AMC12 2016 B

AMC12 2016 B · Q21

AMC12 2016 B · Q21. It mainly tests Functions basics, Fractions.

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}$. For $i=1,2,\ldots$, let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD}$, and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}$. What is $\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_iP_i$?

设$ABCD$为单位正方形。设$Q_1$为线段$\overline{CD}$的中点。对$i=1,2,\ldots$，令$P_i$为$\overline{AQ_i}$与$\overline{BD}$的交点，并令$Q_{i+1}$为从$P_i$向$\overline{CD}$作垂线的垂足。求 $\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_iP_i$。

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{1}{4}$ $\frac{1}{4}$

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) 1 1

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): For any point $P$ between $B$ and $D$, let $Q$ be the foot of the perpendicular from $P$ to $\overline{CD}$, let $P'$ be the intersection of $\overline{AQ}$ and $\overline{BD}$, and let $Q'$ be the foot of the perpendicular from $P'$ to $\overline{CD}$. Let $x = PQ$ and $y = P'Q'$. Because $\triangle PQD$ and $\triangle P'Q'D$ are isosceles right triangles, $DQ = x$ and $DQ' = y$. Because $\triangle ADQ$ is similar to $\triangle P'Q'Q$, $\frac{1}{x} = \frac{y}{x-y}$. Solving for $y$ gives $y = \frac{x}{1+x}$. Now let $P_0$ be the midpoint of $\overline{BD}$. Then $P_0Q_1 = DQ_1 = \frac{1}{2}$. It follows from the analysis above that $P_1Q_2 = DQ_2 = \frac{1}{3}$, $P_2Q_3 = DQ_3 = \frac{1}{4}$, and in general $P_iQ_{i+1} = DQ_{i+1} = \frac{1}{i+2}$. The area of $\triangle DQ_iP_i$ is $$ \frac{1}{2}\cdot DQ_i \cdot P_iQ_{i+1} = \frac{1}{2}\cdot \frac{1}{i+1}\cdot \frac{1}{i+2} = \frac{1}{2}\left(\frac{1}{i+1}-\frac{1}{i+2}\right). $$ The requested infinite sum telescopes: $$ \sum_{i=1}^{\infty}\text{Area of }\triangle DQ_iP_i = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots\right). $$ Its value is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.

答案（B）：对任意位于 $B$ 与 $D$ 之间的点 $P$，令 $Q$ 为从 $P$ 到 $\overline{CD}$ 的垂足，令 $P'$ 为 $\overline{AQ}$ 与 $\overline{BD}$ 的交点，并令 $Q'$ 为从 $P'$ 到 $\overline{CD}$ 的垂足。设 $x=PQ$、$y=P'Q'$。由于 $\triangle PQD$ 与 $\triangle P'Q'D$ 是等腰直角三角形，所以 $DQ=x$ 且 $DQ'=y$。又因为 $\triangle ADQ$ 与 $\triangle P'Q'Q$ 相似，有 $\frac{1}{x}=\frac{y}{x-y}$。解得 $y=\frac{x}{1+x}$。现在令 $P_0$ 为 $\overline{BD}$ 的中点，则 $P_0Q_1=DQ_1=\frac{1}{2}$。由上面的分析可得 $P_1Q_2=DQ_2=\frac{1}{3}$，$P_2Q_3=DQ_3=\frac{1}{4}$，一般地 $P_iQ_{i+1}=DQ_{i+1}=\frac{1}{i+2}$。$\triangle DQ_iP_i$ 的面积为 $$ \frac{1}{2}\cdot DQ_i \cdot P_iQ_{i+1} = \frac{1}{2}\cdot \frac{1}{i+1}\cdot \frac{1}{i+2} = \frac{1}{2}\left(\frac{1}{i+1}-\frac{1}{i+2}\right)。 $$ 所求无穷和为望远镜求和： $$ \sum_{i=1}^{\infty}\text{面积 } \triangle DQ_iP_i = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots\right)。 $$ 其值为 $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$。

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