1977 AMC 12 Problem 19

Question

Let

E

be the point of intersection of the diagonals of convex quadrilateral

ABCD,

and let

P,

Q,

R

and

S

be the centers of the circles circumscribing triangles

ABE,

BCE,

CDE

and

ADE,

respectively. Then

PQRS

is a parallelogram

PQRS

is a parallelogram if and only if

ABCD

is a rhombus

PQRS

is a parallelogram if and only if

ABCD

is a rectangle

PQRS

is a parallelogram if and only if

ABCD

is a parallelogram

none of the above are true

Answer: A

Concepts:circumcircle, circumcenter, and circumradius perpendicular bisector parallelogram

Difficulty rating: 2220

Small hint:

Two adjacent circumcenters lie on the perpendicular bisector of the side shared by their triangles

Big hint:

Use the fact that

A,

E,

C

are collinear and

B,

E,

D

are collinear

Solution:

Both

P

and

Q

lie on the perpendicular bisector of

BE,

while both

R

and

S

lie on the perpendicular bisector of

DE.

Since

B,

E,

D

are collinear, these bisectors are parallel, so

PQ\parallel RS.

Likewise,

Q,

R

lie on the perpendicular bisector of

CE,

and

S,

P

lie on that of

AE.

Since

A,

E,

C

are collinear,

QR\parallel SP.

Therefore

PQRS

is always a parallelogram.

Therefore, the correct answer is A.

1977 AMC 12 Problem 19

Answer: A

Difficulty rating: 2220

Small hint:

Big hint:

Solution:

Problem 19 in other years