AMC12 2011 A

AMC12 2011 A · Q24

AMC12 2011 A · Q24. It mainly tests Triangles (properties), Circle theorems.

Consider all quadrilaterals $ABCD$ such that $AB=14$, $BC=9$, $CD=7$, and $DA=12$. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

考虑所有满足 $AB=14$、$BC=9$、$CD=7$、$DA=12$ 的四边形 $ABCD$。在这样的四边形内部或边界上能放入的最大圆的半径是多少？

(A) \sqrt{15} \sqrt{15}

(B) \sqrt{21} \sqrt{21}

(D) 5 5

(E) 2\sqrt{7} 2\sqrt{7}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Note as above that ABCD must be tangential to obtain the circle with maximal radius. Let $E$, $F$, $G$, and $H$ be the points on $AB$, $BC$, $CD$, and $DA$ respectively where the circle is tangent. Let $\theta=\angle BAD$ and $\alpha=\angle ADC$. Since the quadrilateral is cyclic(because we want to maximize the circle, so we set the quadrilateral to be cyclic), $\angle ABC=180^{\circ}-\alpha$ and $\angle BCD=180^{\circ}-\theta$. Let the circle have center $O$ and radius $r$. Note that $OHD$, $OGC$, $OFB$, and $OEA$ are right angles. Hence $FOG=\theta$, $GOH=180^{\circ}-\alpha$, $EOH=180^{\circ}-\theta$, and $FOE=\alpha$. Therefore, $AEOH\sim OFCG$ and $EBFO\sim HOGD$. Let $x=CG$. Then $CF=x$, $BF=BE=9-x$, $GD=DH=7-x$, and $AH=AE=x+5$. Using $AEOH\sim OFCG$ and $EBFO\sim HOGD$ we have $r/(x+5)=x/r$, and $(9-x)/r=r/(7-x)$. By equating the value of $r^2$ from each, $x(x+5)=(7-x)(9-x)$. Solving we obtain $x=3$ so that $r=\boxed{\textbf{(C)}\ 2\sqrt{6}}$.

注意如上，为得到半径最大的圆，$ABCD$ 必须为可切四边形。设 $E$、$F$、$G$、$H$ 分别为圆与 $AB$、$BC$、$CD$、$DA$ 的切点。令 $\theta=\angle BAD$，$\alpha=\angle ADC$。由于四边形为圆内接（因为我们要最大化圆，所以令四边形为圆内接），有 $\angle ABC=180^{\circ}-\alpha$ 且 $\angle BCD=180^{\circ}-\theta$。设圆心为 $O$，半径为 $r$。注意 $OHD$、$OGC$、$OFB$、$OEA$ 均为直角。因此 $FOG=\theta$，$GOH=180^{\circ}-\alpha$，$EOH=180^{\circ}-\theta$，$FOE=\alpha$。因此，$AEOH\sim OFCG$ 且 $EBFO\sim HOGD$。令 $x=CG$。则 $CF=x$，$BF=BE=9-x$，$GD=DH=7-x$，$AH=AE=x+5$。由 $AEOH\sim OFCG$ 与 $EBFO\sim HOGD$ 得 $r/(x+5)=x/r$，以及 $(9-x)/r=r/(7-x)$。将两式得到的 $r^2$ 相等，得 $x(x+5)=(7-x)(9-x)$。解得 $x=3$，从而 $r=\boxed{\textbf{(C)}\ 2\sqrt{6}}$。

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