1985 AMC 12 Problem 3

Attempt Problem 3 of the 1985 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 1985 AMC 12 solutions, or check the answer key.

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

In right

\triangle ABC

with legs

5

and

12,

arcs of circles are drawn, one with center

A

and radius

12,

the other with center

B

and radius

5.

They intersect the hypotenuse in

M

and

N.

Then

MN

has length

2

\frac{13}{5}

3

4

\frac{24}{5}

First find the hypotenuse of the

5

12

-right triangle

Measure the positions of both arc intersections from

A

The hypotenuse has length

13.

Since

AM=12,

point

M

12

units from

A.

Also

BN=5,

AN=AB-BN=13-5=8.

Therefore

MN=AM-AN

=12-8=4.

Thus the correct answer is D.

Problem 3 in other years