例1 如图,在$\triangle ABC$中,$\angle C = 90^{\circ}$,$AC = 4\mathrm{cm}$,$BC = 3\mathrm{cm}$.动点$M$、$N$从点$C$同时出发,均以$1\mathrm{cm}/\mathrm{s}$的速度分别沿$CA$、$CB$向终点$A$、$B$移动,同时动点$P$从点$B$出发,以$2\mathrm{cm}/\mathrm{s}$的速度沿$BA$向终点$A$移动,连接$PM$、$PN$,设移动时间为$t\mathrm{s}(0 ＜ t ＜ 2.5)$.
(1)$t$为何值时,以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似?
(2)是否存在某一时刻$t$,使四边形$APNC$的面积$S$获得最小值?若存在,求$S$的最小值;若不存在,请说明理由.
分析:根据勾股定理,求得$AB = 5\mathrm{cm}$.第(1)小题分$\triangle AMP \backsim \triangle ABC$和$\triangle APM \backsim \triangle ABC$两种情况讨论,利用相似三角形的对应边成比例求$t$的值;第(2)小题用含$t$的代数式表示$\triangle BNP$的边$BN$上的高,然后根据“$S = S_{\triangle ABC} - S_{\triangle BNP}$”求出$S$与$t$的函数表达式,再根据二次函数表达式,对$S$是否存在最小值进行判断.
解:在$\triangle ABC$中,$\angle C = 90^{\circ}$,$AC = 4\mathrm{cm}$,$BC = 3\mathrm{cm}$,
$\therefore AB = \sqrt{AC^{2} + BC^{2}} = 5\mathrm{cm}$.
(1)以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似,分两种情况:
①当$\triangle AMP \backsim \triangle ABC$时,可得$\frac{AP}{AC} = \frac{AM}{AB}$,即$\frac{5 - 2t}{4} = \frac{4 - t}{5}$.
解得$t = \frac{3}{2}$.
②当$\triangle APM \backsim \triangle ABC$时,可得$\frac{AM}{AC} = \frac{AP}{AB}$,即$\frac{4 - t}{4} = \frac{5 - 2t}{5}$.
解得$t = 0$(不合题意,舍去).
$\therefore$当$t = \frac{3}{2}$时,以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似.
(2)作$PH ⊥ BC$,垂足为$H$,则$PH // AC$.
$\therefore \frac{PH}{AC} = \frac{BP}{BA}$,即$\frac{PH}{4} = \frac{2t}{5}$,$PH = \frac{8t}{5}$.
$\therefore S = S_{\triangle ABC} - S_{\triangle BNP}$
$= \frac{1}{2} × 3 × 4 - \frac{1}{2}(3 - t) · \frac{8t}{5}$
$= \frac{4}{5}t^{2} - \frac{12}{5}t + 6$
$= \frac{4}{5}(t - \frac{3}{2})^{2} + \frac{21}{5}(0 ＜ t ＜ 2.5)$.
$\because S$是$t$的二次函数,且$\frac{4}{5} ＞ 0$,
$\therefore S$有最小值.当$t = \frac{3}{2}$时,$S_{最小值} = \frac{21}{5}$.
想一想:在本例的条件下,四边形$CMPN$的面积有最小值或最大值吗?
(1)$t$为何值时,以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似?
(2)是否存在某一时刻$t$,使四边形$APNC$的面积$S$获得最小值?若存在,求$S$的最小值;若不存在,请说明理由.
分析:根据勾股定理,求得$AB = 5\mathrm{cm}$.第(1)小题分$\triangle AMP \backsim \triangle ABC$和$\triangle APM \backsim \triangle ABC$两种情况讨论,利用相似三角形的对应边成比例求$t$的值;第(2)小题用含$t$的代数式表示$\triangle BNP$的边$BN$上的高,然后根据“$S = S_{\triangle ABC} - S_{\triangle BNP}$”求出$S$与$t$的函数表达式,再根据二次函数表达式,对$S$是否存在最小值进行判断.
解:在$\triangle ABC$中,$\angle C = 90^{\circ}$,$AC = 4\mathrm{cm}$,$BC = 3\mathrm{cm}$,
$\therefore AB = \sqrt{AC^{2} + BC^{2}} = 5\mathrm{cm}$.
(1)以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似,分两种情况:
①当$\triangle AMP \backsim \triangle ABC$时,可得$\frac{AP}{AC} = \frac{AM}{AB}$,即$\frac{5 - 2t}{4} = \frac{4 - t}{5}$.
解得$t = \frac{3}{2}$.
②当$\triangle APM \backsim \triangle ABC$时,可得$\frac{AM}{AC} = \frac{AP}{AB}$,即$\frac{4 - t}{4} = \frac{5 - 2t}{5}$.
解得$t = 0$(不合题意,舍去).
$\therefore$当$t = \frac{3}{2}$时,以$A$、$P$、$M$为顶点的三角形与$\triangle ABC$相似.
(2)作$PH ⊥ BC$,垂足为$H$,则$PH // AC$.
$\therefore \frac{PH}{AC} = \frac{BP}{BA}$,即$\frac{PH}{4} = \frac{2t}{5}$,$PH = \frac{8t}{5}$.
$\therefore S = S_{\triangle ABC} - S_{\triangle BNP}$
$= \frac{1}{2} × 3 × 4 - \frac{1}{2}(3 - t) · \frac{8t}{5}$
$= \frac{4}{5}t^{2} - \frac{12}{5}t + 6$
$= \frac{4}{5}(t - \frac{3}{2})^{2} + \frac{21}{5}(0 ＜ t ＜ 2.5)$.
$\because S$是$t$的二次函数,且$\frac{4}{5} ＞ 0$,
$\therefore S$有最小值.当$t = \frac{3}{2}$时,$S_{最小值} = \frac{21}{5}$.
想一想:在本例的条件下,四边形$CMPN$的面积有最小值或最大值吗?
答案
