《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10分)
【折一折】
将边长为2的正方形纸片ABCD折叠,使边AB,AD都落在对角线AC上,展开得折痕AE,AF,连接EF,如图(1).
(1) ∠EAF =45°;点A到EF的距离是2.
【转一转】
(2) 将图(1)中的∠EAF绕点A旋转,使它的两边分别交边BC,CD于点P,Q,连接PQ,如图(2),点A到PQ的距离是否发生变化? 请说明理由.
【探一探】
(3) 连接正方形对角线BD,若图(2)中的∠PAQ的边AP,AQ分别交对角线BD于点M,N,如图(3),当点Q是CD边的三等分点时,求BM的长.

答案：

(1)$45^{\circ}$ 2
不发生变化.
理由:如图,延长$CB$至点$T$,使得$BT = DQ$.

$\because$四边形$ABCD$是正方形,$\therefore AD = AB$,$\angle ABC = \angle D = 90^{\circ}$,$\therefore \angle ABT = 90^{\circ}$,$\therefore \angle ABT = \angle D$,$\therefore \triangle ADQ \cong \triangle ABT(SAS)$,$\therefore AT = AQ$,$\angle DAQ = \angle BAT$,$\because \angle PAQ=45^{\circ}$,$\therefore \angle PAT = \angle BAP + \angle BAT = \angle BAP + \angle DAQ =45^{\circ}$,$\therefore \angle PAT = \angle PAQ$.
又$\because AP = AP$,$\therefore \triangle PAT \cong \triangle PAQ(SAS)$,$\therefore$点$A$到$PT$的距离等于点$A$到$PQ$的距离.
$\because$点$A$到$PT$的距离为2,$\therefore$点$A$到$PQ$的距离为2,距离不发生变化. (6分)
(3)$\because$四边形$ABCD$是正方形,$\therefore \angle ABM = \angle ACQ = 45^{\circ}$,$AC = \sqrt{2}AB$.
$\because \angle BAC = \angle PAQ = 45^{\circ}$,$\therefore \angle BAM = \angle CAQ$,$\therefore \triangle CAQ \backsim \triangle BAM$,$\therefore \frac{CQ}{BM} = \frac{AC}{AB} = \sqrt{2}$.
分两种情况讨论:
①当$CQ = \frac{1}{3}CD$时,$BM = \frac{\sqrt{2}}{3}$;
②当$CQ = \frac{2}{3}CD$时,$BM = \frac{2\sqrt{2}}{3}$.
综上,当点$Q$是$CD$边的三等分点时,$BM$的长为$\frac{\sqrt{2}}{3}$或$\frac{2\sqrt{2}}{3}$. (10分)

23. (本小题满分13分)
已知抛物线y = ax² - 2ax + b(a,b是常数,且a ≠ 0)经过点A(-1,0).
(1) 若b = 3,求该抛物线的表达式及顶点坐标.
(2) 当-2 ≤ x ≤ 2时,函数y有最小值-9,求a的值.
(3) 如图,在(1)的条件下,抛物线与y轴交于点C,与x轴交于另一点B,连接BC. 若动直线l // BC,直线l与抛物线交于D,E两点(点D在点E左侧),直线CD与直线BE相交于点M,求点M到y轴的距离.

答案：

(1)$\because b = 3$,$\therefore y = ax^{2} - 2ax + 3$.
$\because$抛物线$y = ax^{2} - 2ax + 3$经过点$A(-1,0)$,$\therefore a + 2a + 3 = 0$,$\therefore a = -1$,$\therefore$该抛物线的表达式为$y = -x^{2} + 2x + 3$. (2分)
$\because y = -x^{2} + 2x + 3 = -(x - 1)^{2} + 4$,$\therefore$抛物线的顶点坐标为$(1,4)$. (3分)
(2)将$A(-1,0)$代入$y = ax^{2} - 2ax + b$,得$0 = a + 2a + b$,$\therefore b = -3a$,$\therefore y = ax^{2} - 2ax - 3a$,
$\therefore$抛物线的对称轴为直线$x = -\frac{-2a}{2a} = 1$.
分两种情况讨论:
①当$a ＞ 0$时,抛物线开口向上.
$\because -2 \leqslant x \leqslant 2$,$\therefore$当$x = 1$时,$y$取得最小值$-9$,$\therefore a - 2a - 3a = -9$,解得$a = \frac{9}{4}$. (5分)
②当$a ＜ 0$时,抛物线开口向下.
$\because -2 \leqslant x \leqslant 2$,$1 - (-2) ＞ 2 - 1$,$\therefore$当$x = -2$时,$y$取得最小值$-9$(点拨:当抛物线开口向下时,抛物线上的点离对称轴越远对应的函数值越小),$\therefore 4a + 4a - 3a = -9$,解得$a = - \frac{9}{5}$.
综上可知,$a$的值为$-\frac{9}{5}$或$\frac{9}{4}$. (7分)
(3)由
(1)知$y = -x^{2} + 2x + 3$.
对于$y = -x^{2} + 2x + 3$,当$x = 0$时,$y = 3$,$\therefore C(0,3)$.
对于$y = -x^{2} + 2x + 3$,当$y = 0$时,$-x^{2} + 2x + 3 = 0$,解得$x_{1} = -1$,$x_{2} = 3$,$\therefore B(3,0)$.
设直线$BC$的表达式为$y = kx + m$,
将$B(3,0)$,$C(0,3)$分别代入,得$\begin{cases}3k + m = 0, \\m = 3,\end{cases}$解得$\begin{cases}k = -1, \\m = 3,\end{cases}$$\therefore$直线$BC$的表达式为$y = -x + 3$.
$\because$直线$l // BC$,$\therefore$可设直线$l$的表达式为$y = -x + t$,
设$D(x_{1},y_{1})$,$E(x_{2},y_{2})$,$M(x_{0},y_{0})$,
令$-x + t = -x^{2} + 2x + 3$,整理得$x^{2} - 3x + t - 3 = 0$,
$\therefore x_{1} + x_{2} = 3$(依据:一元二次方程根与系数的关系),
$\therefore x_{2} = 3 - x_{1}$. (10分)
如图,过点$M$作$PM // x$轴,交$y$轴于点$P$,分别过点$D$,$E$,$B$作$DS \perp PM$,$ET \perp PM$,$BQ \perp PM$,垂足分别为$S$,$T$,$Q$,则$\triangle MSD \backsim \triangle MPC$,$\triangle MET \backsim \triangle MBQ$,$\therefore \frac{MS}{MP} = \frac{MD}{MC}$,$\frac{MT}{MQ} = \frac{ME}{MB}$,$\therefore \frac{x_{0} - x_{1}}{x_{0}} = \frac{x_{2} - x_{0}}{3 - x_{0}}$,
$\therefore (x_{0} - x_{1})(3 - x_{0}) = x_{0}(x_{2} - x_{0})$,
整理得$3x_{0} - 3x_{1} + x_{0}x_{1} - x_{0}^{2} = x_{0}x_{2} - x_{0}^{2}$,
将$x_{2} = 3 - x_{1}$代入,得$3x_{0} - 3x_{1} + x_{0}x_{1} - x_{0}(3 - x_{1}) = 0$,
$\therefore 2x_{0}x_{1} = 3x_{1}$,$\therefore x_{1} = \frac{3}{2}$,
$\therefore$点$M$到$y$轴的距离为$\frac{3}{2}$. (13分)