1987 AMC 12 Problem 25

Attempt Problem 25 of the 1987 AMC 12 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 AMC 12 solutions, or check the answer key.

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25.

ABC

is a triangle:

A=(0,0),

B=(36,15),

and both the coordinates of

C

are integers. What is the minimum area

\triangle ABC

can have?

\frac12

1

\frac32

\frac{13}{2}

there is no minimum

Write

C=(u,v)

and use the determinant formula for triangle area

Find the smallest positive value of

|36v-15u|

using

\gcd(36,15)

For

C=(u,v),

the area is

\frac12|36v-15u|.

Since

\gcd(36,15)=3,

the smallest possible positive value of the determinant is at least

3.

It is attained, for example, by

u=7,\ v=3,

since

36(3)-15(7)=3.

Therefore the minimum area is

\frac{3}{2}.

Thus the correct answer is C.

Problem 25 in other years