Skip to main content

2012 AMC 12B Problem 10

Problem 10 of 25EasierAlgebraGeometry

What is the area of the polygon whose vertices are the points of intersection of the curves x2+y2=25x^2 + y^2 = 25 and (x4)2+9y2=81?(x - 4)^2 + 9y^2 = 81?

Answer choices

Show solution

Solution

From x2+y2=25x^2+y^2=25 we get y2=25x2.y^2=25-x^2. Substituting into (x4)2+9y2=81(x-4)^2+9y^2=81 gives x2+x20=0,x^2+x-20=0, so x=4x=4 or x=5.x=-5. The intersection points are (5,0),(-5,0), (4,3),(4,3), and (4,3).(4,-3). The vertical side from (4,3)(4,3) to (4,3)(4,-3) has length 6,6, and the horizontal distance to (5,0)(-5,0) is 9,9, so the area is 1269=27.\tfrac12\cdot6\cdot9=27. Thus, the correct answer is B.

More practice

Concepts: system of equations · substitution · triangle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.