24. (12 分)如图1,在平面直角坐标系中,长方形ABCD的边AD//y轴.已知A(a, -a),C(c,c),且a,c满足$\sqrt{a + 2} + |2a + c| = 0$.
(1)直接写出a,c的值;
(2)如图2,点P为x轴上一点.连接AP,CP和AC.若三角形ACP的面积等于长方形ABCD的面积,求出点P的坐标;
(3)如图3,将AC向下平移m个单位长度(m＞0)得到MN.射线CO交线段MN于点H,AC交y轴于点K.若$S_{△ AHC} ≤ \frac{6}{5}S_{△ HKN}$,求出m的取值范围.(用$S_{△ AHC}$表示三角形AHC的面积)
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24. 【点拨】本题考查三角形的面积,非负数的性质,平移的性质,运用分类讨论的思想及用参数构建方程是解题的关键.【解析】(1)
∵$\sqrt{a + 2} + |2a + c| = 0$,
∴$a + 2 = 0$,$2a + c = 0$,
∴$a = -2$,$c = 4$.(2)由(1)知$a = -2$,$c = 4$,
∴$-a = 2$,
∴$A(-2,2)$,$C(4,4)$.
∵$AD// y$轴,四边形ABCD是长方形,
∴$AB// x$轴,
∴$B(4,2)$,
∴$AB = 4 - (-2) = 6$,$BC = 4 - 2 = 2$,
∴$S_{长方形ABCD} = AB · BC = 6 × 2 = 12$.
∵三角形 ACP 的面积等于长方形 ABCD 的面积,
∴$S_{△ ACP} = 12$.设$P(p,0)$,当$p ＜ -2$时,若 PC 在点 A 下方时,如图1,延长 DA,CB 分别交 x 轴于点 G,H,则$G(-2,0)$,$H(4,0)$,
∴$AG = 2$,$CH = 4$,$GH = 6$,$PG = -2 - p$,$PH = 4 - p$,
∴$S_{△ ACP} = S_{△ APG} + S_{梯形AGHC} - S_{△ PHC} = 12$,
∴$\dfrac{1}{2}AG · PG + \dfrac{1}{2}GH(AG + CH) - \dfrac{1}{2}PH · CH = 12$,
∴$\dfrac{1}{2} × 2(-2 - p) + \dfrac{1}{2} × 6 × (2 + 4) - \dfrac{1}{2} × 4(4 - p) = 12$,解得$p = 4 ＞ -2$,不符合题意.若 PC 在点 A 上方时,如图2,
∴$S_{△ ACP} = S_{△ PHC} - S_{△ APG} - S_{梯形AGHC} = 12$,
∴$\dfrac{1}{2}PH · CH - \dfrac{1}{2}AG · PG - \dfrac{1}{2}GH(AG + CH) = 12$,
∴$\dfrac{1}{2} × 4(4 - p) - \dfrac{1}{2} × 2(-2 - p) - \dfrac{1}{2} × 6 × (2 + 4) = 12$,解得$p = -20$,符合题意,
∴$P(-20,0)$.当$-2 ≤ p ≤ 4$时,如图3,
∴$S_{△ ACP} = S_{梯形AGHC} - S_{△ PHC} - S_{△ APG} = 12$,
∴$\dfrac{1}{2}GH(AG + CH) - \dfrac{1}{2}PH · CH - \dfrac{1}{2}AG · PG = 12$,
∴$\dfrac{1}{2} × 6 × (2 + 4) - \dfrac{1}{2} × 4(4 - p) - \dfrac{1}{2} × 2(p + 2) = 12$,解得$p = 4$,符合题意,
∴$P(4,0)$.当$p ＞ 4$时,如图4,
∴$S_{△ ACP} = S_{梯形AGHC} + S_{△ PHC} - S_{△ APG} = 12$,
∴$\dfrac{1}{2}GH(AG + CH) + \dfrac{1}{2}PH · CH - \dfrac{1}{2}AG · PG = 12$,
∴$\dfrac{1}{2} × 6 × (2 + 4) + \dfrac{1}{2} × 4(p - 4) - \dfrac{1}{2} × 2(p + 2) = 12$,解得$p = 4$,不符合题意.综上所述,点 P 的坐标为$(-20,0)$或$(4,0)$.(3)
∵点 H 在射线 CO 上,$C(4,4)$,
∴射线 CO 上点的横、纵坐标数值相等.设$H(n,n)$,如图5,过 M 作$MF ⊥ CN$交 CN 延长线于 F,连接 HF,则$M(-2,2 - m)$,$N(4,4 - m)$,$F(4,2 - m)$.
∵$S_{△ HMF} + S_{△ HFN} = S_{△ MFN}$,
∴$\dfrac{1}{2} × 6[n - (2 - m)] + \dfrac{1}{2}[(4 - m) - (2 - m)](4 - n) = \dfrac{1}{2} × 6[(4 - m) - (2 - m)]$,
∴$2n + 3m = 8$,
∴$n = 4 - \dfrac{3}{2}m$.
∵$AC// MN$,
∴$S_{四边形AMNC} = \dfrac{1}{2} × 6 × [2 - (2 - m) + 4 - (2 - m)] - \dfrac{1}{2} × 6[(4 - m) - (2 - m)] = 6m$,
∴$S_{△ AHC} = \dfrac{1}{2}S_{四边形AMNC} = 3m$,
∴$S_{△ CNH} = S_{四边形AMNC} - S_{△ AMH} - S_{△ AHC} = 6m - \dfrac{1}{2}m × (n + 2) - 3m = 6m - \dfrac{1}{2}m × (4 - \dfrac{3}{2}m + 2) - 3m = \dfrac{3}{4}m^2$.
∵$S_{△ AHC} ≤ \dfrac{6}{5}S_{△ HKN}$,$S_{△ HKN} = S_{△ CNH}$,
∴$3m ≤ \dfrac{6}{5} × \dfrac{3}{4}m^2$,即$3 ≤ \dfrac{9}{10}m$,
∴$m ≥ \dfrac{10}{3}$.又
∵H 在线段 MN 上,
∴$4 ≥ 4 - \dfrac{3}{2}m ≥ -2$,
∴$0 ≤ m ≤ 4$,
∴$m$的取值范围是$\dfrac{10}{3} ≤ m ≤ 4$.