AIME geometry walkthrough
Hard AIME Geometry Problems Solved: A Worked Example
When a geometry diagram feels complicated, coordinates can turn lengths into simple area calculations. Try this AIME-style problem before reading the solution. It is an original practice example, not a past contest question.
Problem
Square ABCD has side length 10. Point E lies on side AB with AE = 6, and point F lies on side AD with AF = 8. Find the area of triangle CEF.
Solution 1: Set coordinates
Place A = (0, 0), B = (10, 0), D = (0, 10), and C = (10, 10). Then E = (6, 0) and F = (0, 8). Using vectors from E, EC = (4, 10) and EF = (−6, 8).
The absolute value of their 2-by-2 determinant is the area of the parallelogram they span:
|4·8 − 10·(−6)| = 92.
Triangle CEF is half that parallelogram, so its area is 92/2 = 46. The answer is 46.
Solution 2: Subtract the corner triangles
Subtract the three regions outside CEF from the square. Triangle AEF has area (1/2)(6)(8) = 24. Triangle BCE has base BE = 4 and height 10, so its area is 20. Triangle DCF has base DC = 10 and height 10 − 8 = 2, so its area is 10. Therefore, the area of CEF is 100 − 24 − 20 − 10 = 46.
What to learn from the setup
A coordinate choice is useful when the figure has perpendicular sides, known lengths, and an area target. Put the right-angle corner at the origin and align the square with the axes; this makes each point’s coordinates immediate. Check the result by estimating from the diagram: 46 is a little under half the square’s area, which is plausible here.