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考题17-18 2015 AMC 10A

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数学竞赛真题以及官方解答吧：

Problem 17

A line that passes through the origin intersects both the line $x = 1$ and the line $y=1+ \frac{\sqrt{3}}{3} x$ . The three lines create an equilateral triangle. What is the perimeter of the triangle?

$考题17-18 2015 AMC 10A$

Solution 1

Since the triangle is equilateral and one of the sides is a vertical line, the triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is $\frac{\sqrt{3}}{3}$ so the third must be $-\frac{\sqrt{3}}{3}$ . Since this third line passes through the origin, its equation is simply $y = -\frac{\sqrt{3}}{3}x$ . To find two vertices of the triangle, plug in $x=1$ to both the other equations.

$y = -\frac{\sqrt{3}}{3}$

$y = 1 + \frac{\sqrt{3}}{3}$

We now have the coordinates of two vertices, $\left(1, -\frac{\sqrt{3}}{3}\right)$ and $考题17-18 2015 AMC 10A$ . The length of one side is the distance between the y-coordinates, or $1 + \frac{2\sqrt{3}}{3}$ .

The perimeter of the triangle is thus $考题17-18 2015 AMC 10A$ , so the answer is $\boxed{\textbf{(D) }3 + 2\sqrt{3}}$

Solution 2

Draw a line from the y-intercept of the equation $y=1+ \frac{\sqrt{3}}{3} x$ perpendicular to the line $x=1$ . There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is 2 $\left(\frac{1}{\sqrt{3}}\right)$ + 1. After multiplying the side length by 3 and rationalizing, you get $考题17-18 2015 AMC 10A$ .

Problem 18

Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$ . Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$ ?

$考题17-18 2015 AMC 10A$

Solution

Notice that $1000$ is $3E8$ in hexadecimal. We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.

The first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0 - 9$ , giving $4 \cdot 10 \cdot 10 = 400$ combinations. However, this includes $000,$ so this number must be diminished by $1.$ Therefore, there are $399$ valid $n$ corresponding to those $399$ positive integers less than $1000$ that consist of only numeric digits. (Notice that $399 < 3E8$ in hexadecimal.) Therefore, our answer is $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$