。当$\sqrt{3} ＜ t \leq \frac{3\sqrt{3}}{2}$时，显然$S$随着$t$的增大而减小，$\therefore$当$t = \frac{3\sqrt{3}}{2}$时，$S$有最小值。如图③，连接$CO'$，记$BC$交$x$轴于点$N$，交$F''O'$于点$Q$，$AC$与$F''O'$交于点$P$，与$E'F''$交于点$K$，其中$F''$为图②中点$F'$平移后的对应点，连接$F'F''$。此时易知$CO' \perp x$轴，重叠部分为五边形$E'KPQN$。$\because ON = OB · \tan 60° = \sqrt{3}$，$\therefore$易知$O'N = \frac{3\sqrt{3}}{2} - \sqrt{3} = \frac{\sqrt{3}}{2}$。$\because \angle CNO' = \angle BNO = 90° - \angle ABC = 30°$，$\angle E'O'F'' = 60°$，$\therefore \angle NQO' = 90°$。$O'Q = \frac{1}{2}O'N = \frac{\sqrt{3}}{4}$，$QN = \sqrt{3}O'Q = \frac{3}{4}$。$\therefore S_{\triangle ONQ} = \frac{1}{2} × \frac{\sqrt{3}}{4} × \frac{3}{4} = \frac{3\sqrt{3}}{32}$。$\because \angle ACB = 60°$，$\angle CQP = \angle NQO' = 90°$，$\therefore \angle F''PK = \angle CPQ = 180° - 30° - 60° = 90°$。由平移，可得$F'F'' = NO' = \frac{\sqrt{3}}{2}$，$F'F'' // NO'$，$\therefore \angle F''F'K = \angle O'E'F'' = 60°$。$\therefore \angle F''F'K = 30° = \angle F''PK$。$\therefore F''P = F'F'' = \frac{\sqrt{3}}{2}$。$\therefore$易得$F''K = \frac{\sqrt{3}}{4}$，$PK = \frac{3}{4}$。$\therefore S_{\triangle FKP} = \frac{1}{2} × \frac{\sqrt{3}}{4} × \frac{3}{4} = \frac{3\sqrt{3}}{32}$。$\therefore S = S_{\triangle O'E'F''} - S_{\triangle ONQ} - S_{\triangle FKP} = \frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{32} - \frac{3\sqrt{3}}{32} = \frac{9\sqrt{3}}{16}$。$\therefore$当$\sqrt{3} ＜ t \leq \frac{3\sqrt{3}}{2}$时，$\frac{9\sqrt{3}}{16} \leq S ＜ \frac{3\sqrt{3}}{4}$。综上所述，$S$的取值范围是$\frac{9\sqrt{3}}{16} \leq S \leq \frac{3\sqrt{3}}{4}$。

16. (2025·甘肃)已知四边形ABCD是正方形，E是边AD上一动点(点D除外)，$\triangle EFG$是直角三角形，$EG = EF$，点G在CD的延长线上。
(1) 如图①，当点E与点A重合，且点F在边BC上时，写出BF和DG之间的数量关系，并说明理由；
(2) 如图②，当点E与点A不重合，且点F在正方形ABCD内部时，FE的延长线与BA的延长线交于点P，如果$EF = EP$，写出AE和DG之间的数量关系，并说明理由；
(3) 如图③，在(2)的条件下，连接BF，写出BF和DG之间的数量关系，并说明理由。

答案：
(1) $BF = DG$ 理由：$\because$四边形$ABCD$是正方形，$\therefore AB = AD$，$\angle BAD = 90°$。$\because \triangle EFG$是直角三角形，$EG = EF$，$\therefore \angle FEG = 90°$。当点$E$与点$A$重合时，则$\angle FAG = 90° = \angle BAD$。$\therefore \angle DAG = \angle BAF = 90° - \angle DAF$。又$\because AD = AD$，$AF = AG$，$\therefore \triangle ABF \cong \triangle ADG$。$\therefore BF = DG$。
(2) $AE = DG$ 理由：$\because$四边形$ABCD$是正方形，$\therefore \angle ADC = \angle DAB = 90°$。$\because$点$G$在$CD$的延长线上，$FE$的延长线与$BA$的延长线交于点$P$，$\therefore \angle PAE = \angle EDG = 90°$。$\because \angle P + \angle AEP = 90°$。$\because \angle FEG = \angle DEF + \angle DEG = 90°$，$\angle AEP = \angle DEF$，$\therefore \angle P = \angle DEG$。$\because EG = EF$，$EP = EP$，$\therefore$在$\triangle PAE$和$\triangle EDG$中，$\begin{cases} \angle P = \angle DEG \\ EP = EG \end{cases}$，$\therefore \triangle PAE \cong \triangle EDG$。$\therefore AE = DG$。
(3) $BF = \sqrt{5}DG$ 理由：由
(2)，可知$\triangle PAE \cong \triangle EDG$，$\therefore AE = DG$，$AP = DE$。过点$F$作$FH \perp AB$于点$H$，则$\angle FHB = \angle FHA = 90° = \angle PAE$，$\therefore AE // FH$。$\therefore \frac{PA}{AH} = \frac{PE}{EF} = 1$。$\therefore PA = AH$。$\because PE = EF$，$\therefore AE$为$\triangle PHF$的中位线。$\therefore HF = 2AE$。$\because AP = DE$，$PA = AH$，$\therefore DE = AH$。又$\because AD = AB$，$\therefore$易得$AE = BH$。在$Rt \triangle BHF$中，由勾股定理，得$BF = \sqrt{HF^2 + BH^2} = \sqrt{5}AE$。$\because AE = DG$，$\therefore BF = \sqrt{5}DG$。