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考题19-20 2015 AMC 10B

2018-08-06 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的官方真题以及官方解答吧：

Problem 19

In $\triangle{ABC}$ , $\angle{C} = 90^{\circ}$ and $AB = 12$ . Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$ , and $W$ lie on a circle. What is the perimeter of the triangle?

$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32$

Solution

The center of the circle lies on the perpendicular bisectors of both chords $ZW$ and $YX$ . Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$ . Draw perpendiculars to $ZW$ and $YX$ from $O$ , and connect $OZ$ and $OY$ . $OY^2=6^2+12^2=180$ . Let $AC=a$ and $BC=b$ . Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$ . Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$ . But by Pythagorean Theorem on $\triangle ABC$ , we know $a^2+b^2=144$ , because $AB=12$ . Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$ . So our equation simplifies further to $a^2+ab=144$ . However $a^2+b^2=144$ , so $a^2+ab=a^2+b^2$ , which means $ab=b^2$ , or $a=b$ . Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$ , and thus the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$ .

$考题19-20 2015 AMC 10B$

Problem 20

Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions? $\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}$

Solution

$考题19-20 2015 AMC 10B$

We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$ .

If we define a "move" as each time Erin crawls along a single edge from 1 vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be 1 unit away from $A$ (since Erin cannot crawl back to $A$ ), this vertex must be $4$ .

Therefore, we now just need to count the number of paths from $A$ to $4$ . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$ , and 2 choices for which vertex Erin was at 2 moves before $4$ . All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \cdot 2 = \boxed{\textbf{(A)}\; 6}$ .