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AMC数学竞赛真题2017年10B 23-24

2018-08-30 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 23

Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$ ?

$amc真题$

Solution

We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, $N \equiv 4 \text{ (mod 5)}$ . The remainder when $N$ is divided by $9$ is $amc数学竞赛$ , but since $10 \equiv 1 \text{ (mod 9)}$ , we can also write this as $美国数学竞赛真题$ , which has a remainder of 0 mod 9. Therefore, by inspection, the answer is $\boxed{\textbf{(C) } 9}$ .

Note: the sum of the digits of $N$ is $270$ .

Solution 2

Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by $9$ . From $1$ thru $9$ , the sum is $45$ . $10$ thru $19$ , the sum is $55$ , $20$ thru $29$ is $65$ , and $30$ thru $39$ is $75$ . Thus the sum of the digits is $amc竞赛$ , and thus $N$ is divisible by $9$ . Now, refer to the above solution. $N \equiv 4 \text{ (mod 5)}$ and $N \equiv 0 \text{ (mod 9)}$ . From this information, we can conclude that $N \equiv 54 \text{ (mod 5)}$ and $N \equiv 54 \text{ (mod 9)}$ . Therefore, $N \equiv 54 \text{ (mod 45)}$ and $N \equiv 9 \text{ (mod 45)}$ so the remainder is $\boxed{\textbf{(C) }9}$

Problem 24

The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution

WLOG, let the centroid of $\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $美国数学竞赛$ , so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$ , then by Law of Cosines, $AB = 2\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$ . Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$ , so the square of the area of the triangle is $\boxed{\textbf{(C) } 108}$ .

Solution 2

WLOG, let the centroid of $\triangle ABC$ be $G = (-1,-1)$ . Then, one of the vertices must be the other curve of the hyperbola. WLOG, let $A = (1,1)$ . Then, point $B$ must be the reflection of $C$ across the line $y=x$ , so let $B = (a,\frac{1}{a})$ and $C=(\frac{1}{a},a)$ , where $a <-1$ . Because $G$ is the centroid, the average of the $x$ -coordinates of the vertices of the triangle is $-1$ . So we know that $a + 1/a+ 1 = -3$ . Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$ . So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$ . So $BC=2\sqrt{6}$ , and finding the square of the area gives us $\boxed{\textbf{(C) } 108}$ .