AMC12 2014 A

AMC12 2014 A · Q20

AMC12 2014 A · Q20. It mainly tests Angle chasing, Transformations.

In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$?

在$\triangle BAC$中，$\angle BAC=40^\circ$，$AB=10$，$AC=6$。点$D$和$E$分别在$\overline{AB}$和$\overline{AC}$上。$BE+DE+CD$的最小可能值是多少？

(A) 6\sqrt 3+3 6\sqrt 3+3

(B) \dfrac{27}2 \dfrac{27}2

(D) 14 14

(E) 3\sqrt 3+9\qquad 3\sqrt 3+9\qquad

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $C_1$ be the reflection of $C$ across $\overline{AB}$, and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$. Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$. (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As $A$ lies on both $AB$ and $AC$, we have $C_2A=C_1A=CA=6$. Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$. Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}.\]

设$C_1$是$C$关于$\overline{AB}$的反射，$C_2$是$C_1$关于$\overline{AC}$的反射。那么众所周知，$BE+DE+CD$的最小值等于$C_2B$。（证明这是三角不等式的简单应用；简单情形的例子见Heron's Shortest Path Problem。）由于$A$位于$AB$和$AC$上，我们有$C_2A=C_1A=CA=6$。此外，由反射性质，$\angle CAC_1=2\angle CAB=80^\circ$，所以$\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$。因此由余弦定律 \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}\]。

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