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AMC数学竞赛真题2016年12A 15-16

2018-09-01 重点归纳

AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 15

Circles with centers $P, Q$ and $R$ , having radii $1, 2$ and $3$ , respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$ , respectively, with $Q'$ between $P'$ and $R'$ . The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$ ?

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Solution 1

$美国数学竞赛$ Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR']$ , so $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']$ $\break$

$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}$ . Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$ . Therefore, $amc竞赛$ . Similarly, $[Q'QRR']=5\sqrt6$ . We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$ . $[P'PRR']=4\sqrt{2}+4\sqrt{6}$ . $\newline$

Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]$ . $\newline$

$[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}$ .

Solution 2

Let the center of the first circle of radius 1 be at (0, 1).

Draw the trapezoid $PQQ'P'$ and using the Pythagorean Theorem, we get that $P'Q' = 2\sqrt{2}$ so the center of the second circle of radius 2 is at $(2\sqrt{2}, 2)$ .

Draw the trapezoid $QRR'Q'$ and using the Pythagorean Theorem, we get that $Q'R' = 2\sqrt{6}$ so the center of the third circle of radius 3 is at $(2\sqrt{2}+2\sqrt{6}, 3)$ .

Now, we may use the Shoelace Theorem!

$(0,1)$

$(2\sqrt{2}, 2)$

$(2\sqrt{2}+2\sqrt{6}, 3)$

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$= \sqrt{6}-\sqrt{2}$ $\fbox{D}$ .

Problem 16

The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?

$amc 真题$

Solution

Setting the first two equations equal to each other, $\log_3 x = \log_x 3$ .

Solving this, we get $\left(3, 1\right)$ and $amc美国数学竞赛$ .

Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$ .

Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$ .

Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$ .

Pairing the second and fourth equations will yield $x = 1$ , but since you can't divide by $\log 1 = 0$ , it doesn't work.

After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{\textbf{(D)} 5}$

Solution 2

Note that $\log_b a =\log_c a / \log_c b$ .

Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$

$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$

$\log_\frac{1}{b} a = -\log_a b$

Therefore, the system of equations can be simplified to:

$y = t$

$y = -t$

$y = \frac{1}{t}$

$y = -\frac{1}{t}$

where $t = \log_3 x$ . Note that all values of $t$ correspond to exactly one positive $x$ value, so all $(t,y)$ intersections will correspond to exactly one $(x,y)$ intersection in the positive-x area.