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考题24-25 2015 AMC 12B

2018-08-06 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 24

Four circles, no two of which are congruent, have centers at $A$ , $B$ , $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $AR+BR+CR+DR$ ?

$考题24-25 2015 AMC 12B$

Solution

First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$ , with $O_1$ and $O_2$ lying on the same side of $PQ$ , and $O_1 O_2=39$ . Let $x=O_1 R$ , $y=O_2 R$ , and suppose that the radius of circle $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$ .

Then the power of point $R$ with respect to $C_1$ is

$考题24-25 2015 AMC 12B$

and the power of point $R$ with respect to $C_2$ is

$考题24-25 2015 AMC 12B$

Also, note that $x-y=39$ .

Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$ . As $x-y=39$ , we find that $r^2=64(x+y) = 64(2y+39)$ . Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$ . This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$ , and as $y$ is a length, it is positive, hence $y=57$ , and $x=y+39=96$ . This is the only possibility if the two centers lie on the same same of their radical axis.

On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that $O_1 R + O_2 R = O_1 O_2 = 39$ . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{\textbf{(D)}\; 192}$ .

Number-Intensive Solution

Start by drawing $PQ$ first, because trying to get all four circles down will take you a few years. Next, because all circles have $P$ and $Q$ on them, and since all points on a circle are equidistant from the center, all circle centers lie on the perpendicular bisector of $PQ$ , and point $R$ is on this bisector.

In order for all the circle radii to be different (because the circles can't be congruent), two circle centers are on the same side of $PQ$ , and two are straddling it. For the latter two circles- just call them $A$ and $B$ - clearly $AR+BR$ is 39.

Now, let's take the next case. Then $C$ and $D$ lie on the same side. Construct the triangles from your picture, and use the Pythagorean Theorem (maybe divide all lengths by 3 to lessen big numbers) and then you get that the distance from $R$ to the closest circle center is $57$ . Therefore, the answer is $39+2*57+39=192 \boxed{(D)}$ .

Problem 25

A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ , $b$ , $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?

$考题24-25 2015 AMC 12B$

Solution 1

Let $x = e^{i \pi / 6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:

$1 + 2x + 3x^2 + \cdots + (k+1)x^k$

We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetic/geometric series.

$考题24-25 2015 AMC 12B$

We want to find $|S|$ . First, note that $x^{2015} = x^{11} = x^{-1}$ because $x^{12} = 1$ . Therefore

$考题24-25 2015 AMC 12B$

Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$

Now we just have to find $|1-x|$ . This can just be computed directly:

$1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i$

$考题24-25 2015 AMC 12B$

Therefore $|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.$

Thus the answer is $考题24-25 2015 AMC 12B$

Solution 2

Here is an alternate solution that does not use complex numbers:

We will calculate the distance from $P_{2015}$ to $P_0$ using the Pythagorean theorem. Assume $P_0$ lies at the origin, so we will calculate the distance to $P_{2015}$ by calculating the distance traveled in the x-direction and the distance traveled in the y-direction. We can calculate this by summing each movement:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+\cdot \cdot \cdot+2011\cos{180}+2012\cos {210}+2013\cos{240}+2014\cos{270}+2015\cos{300}$

A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than 180 as follows:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdot \cdot \cdot -2015 \cos{120}$

Now we group terms with like-cosines and factor out the cosines:

$考题24-25 2015 AMC 12B$

Each sum in the parentheses has 336 terms (except the very last one, which has 335), so by pairing each term, we can see that there are $\frac {336}{2}$ pairs of $-6$ . So each sum evaluates to $168\cdot -6=-1008$ , except the very last sum, which has 167 pairs of $-6$ and an extra 2010, so it evaluates to $167\cdot -6+2010=1008$ . Plugging in these values:

$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $考题24-25 2015 AMC 12B$

Now that we have how far was traveled in the x-direction, we need to find how far was traveled in the y-direction. Using the same logic as above, we arrive at the sum:

$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$

$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$

The last step is to use the Pythagorean to find the distance from $P_0$ . This distance is given by:

$考题24-25 2015 AMC 12B$

Multiplying out, we have $1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6= \boxed {\bold {(B)}\; 2024}$ .

Solution 3

We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?

First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\overline{P_0P_1}$ and $\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\overline{P_3C_1}$ . We will also set the distance $\overline{P_0P_1} = n$ and thus $\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}$ and the length $\overline{C_1C_2} = \frac{n+1}{2}$ . We can also see that the triangle $\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\overline{C_0C_1} = \frac{n+2}{2}$ and $\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length

$考题24-25 2015 AMC 12B$

$\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}$

$\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}$

, and finally

$\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}$

Here the point $C_4$ is defined as the intersection of lines $\overline{P_3P_4}$ and $\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\overline{P_6P_7}$ and $\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\overline{P_{9}P_{10}}$ and $\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\sqrt{3})(\sqrt{2})$ which gives us a final answer of $\boxed{2024}$ .