1969 AMC 12 Problem 11

Attempt Problem 11 of the 1969 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 1969 AMC 12 solutions, or check the answer key.

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

11.

Given points

P(-1,-2)

and

Q(4,2)

in the

xy

-plane, point

R(1,m)

is taken so that

PR+RQ

is a minimum. Then

m

equals:

-\dfrac35

-\dfrac25

-\dfrac15

\dfrac15

either

-\dfrac15

\dfrac15

The triangle inequality is sharp when

P,

R,

and

Q

are collinear

Find the point on line

PQ

whose

x

-coordinate is

1

By the triangle inequality,

PR+RQ\ge PQ,

with equality when

R

lies on segment

PQ.

The slope of

PQ

\frac{2-(-2)}{4-(-1)}=\frac{4}{5}.

Moving from

x=-1

x=1

raises the

y

-coordinate by

(\frac{4}{5})(2)=\frac{8}{5},

m=-2+\frac85=-\frac25.

Therefore, the correct answer is B.

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