2017 AMC 10B Problems/Problem 24
Contents
Problem 24
The vertices of an equilateral triangle lie on the hyperbola , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
Solution 1
WLOG, let the centroid of be . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, , so , so since is isosceles and , then by Law of Cosines, . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to . Therefore, the area of the triangle is , so the square of the area of the triangle is .
Solution 2
WLOG, let the centroid of be . Then, one of the vertices must be the other curve of the hyperbola. WLOG, let . Then, point must be the reflection of across the line , so let and , where . Because is the centroid, the average of the -coordinates of the vertices of the triangle is . So we know that . Multiplying by and solving gives us . So and . So , and finding the square of the area gives us .
Solution 3
WLOG, let the centroid of be and let point be . It is known that the centroid is equidistant from the three vertices of . Because we have the coordinates of both and , we know that the distance from to any vertice of is . Therefore, . It follows that from , where and , using the formula for the area of a triangle with sine . Because and are congruent to , they also have an area of . Therefore, . Squaring that gives us the answer of .
Solution 4 (5-second solution)
WLOG, let the centroid of the triangle be . By symmetry, the other vertex is . The distance between these two points is , so the height of the triangle is , the side length is , and the area is , yielding an answer of . -Stormersyle
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.