1985 AIME Problem 11

Attempt Problem 11 of the 1985 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

An ellipse has foci at (9,20)(9,20) and (49,55)(49,55) in the xyxy-plane and is tangent to the xx-axis. What is the length of its major axis?

Answer: 85
Concepts:ellipsereflection (geometry)distance formula
Difficulty rating: 2360
Small Hint:

At tangency, the constant sum of distances is the minimum such sum for a point on the xx-axis

Big Hint:

Reflect one focus across the xx-axis to turn the broken path into a straight segment

Solution:

Reflect (49,55)(49,55) across the xx-axis to (49,55).(49,-55). For a point PP on the xx-axis, the sum of its distances to the original foci equals the length of a broken path from (9,20)(9,20) through PP to (49,55).(49,-55). Its minimum is the straight-line distance (499)2+(5520)2=402+752=85. \begin{aligned} &\sqrt{(49-9)^2+(-55-20)^2}\\ &\qquad{}=\sqrt{40^2+75^2}=85. \end{aligned} Tangency means that the ellipse’s constant distance sum equals this minimum. That sum is the major-axis length, so the answer is 85.85.

← Problem 10#10
Full Exam

Problem 11 in Other Years