AMC12 2016 A

AMC12 2016 A · Q21

AMC12 2016 A · Q21. It mainly tests Circle theorems, Trigonometry (basic).

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of its fourth side?

一个四边形内接于半径为 $200\sqrt{2}$ 的圆。这个四边形的三条边长度为 $200$。它的第四条边的长度是多少？

(A) 200 200

(B) $200\sqrt{2}$ $200\sqrt{2}$

(D) $300\sqrt{2}$ $300\sqrt{2}$

(E) 500 500

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $ABCD$ be the given cyclic quadrilateral with $AB=BC=CD=200$, and let $E$ and $F$ be the feet of the perpendicular segments from $B$ and $C$, respectively, to $AD$, as shown in the figure. Let the center of the circle be $O$, and let $\angle AOB=\angle BOC=\angle COD=\theta$. Because inscribed $\angle BAD$ is half the size of central $\angle BOD=2\theta$, it follows that $\angle BAD=\theta$. Let $M$ be the midpoint of $AB$. Then $\sin\left(\frac{\theta}{2}\right)=\frac{AM}{AO}=\frac{100}{200\sqrt{2}}=\frac{1}{2\sqrt{2}}$. Then $\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)=\frac{3}{4}$. Hence $AE=AB\cos\theta=200\cdot\frac{3}{4}=150$, and $FD=150$ as well. Because $EF=BC=200$, the remaining side $AD=AE+EF+FD=150+200+150=500$.

答案（E）：设$ABCD$为所给的圆内接四边形，且$AB=BC=CD=200$。如图，$E$与$F$分别为从$B$、$C$向$AD$作垂线的垂足。设圆心为$O$，并令$\angle AOB=\angle BOC=\angle COD=\theta$。由于圆周角$\angle BAD$等于对应圆心角$\angle BOD=2\theta$的一半，故$\angle BAD=\theta$。设$M$为$AB$的中点，则$\sin\left(\frac{\theta}{2}\right)=\frac{AM}{AO}=\frac{100}{200\sqrt{2}}=\frac{1}{2\sqrt{2}}$。于是$\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)=\frac{3}{4}$。因此$AE=AB\cos\theta=200\cdot\frac{3}{4}=150$，同理$FD=150$。又因为$EF=BC=200$，所以剩余边$AD=AE+EF+FD=150+200+150=500$。

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