1998 AMC 12 Problem 25

Attempt Problem 25 of the 1998 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 1998 AMC 12 solutions, or check the answer key.

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25.

A piece of graph paper is folded once so that

(0,2)

is matched with

(4,0),

and

(7,3)

is matched with

(m,n).

Find

m+n.

6.7

6.8

6.9

7.0

8.0

The crease is the perpendicular bisector of the segment joining

(0,2)

and

(4,0)

Find the line

y=2x-3

, then reflect

(7,3)

across it

The crease passes through

(2,1)

with slope

2,

so it is

y=2x-3.

The perpendicular through

(7,3)

has slope

-\frac{1}{2}

and meets the crease at

(\frac{19}{5},\frac{23}{5}).

This intersection is the midpoint of

(7,3)

and

(m,n),

giving

m=\frac{3}{5},\ n=\frac{31}{5}.

Hence

m+n=\frac{34}{5}=6.8,

so B is correct.

Problem 25 in other years