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 n 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
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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