AMC12 2019 B

AMC12 2019 B · Q12

AMC12 2019 B · Q12. It mainly tests Triangles (properties), Trigonometry (basic).

Right triangle $ACD$ with right angle at $C$ is constructed outward on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length 1, as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$?

在等腰直角三角形 $ABC$（腿长为1）的斜边 $\overline{AC}$ 上向外构造直角三角形 $ACD$，直角在 $C$，使得两个三角形的周长相等，如图所示。求 $\sin(2\angle BAD)$？

(A) $\frac{1}{3}$ $\frac{1}{3}$

(B) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$

(D) $\frac{7}{9}$ $\frac{7}{9}$

(E) $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $\theta=\angle CAD$, $x=CD$, and $y=DA$. It follows from the given information that $AC=\sqrt{2}$ and $\angle BAC=45^\circ$. Because the perimeters of the two triangles are equal, $x+y=2$. By the Pythagorean Theorem, $2+x^2=y^2$. Solving these two equations simultaneously gives $x=\frac{1}{2}$ and $y=\frac{3}{2}$. It follows that $\sin\theta=\frac{1}{3}$ and $\cos\theta=\frac{2}{3}\sqrt{2}$. Then $$ \begin{array}{rcl} \sin(2\angle BAD) & = & 2\sin(\angle BAD)\cdot \cos(\angle BAD)\\ & = & 2\sin(\theta+45^\circ)\cdot \cos(\theta+45^\circ)\\ & = & 2(\sin\theta\cos45^\circ+\cos\theta\sin45^\circ)\cdot(\cos\theta\cos45^\circ-\sin\theta\sin45^\circ)\\ & = & 2\left(\frac{4+\sqrt{2}}{6}\right)\left(\frac{4-\sqrt{2}}{6}\right)\\ & = & \frac{7}{9}. \end{array} $$

答案（D）：设$\theta=\angle CAD$，$x=CD$，$y=DA$。由已知信息可得$AC=\sqrt{2}$且$\angle BAC=45^\circ$。因为两个三角形的周长相等，所以$x+y=2$。由勾股定理，$2+x^2=y^2$。联立解这两个方程得$x=\frac{1}{2}$且$y=\frac{3}{2}$。因此$\sin\theta=\frac{1}{3}$且$\cos\theta=\frac{2}{3}\sqrt{2}$。于是 $$ \begin{array}{rcl} \sin(2\angle BAD) & = & 2\sin(\angle BAD)\cdot \cos(\angle BAD)\\ & = & 2\sin(\theta+45^\circ)\cdot \cos(\theta+45^\circ)\\ & = & 2(\sin\theta\cos45^\circ+\cos\theta\sin45^\circ)\cdot(\cos\theta\cos45^\circ-\sin\theta\sin45^\circ)\\ & = & 2\left(\frac{4+\sqrt{2}}{6}\right)\left(\frac{4-\sqrt{2}}{6}\right)\\ & = & \frac{7}{9}. \end{array} $$

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