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考题11-12 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 11

The ratio of the length to the width of a rectangle is $4$ : $3$ . If the rectangle has diagonal of length $d$ , then the area may be expressed as $kd^2$ for some constant $k$ . What is $k$ ?

$考题11-12 2015 AMC 10A$

Solution

Let the rectangle have length $4x$ and width $3x$ . Then by $3-4-5$ triangles (or the Pythagorean Theorem), we have $d = 5x$ , and so $x = \dfrac{d}{5}$ . Hence, the area of the rectangle is $3x \cdot 4x = 12x^2 = \dfrac{12d^2}{25}$ , so the answer is $\boxed{\textbf{(C) }\frac{12}{25}}$

Problem 12

Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$ ?

$考题11-12 2015 AMC 10A$

Solution 1

Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$ .

$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$

$考题11-12 2015 AMC 10A$

$y^2 - 2\pi y + \pi^2 = 1$

$(y-\pi)^2 = 1$

$y-\pi = \pm 1$

$y = \pi + 1$

$y = \pi - 1$

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$ . The order does not matter because of the absolute value sign.

$| (\pi + 1) - (\pi - 1) | = 2$

The answer is $\boxed{\textbf{(C) }2}$

Solution 2

This solution is very related to Solution #1 but just simplifies the problem earlier to make it easier.

$y^2 + x^4 = 2x^2 y + 1$ can be written as $考题11-12 2015 AMC 10A$ . Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$ . This gives us two equations: