2003 AIME II Problems/Problem 14
Contents
Problem
Let and be points on the coordinate plane. Let be a convex equilateral hexagon such that and the y-coordinates of its vertices are distinct elements of the set The area of the hexagon can be written in the form where and are positive integers and is not divisible by the square of any prime. Find
Solution 1
The y-coordinate of must be . All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting , and knowing that , we can use rewrite using complex numbers: . We solve for and and find that and that .
The area of the hexagon can then be found as the sum of the areas of two congruent triangles ( and , with height and base ) and a parallelogram (, with height and base ).
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Thus, .
Solution 2
From this image, we can see that the y-coordinate of F is 4, and from this, we can gather that the coordinates of E, D, and C, respectively, are 8, 10, and 6.
Let the angle between the -axis and segment be , as shown above. Thus, as , the angle between the -axis and segment is , so . Expanding, we have
Isolating we see that , or . Using the fact that , we have , or . Letting the side length of the hexagon be , we have . After simplification we find that that .
In particular, note that by the Pythagorean theorem, , hence . Also, if , then , hence and thus . Using similar methods (or symmetry), we determine that , , and . By the Shoelace theorem,
Hence the answer is .
Note
By symmetry the area of is twice the area of . Therefore, you only need to calculate the coordinates of , , and .
Solution 3
This is similar to solution 2 but faster and easier. First off we see that the y coordinate of F must be 4, the y coordinate of E must be 8, the y coordinate of D must be 10, and the y coordinate of C must be 6 (from the parallel sides of the hexagon). We then use the sine sum angle formula to find the x coordinate of B (lets call it ): . Now that we know we can find the x coordinate of F in multiple ways, including using the cosine sum angle formula or using the fact that triangle AFE is isosceles and AE is on the y axis. Either way, we find that the x coordinate of F is . Now, divide ABCDEF into two congruent triangles and a parallelogram: AFE, BCD, and ABDE. The areas of AFE and BCD are each . The area of ABDE is . The total area of the hexagon is
Solution 4 (No Trig)
First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A and , respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B . This tells us that the area of the entire rectangle is , since the opposite sides are parallel and thus the length of the rectangle is . Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as . However, noticing that , the area of ABCDEF can also be expressed as . Now we just need to find . Since and degrees, . However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY). From triangle ABX we have that , so . Since AB=AF, we can also form the equation . We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us . Setting our two values of BF equal and substituting as and simplifying, we get the equation . Now we can use the quadratic formula to get that or , so . Plugging this value back into the equation , we get that . Now we get that is , so the area of the hexagon is , so the answer is
~ant08 and sky2025
Video Solution by Sal Khan
https://www.youtube.com/watch?v=Ec-BKdC8vOo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=4 - AMBRIGGS
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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