1993 AMC 12 Problem 25

Question

Let

S

be the set of points on the rays forming the sides of a

120^\circ

angle, and let

P

be a fixed point inside the angle on the angle bisector. Consider all distinct equilateral triangles

PQR

with

Q

and

R

in

S.

(Points

Q

and

R

may be on the same ray, and switching the names of

Q

and

R

does not create a distinct triangle.) There are

exactly

2

such triangles.

exactly

3

such triangles.

exactly

7

such triangles.

exactly

15

such triangles.

more than

15

such triangles.

Answer: E

Concepts:equilateral triangles congruence infinite family

Difficulty rating: 2310

Small hint:

Call the angle’s vertex

O

and choose

A

on one ray so that

\triangle AOP

is equilateral

Big hint:

Vary a point

Q

continuously on

\overline{AO}

and construct a matching point

R

on the other ray

Solution:

Let

O

be the vertex. Choose

A

on one ray so that

OA=OP;

because

OP

bisects the

120^\circ

angle,

\triangle AOP

is equilateral. For any

Q

on

\overline{AO},

choose

R

on the other ray with

OR=AQ.

Then

\triangle POR\cong\triangle PAQ

by SAS. Consequently

PQ=PR

and

\angle QPR=\angle APO=60^\circ,

so

\triangle PQR

is equilateral. Continuously many choices of

Q

give distinct triangles, hence certainly more than

15.

Thus the correct answer is E.

1993 AMC 12 Problem 25

Answer: E

Difficulty rating: 2310

Small hint:

Big hint:

Solution:

Problem 25 in other years