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AMC数学竞赛真题2017年10B 7-8

2018-08-29 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 7

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$美国数学竞赛$

Solution 1

Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\frac{2x}{2}$ , or $x$ . $\[\]$ She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\frac{x}{5}$ hours. $\[\]$ The total time is $\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}$ . This is equal to $amc数学竞赛$ of an hour. Solving for $x$ , we have: $\[\]$ $\frac{22x}{85} = \frac{11}{15}$ $\frac{2x}{85} = \frac{1}{15}$ $30x = 85$ $6x = 17$ $\[\]$ Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\boxed{\bold{(C)} 2.8}$ .

Solution 2 (Quick, Make Approximations)

Notice that Samia walks $\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is $\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}$ Then, multiply this by the time34 minutes is a little greater than $\frac {1}{2}$ of an hour so Samia traveled $\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}$ The answer choice a little greater than 2.5 is $\boxed{\bold{(C)} 2.8}$ . (Note that we could've multiplied $\frac {34}{60}=\frac {17}{30}$ by $5$ and gotten the exact answer as well)

Problem 8

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$ ?

$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solutions

$[asy] pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); [/asy]$

Solution 1

Since $AB = AC$ , then $\triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}$ .

Solution 2

Calculating the equation of the line running between points $B$ and $D$ , $y = -2x + 1$ . The only coordinate of $C$ that is also on this line is $\boxed{\textbf{(C) } (-4,9)}$ .