AMC10 2018 B

AMC10 2018 B · Q17

AMC10 2018 B · Q17. It mainly tests Quadratic equations, Triangles (properties).

In rectangle $PQRS$, $PQ = 8$ and $QR = 6$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, points $E$ and $F$ lie on $\overline{RS}$, and points $G$ and $H$ lie on $\overline{SP}$ so that $AP = BQ < 4$ and the convex octagon $ABCDEFEGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k + m\sqrt{n}$, where $k, m,$ and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$?

在矩形 $PQRS$ 中，$PQ = 8$，$QR = 6$。点 $A$ 和 $B$ 在 $\overline{PQ}$ 上，点 $C$ 和 $D$ 在 $\overline{QR}$ 上，点 $E$ 和 $F$ 在 $\overline{RS}$ 上，点 $G$ 和 $H$ 在 $\overline{SP}$ 上，使得 $AP = BQ < 4$，且凸八边形 $ABCDEFEGH$ 是等边的。这个八边形的边长可以表示为 $k + m\sqrt{n}$ 的形式，其中 $k, m, n$ 是整数，且 $n$ 不能被任何质数的平方整除。求 $k+m+n$？

(A) 1 1

(B) 7 7

(D) 92 92

(E) 106 106

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because $AP<4=\frac12 PQ$, it follows that $A$ is closer to $P$ than it is to $Q$ and that $A$ is between points $P$ and $B$. Because $AP=BQ$, $AH=BC$, and angles $APH$ and $BQC$ are right angles, $\triangle APH\cong\triangle BQC$. Thus $PH=QC$, and $PQCH$ is a rectangle. Because $CD=HG$, it follows that $HCDG$ is also a rectangle. Thus $GDRS$ is a rectangle and $DR=GS$, and it follows that $\triangle ERD\cong\triangle FSG$. Therefore segment $EF$ is centered in $RS$ just as congruent segment $AB$ is centered in $PQ$. Therefore $\triangle ERD\cong\triangle BQC$, and $CD$ is also centered in $QR$. Let $2x$ be the side length $AB=BC=CD=DE=EF=FG=GH=HA$ of the regular octagon; then $AP=BQ=4-x$ and $QC=RD=3-x$. Applying the Pythagorean Theorem to $\triangle BQC$ yields $(4-x)^2+(3-x)^2=(2x)^2$, which simplifies to $2x^2+14x-25=0$. Thus $x=\frac12\cdot(-7\pm3\sqrt{11})$, and because $x>0$, it follows that $2x=-7+3\sqrt{11}$. Hence $k+m+n=-7+3+11=7$.

答案（B）：因为$AP<4=\frac12 PQ$，可知$A$到$P$比到$Q$更近，并且$A$位于点$P$与$B$之间。由于$AP=BQ$、$AH=BC$，且$\angle APH$与$\angle BQC$都是直角，所以$\triangle APH\cong\triangle BQC$。因此$PH=QC$，且$PQCH$是矩形。又因为$CD=HG$，可知$HCDG$也是矩形。于是$GDRS$是矩形且$DR=GS$，从而$\triangle ERD\cong\triangle FSG$。因此线段$EF$在$RS$中居中，正如全等线段$AB$在$PQ$中居中一样。因此$\triangle ERD\cong\triangle BQC$，并且$CD$也在$QR$中居中。设正八边形的边长$AB=BC=CD=DE=EF=FG=GH=HA$为$2x$；则$AP=BQ=4-x$，$QC=RD=3-x$。对$\triangle BQC$应用勾股定理得$(4-x)^2+(3-x)^2=(2x)^2$，化简为$2x^2+14x-25=0$。因此$x=\frac12\cdot(-7\pm3\sqrt{11})$，由于$x>0$，可得$2x=-7+3\sqrt{11}$。所以$k+m+n=-7+3+11=7$。

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