【答案】
分析:作出OQ關(guān)于OP的對(duì)稱(chēng)射線(xiàn)OM,在射線(xiàn)OM上找出A關(guān)于OP的對(duì)稱(chēng)點(diǎn)A′,根據(jù)對(duì)稱(chēng)性質(zhì)可得AA
1=A′A
1,把要求的AA
1+A
1A
2轉(zhuǎn)化為A′A
1+A
1A
2,然后根據(jù)“兩點(diǎn)之間線(xiàn)段最短”,只有當(dāng)A′,A
1,A
2在一條直線(xiàn)上時(shí),滿(mǎn)足AA
1+A
1A
2最小值等于A′A
2;找出射線(xiàn)OM關(guān)于OQ的對(duì)稱(chēng)射線(xiàn),在射線(xiàn)ON上找出A′關(guān)于OQ的對(duì)稱(chēng)點(diǎn)A″,同理只有當(dāng)A″,A
2,B在一條直線(xiàn)上時(shí),滿(mǎn)足A′A
2+A
2B最小值等于A″B,然后根據(jù)對(duì)稱(chēng)性質(zhì)求出OA″的長(zhǎng)及∠BOA″,以及OB,判斷可得三角形OBA″為直角三角形,由OB和OA″的長(zhǎng),根據(jù)勾股定理求出A″B的長(zhǎng),即為l的最小值.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/images0.png)
解:作OQ關(guān)于OP的對(duì)稱(chēng)射線(xiàn)OM,A關(guān)于OP的對(duì)稱(chēng)點(diǎn)A′,
∴AA
1=A′A
1,
則AA
1+A
1A
2=A′A
1+A
1A
2,
根據(jù)“兩點(diǎn)之間線(xiàn)段最短”,
當(dāng)A′,A
1,A
2在一條直線(xiàn)時(shí),AA
1+A
1A
2最小=A′A
2,
同理,作OM關(guān)于OQ的對(duì)稱(chēng)射線(xiàn)ON,A′關(guān)于OQ的對(duì)稱(chēng)點(diǎn)A″,
∴A′A
2=A″A
2,
則A
2B=A″A
2+A
2B,
根據(jù)“兩點(diǎn)之間線(xiàn)段最短”,
當(dāng)A″,A
2,B在一條直線(xiàn)上時(shí),A′A
2+A
2B最小=A″B,
由對(duì)稱(chēng)可知:∠POQ=∠POM=20°,即∠MOQ=40°,
再由對(duì)稱(chēng)可知:∠NOQ=∠MOQ=40°,且OA=OA′=OA″=1,
在△OA″B,∠A″OB=∠POQ+∠NOQ=20°+40°=60°,
取OB的中點(diǎn)E,連接A″E,如圖所示:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/images1.png)
則OA″=OE=BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/0.png)
OB=1,
又∠A″OB=60°,
∴△OA″E為等邊三角形,
∴∠OEA″=60°,A″E=1,即A″E=BE,
∴∠BA″E=∠B,
又∠OEA″是△A″EB的外角,
∴∠OEA″=∠BA″E+∠B=2∠B=60°,
∴∠B=30°,
∴∠OA″B=180°-60°-30°=90°,
∴△OA″B為直角三角形,
A″B=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/2.png)
,
則l=AA
1+A
1A
2+A
2B的最小值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163440861314306/SYS201310221634408613143011_DA/3.png)
.
點(diǎn)評(píng):此題考查了利用軸對(duì)稱(chēng)求最短路線(xiàn)的問(wèn)題,涉及的知識(shí)有對(duì)稱(chēng)線(xiàn)段的性質(zhì),軸對(duì)稱(chēng)的性質(zhì),線(xiàn)段公理等,利用了數(shù)形結(jié)合及轉(zhuǎn)化的思想,此類(lèi)題往往利用的是“情理結(jié)合法”,即由實(shí)情聯(lián)想原理,再由原理解決問(wèn)題.能正確畫(huà)圖和根據(jù)畫(huà)圖條件進(jìn)行推理是解本題的關(guān)鍵.