1977 AHSME Problems/Problem 19

Problem 19

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

$\textbf{(A) }PQRS\text{ is a parallelogram}\\ \textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}\\ \textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}\\ \textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}\\ \textbf{(E) }\text{none of the above are true}$

Solution

Let $L$ , $M$ , $N$ , and $O$ be the intersections of $AC$ and $PS$ , $BD$ and $PQ$ , $CA$ and $QR$ , and $DB$ and $RS$ respectively. Since the circumcenter of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let $\angle SPQ=\alpha$ . Then $\angle LEM=180^\circ-\alpha$ since $\angle PLC=PMD=90^\circ$ . This means $\angle DEC=180^\circ-\alpha$ , which similarly implies $\angle QRS=\alpha$ . Through a similar method, we find that $\angle PSR=\angle PQR=180^\circ-\alpha$ .

Since the opposite angles of $PQRS$ are congruent, $PQRS$ is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral $ABCD$ was. Therefore, $PQRS$ is always a parallelogram, and so our answer is $\fbox{A}$ .

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1977 AHSME Problems/Problem 19

Problem 19

Solution

See Also