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
WLOG, let the centroid of be . The centroid of an equilateral triangle is the same as the circumcircle. It follows that the circumcircle must hit the graph exactly three times. Therefore, , so , so since is isosceles and , then by Law of Cosines, . 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 vertex 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 .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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