2015 AMC 12B Problems/Problem 25
A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactly inches away from , where , , and are positive integers and and are not divisible by the square of any prime. What is ?
Let , a counterclockwise rotation centered at the origin. Notice that on the complex plane is:
We need to find the magnitude of on the complex plane. This is an arithmetico-geometric series.
We want to find . First, note that because . Therefore
Hence, since , we have
Now we just have to find . This can just be computed directly:
Thus the answer is .
Here is an alternate solution that does not use complex numbers:
The distance from to can be evaluated using the Pythagorean theorem. Assuming lies at the origin, we can calculate the distance the bee traveled to by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:
A movement of units at degrees is the same thing as a movement of units at degrees, so we can adjust all the cosines with arguments greater than as follows:
Grouping terms with like-cosines and factoring out the cosines:
Each sum in the parentheses has terms (except the very last one, which has ). By pairing each term, we see there are pairs of . Therefore, each sum equals except the very last sum, which has pairs of plus an extra 2010 and equals . Plugging in these values:
We can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:
Finally, we use the Pythagorean to find the distance from . This distance is given by:
, so the answer is .
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 to the line intersecting a point . We will also place the point at the intersection of and . In addition, the point is placed at the perpendicular dropped from to the line . We will also set the distance and thus . With this perpendicular we see that the triangle is a 30-60-90 triangle. This means that the length and the length . We can also see that the triangle is a 30-60-90 triangle and thus and . Now if we continue this across all and set the point to the coordinates . 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
, , , , and finally .
Here the point is defined as the intersection of lines and . The point is defined as the intersection of lines and . Finally, the point is defined as the intersection of lines and . Note that our spiral stops at before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point , is . This is an interesting property that the points’ coordinate changes by a constant offset no matter what is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point . Using similar 30-60-90 triangle properties, we see that . 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 which gives us a final answer of .
Suppose that the bee makes a move of distance . After turns it will be facing the opposite direction and move units. Combining these opposite movements gives a total movement of units in the original direction. This means that every moves, the bee will move units in each direction of .
We want to find the displacement vector for every moves. Factoring out the for now (which flips the direction), we draw a quick diagram of one unit in each direction. Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the ) is .
To compute the distance to , we can compute the position of (a multiple of moves) and then subtract the vector from to .
The bee reaches after sets of moves, so the total displacement vector to is .
The bee moves at an angle of from to , so subtracting it means moving an angle of . Since the vector is units long, by a 30-60-90 triangle, it is .
Therefore the total displacement vector to is . The displacement is thus .
Fun fact: The displacement to is the same as for .
Solution 5 (big complex bash)
Let be the origin. East would be the real axis in the positive direction. Then we can assign each a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the th move the value of is . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.
Then we have and this becomes
Simplifying, we have which eventually simplifies to and this is a triangle which has ratios of so the magnitude is and the answer is .
After each 12 moves, the bee will be facing the same direction as it started. Let be the origin and let (with divisible by ) be . We now notice that each of the move pairs with lengths /, /, /, /, /, / will move the bee 6 units in the directions corresponding to the moves with lengths , , , , , and . This equates to moving the bee from to , a move that repeats every 12 moves. Since , we have that . It follows that so the distance to is so the answer is .
|2015 AMC 12B (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25|
|All AMC 12 Problems and Solutions|