Solutions to Problem Set 6, Math 461, Spring 2010
D
G
C
H
F
A
E
B
Problem 1 (1H.3) Let ABCD be an arbitrary quadrilateral. Show that the
midpoints of the four sides are the vertices of a parallelogram
Solution Let E be the midpoint of AB, F the midpoint of BC, and so forth,
as in the figure at left. Construct the segment BD. By Corollary 1.31 in the
text, we see F G k BD and EH k BD. Thus EH k F G. Furthermore, we also
get F G = BD/2 = EH. Theorem 1.8 shows that EF GH is a parallelogram.
A
Problem 2 (1H.7) In △ABC draw line segments AQ and BP where P and Q lie
on sides AC and BC, respectively. Now draw P Y parallel to AQ and QX parallel
to BP , where X and Y lie on AC and BC, as shown in the figure at right. Show
that XY k AB.
Solution Since P Y k AQ, we get △CY P ∼ △CQA. Similarly, △CXQ ∼ △CP B.
Thus we have CY /CQ = CP/CA and CX/CP = CQ/CB. Then we get
CX
CX CP
CQ CY
CY
=
=
=
.
CA
CP CA
CB CQ
CB
P
X
B
Q Y
C
By Theorem 1.33, we get △CXY ∼ △CAB. Hence 6 A ∼
= 6 CXY , so AB k XY .
X
O
Y
O′
Problem 3 (1H.10) Given a circle centered at O and an arbitrary point P , consider
the locus of all points Y that occur as midpoints of segments P X, where X lies on
P the given circle. Show that this locus is a circle with radius half that of the original
circle. Locate the center of the locus.
Solution Let O′ be the midpoint of OP . Let r be the radius of the given circle. Since X is the midpoint
of P Y , we use Corollary 1.31 to see that O′ X = OY /2 = r/2, independent of X. Thus Y lies on a circle
centered at O′ with half the radius of the original circle.
1