Question

正方形 $ABCD$ 的边长为 $13$ 。点 $E$ 和 $F$ 在正方形外部，满足 $BE = DF = 5$ 且 $AE = CF = 12$ 。求 $EF^2$ 。

Square $ABCD$ has side length $13,$ and points $E$ and $F$ are exterior to the square such that $BE = DF = 5$ and $AE = CF = 12.$ Find $EF^2.$

答案

答案：578

知识点：勾股数全等（几何）勾股定理

难度评级：2300

解答：

因为 $5^2 + 12^2 = 13^2$ ，三角形 $AEB$ 和 $CFD$ 分别在 $E$ 和 $F$ 处为直角，且它们全等（边长 $5$ ， $12$ ， $13$ ）。将 $EA$ 延长过 $A$ ，将 $FD$ 延长过 $D$ ，直到两条直线交于 $G$ 。

于是 $\angle GAD = 90^\circ$ $- \angle EAB = \angle ABE$ ，且 $\angle GDA = 90^\circ$ $- \angle FDC = \angle DCF$ $= \angle BAE$ 。这两个角之和为 $90^\circ$ ，所以 $\angle AGD = 90^\circ$ ，三角形 $AGD$ 与 $BEA$ 全等（角相等且斜边 $AD = BA = 13$ ）。因此 $GA = EB = 5$ 且 $GD = EA = 12$ 。

所以 $GE = GA + AE = 5 + 12 = 17$ ， $GF = GD + DF = 12 + 5 = 17$ ，且它们在 $G$ 处成直角，因此 $EF^2 = 17^2 + 17^2 = 578$ 。

Since $5^2 + 12^2 = 13^2,$ triangles $AEB$ and $CFD$ are right-angled at $E$ and $F,$ and they are congruent (sides $5,$ $12,$ $13$ ). Extend $EA$ beyond $A$ and $FD$ beyond $D$ until the two lines meet at $G.$

Then $\angle GAD = 90^\circ$ $- \angle EAB = \angle ABE,$ and $\angle GDA = 90^\circ$ $- \angle FDC = \angle DCF$ $= \angle BAE.$ These two angles sum to $90^\circ,$ so $\angle AGD = 90^\circ,$ and triangle $AGD$ is congruent to $BEA$ (equal angles and hypotenuse $AD = BA = 13$ ). Hence $GA = EB = 5$ and $GD = EA = 12.$

Therefore $GE = GA + AE = 5 + 12 = 17$ and $GF = GD + DF = 12 + 5 = 17,$ with a right angle between them at $G,$ so $EF^2 = 17^2 + 17^2 = 578.$

2007 AIME II 第 3 题

答案：578

难度评级：2300

解答：

其他年份的第 3 题