AMC12 2021 B

AMC12 2021 B · Q17

AMC12 2021 B · Q17. It mainly tests Area & perimeter, Ratios in geometry.

Let $ABCD$ be an isosceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\frac{AB}{CD}?$

设 $ABCD$ 是一个等腰梯形，具有平行底边 $\overline{AB}$ 和 $\overline{CD}$，且 $AB>CD$。从梯形内部一点到顶点的线段将梯形分成四个三角形，其面积分别为 $2, 3, 4,$ 和 $5$，从底边 $\overline{CD}$ 的三角形开始顺时针方向如图所示。$rac{AB}{CD}$ 的比值为多少？

(A) \: 3 \: 3

(B) \: 2+\sqrt{2} \: 2+\sqrt{2}

(D) \: 2\sqrt{3} \: 2\sqrt{3}

(E) \: 3\sqrt{2} \: 3\sqrt{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Without the loss of generality, let $\mathcal T$ have vertices $A$, $B$, $C$, and $D$, with $AB = r$ and $CD = s$. Also denote by $P$ the point in the interior of $\mathcal T$. Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Observe that $PX = \tfrac 8r$ and $PY = \tfrac 4s$. Now using the formula for the area of a trapezoid yields \[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.\] Thus, the ratio $\rho := \tfrac rs$ satisfies $\rho + 2\rho^{-1} = 4$; solving yields $\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}$. (Observe that the given areas of $3$ and $5$ are irrelevant to the ratio $\frac{AB}{CD}$.)

不失一般性，设梯形 $\mathcal T$ 的顶点为 $A$, $B$, $C$, 和 $D$，$AB = r$，$CD = s$。设 $P$ 为梯形内部的点。设 $X$ 和 $Y$ 分别为从 $P$ 到 $AB$ 和 $CD$ 的垂足。注意到 $PX = \tfrac 8r$ 和 $PY = \tfrac 4s$。现在使用梯形面积公式得到 \[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.\] 因此，比值 $\rho := \tfrac rs$ 满足 $\rho + 2\rho^{-1} = 4$；求解得到 $\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}$。（注意到给定的面积 $3$ 和 $5$ 与比值 $\frac{AB}{CD}$ 无关。）

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