《2026年金考卷中考试题汇编45套数学山东专版》

22. (本小题满分11分)
(2025北京)在平面直角坐标系$xOy$中,抛物线$y = ax^{2}+bx + c(a\neq0)$经过点$O$和点$A(3,3a)$.
(1)求$c$的值,并用含$a$的式子表示$b$.
(2)过点$P(t,0)$作$x$轴的垂线,交抛物线于点$M$,交直线$y = ax$于点$N$.
①若$a = 1$,$t = 4$,求$MN$的长;
②已知在点$P$从点$O$运动到点$B(2a,0)$的过程中,$MN$的长随$OP$的长的增大而增大,求$a$的取值范围.

答案：
22
(1)将$(0,0)$代入$y = ax² + bx + c$，得$c = 0$. (2分)
将$(3,3a)$代入$y = ax² + bx$，得$9a + 3b = 3a,\therefore b = -2a$. (4分)
(2)①当$a = 1$时，抛物线及直线的解析式分别为$y = x² - 2x,y = x$.
由题意可知$x_M = x_N = t = 4$，
把$x = 4$代入$y = x² - 2x$，得$y = 8$，
把$x = 4$代入$y = x$，得$y = 4$，$\therefore MN = 8 - 4 = 4$. (7分)
②当$a ＞ 0$时，如图
(1)，结合图象分析可知，若要满足题意，则$y_N ＞ y_M$，
$MN = at - (at² - 2at)= -at² + 3at$，
其对应图象的对称轴为直线$x = \frac{-3a}{-2a}=\frac{3}{2}$
$\because$在$0\leq t\leq2a$的范围内，MN随$t$的增大而增大，
$\therefore2a\leq\frac{3}{2},\therefore a\leq\frac{3}{4},\therefore0 ＜ a\leq\frac{3}{4}$. (9分)
当$a ＜ 0$时，如图
(2)，此时由图易知点P从点O运动到点$B(2a,0)$的过程中，MN的长随OP的长的增大而增大，$\therefore a ＜ 0$符合题意.
综上，$a ＜ 0$或$0 ＜ a\leq\frac{3}{4}$. (11分)

23. (本小题满分11分)
(2025山西)综合与探究
问题情境:
如图(1),在$\triangle ABC$纸片中,$AB＞BC$,点$D$在边$AB$上,$AD＞BD$.沿过点$D$的直线折叠该纸片,使$DB$的对应线段$DB'$与$BC$平行,且折痕与边$BC$交于点$E$,得到$\triangle DB'E$,然后展平.
猜想证明:
(1)判断四边形$BDB'E$的形状,并说明理由.
拓展延伸:
(2)如图(2),继续沿过点$D$的直线折叠该纸片,使点$A$的对应点$A'$落在射线$DB'$上,且折痕与边$AC$交于点$F$,然后展平.连接$A'E$交边$AC$于点$G$,连接$A'F$.
①若$AD = 2BD$,判断$DE$与$A'E$的位置关系,并说明理由;
②若$\angle C = 90^{\circ}$,$AB = 15$,$BC = 9$,当$\triangle A'FG$是以$A'F$为腰的等腰三角形时,请直接写出$A'F$的长.

答案：
23
(1)四边形BDB'E是菱形. (1分)
理由：由折叠可知，$B'D = BD,B'E = BE,\angle B'DE = \angle BDE$.
∵$DB'// BC$，$\therefore \angle B'DE = \angle BED$，
$\therefore \angle BDE = \angle BED,\therefore BD = BE$，
$\therefore B'D = BD = BE = B'E,\therefore$四边形BDB'E是菱形. (4分)
(2)①$DE\perp A'E$. (5分)
理由：如图
(1)，由
(1)得$B'D = BD = B'E$.
$\because B'D = B'E,\therefore \angle1 = \angle2$.

由折叠可知，$A'D = AD$.
∵$AD = 2BD$，$\therefore A'D = 2B'D$，
$\therefore$点$B'$是$A'D$的中点，$\therefore B'D = B'A'$，
$\therefore B'A' = B'E,\therefore \angle3 = \angle4$.
在$\triangle A'DE$中，$\because \angle1 + \angle A'ED + \angle4 = 180°$，
$\therefore \angle1 + \angle2 + \angle3 + \angle4 = 180°,\therefore \angle2 + \angle3 = 90°$，
即$\angle DEA' = 90°,\therefore DE\perp A'E$. (9分)
②5或$\frac{165}{37}$. (11分)
解法提示：$\because \angle C = 90°,AB = 15,BC = 9,\therefore AC = \sqrt{AB² - BC²}=12,\therefore \tan A = \frac{3}{4},\sin A = \frac{3}{5},\cos A = \frac{4}{5}$.
设AC与DA'交于点M，$\because DB'// BC$，$\therefore \angle AMD = \angle C = 90°$.由折叠可知$A'F = AF,\angle FA'D = \angle A,\therefore \tan\angle FA'M = \frac{3}{4}$，
$\therefore$可设$FM = 3m,MA' = 4m$，则$AF = A'F = 5m$，
$\therefore AM = AF + FM = 8m,\therefore AD = \frac{AM}{\cos A}=10m$，
$\therefore BD = AB - AD = 15 - 10m$.
在菱形BDB'E中，$BE = BD = 15 - 10m,\therefore EC = BC - BE = 9 - (15 - 10m)=10m - 6$.
$\because DB'// BE,\therefore \triangle A'MG\backsim\triangle ECG$(点拨：“8”字型相似)，
$\therefore \frac{A'M}{EC}=\frac{MG}{CG}$.
$\triangle A'FG$是以$A'F$为腰的等腰三角形，可分$A'F = GF$和$A'F = A'G$两种情况讨论.
a.当$A'F = GF$时，如图
(2)，$MG = FG - FM = A'F - FM = 2m$，
$\therefore CG = AC - AM - MG = 12 - 10m - 2m = 12 - 12m$.
此时$\frac{4m}{10m - 6}=\frac{2m}{12 - 12m},\therefore m = 1,\therefore A'F = 5m = 5$.

b.当$A'F = A'G$时，如图
(3)，$A'M\perp FG$，则$GM = FM = 3m$，
$CG = AC - AM - MG = 12 - 11m$.
此时$\frac{4m}{10m - 6}=\frac{3m}{12 - 11m},\therefore m = \frac{33}{37},\therefore A'F = 5m=\frac{165}{37}$.

综上，$A'F$的长为5或$\frac{165}{37}$.