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AMC数学竞赛真题2017年10A 21-22

2018-08-27 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：、

Problem 21

A square with side length $x$ is inscribed in a right triangle with sides of length $3$ , $4$ , and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$ , $4$ , and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\tfrac{x}{y}$ ?

$\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}$

Solution

Analyze the first right triangle.

amc数学竞赛

Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$ . This can be written as $\frac{4-x}{x}=\frac{4}{3}$ . Solving, $x = \frac{12}{7}$ .

Now we analyze the second triangle.

amc考试

Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, so $RB' = \frac{4}{3}y$ , and $C'S = \frac{3}{4}y$ . Thus, $C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5$ . Solving for $y$ , we get $y = \frac{60}{37}$ . Thus, $\frac{x}{y} = \boxed{\textbf{(D)}\:\frac{37}{35}}$ .

Problem 22

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$

Solution

amc美国数学竞赛 Let the radius of the circle be $r$ , and let its center be $O$ . Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$ , then $\angle OBA = \angle OCA = 90^{\circ}$ , so $\angle BOC = 120^{\circ}$ . Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$ , then (pick your favorite method) $\overline{BC} = r\sqrt{3}$ . The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$ , and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$ . The area outside of the circle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$ . Therefore, the answer is $\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}$