将军饮马
上传时间：2026-04-09 10:59 · 作者：ytry1983
最短距离：将军饮马 (第10题)
利用“轴对称”原理求解线段之和最小值
📝 数学计算过程
1. **对称性**：作 $A$ 关于河岸的对称点 $A'$，则 $AC = A'C = 400\text{m}$。
2. **最短路径**：连接 $A'B$，根据“两点之间线段最短”，线段 $A'B$ 的长度即为最短路程。
3. **勾股定理**：构造直角三角形，直角边长分别为：
- 水平方向距离 = $CD = 800\text{m}$
- 垂直方向距离 = $AC + BD = 400 + 200 = 600\text{m}$
4. **结果**：最短路程 = $\sqrt{800^2 + 600^2} = 1000\text{m}$。
💡 饮水点 $P$ 的位置
通过相似三角形推导，饮水点 $P$ 距离 $C$ 点的距离为：
$800 \times \frac{400}{400+200} = 800 \times \frac{2}{3} \approx 533.3\text{米}$。
查看解析页源码
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<h1 class="text-2xl font-bold text-white">最短距离：将军饮马 (第10题)</h1>
<p class="mt-2 opacity-90 text-white">利用“轴对称”原理求解线段之和最小值</p>
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<span>📝</span> 数学计算过程
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<p>1. **对称性**：作 $A$ 关于河岸的对称点 $A'$，则 $AC = A'C = 400\text{m}$。</p>
<p>2. **最短路径**：连接 $A'B$，根据“两点之间线段最短”，线段 $A'B$ 的长度即为最短路程。</p>
<p>3. **勾股定理**：构造直角三角形，直角边长分别为：</p>
<ul class="list-disc list-inside ml-4">
<li>水平方向距离 = $CD = 800\text{m}$</li>
<li>垂直方向距离 = $AC + BD = 400 + 200 = 600\text{m}$</li>
</ul>
<p>4. **结果**：最短路程 = $\sqrt{800^2 + 600^2} = 1000\text{m}$。</p>
</div>
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<span>💡</span> 饮水点 $P$ 的位置
</h4>
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通过相似三角形推导，饮水点 $P$ 距离 $C$ 点的距离为：<br>
$800 \times \frac{400}{400+200} = 800 \times \frac{2}{3} \approx 533.3\text{米}$。
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