1987 AIME Problem 4

Attempt Problem 4 of the 1987 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 1987 AIME solutions, or check the answer key.

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Find the area of the region enclosed by the graph of

|x-60|+|y|=|\frac{x}{4}|.

Answer

Solve for

|y|

and determine where its right-hand side is nonnegative

The boundary is a kite whose vertices occur at the breakpoints of the absolute values

We need

|y|=\frac{|x|}{4}-|x-60|\geq0.

This forces

48\leq x\leq80.

x=48,

x=60,

x=80,

the boundary values are respectively

y=0,

y=\pm15,

y=0.

Hence the region is a kite with perpendicular diagonals

80-48=32

and

30,

so its area is

\frac12(32)(30)=480.

Problem 4 in other years