Solutions to Problem Set 7, Math 461, Spring 2010
A=X
Problem 1 (2A.2) Given △ABC and △XY Z, assume AB = XY and that 6 C
and 6 Z are supplementary. Prove without using the extended law of sines that
the two triangles have equal circumradii.
Solution But a suitable rotation and a reflection, if needed, we can get △XY Z
and △ABC juxtaposed as in the figure at left. Then Z must be on the circle, as
C
B=Y
if Z is not on the circle, then 6 Z and 6 C are not supplementary. Hence △XY Z
and △ABC have congruent circumscribing circles, and thus the circumradii are
equal.
Z
A
Problem 2 (2A.3) Given acute angled △ABC, extend the altitudes from A, B,
Y
and C to meet the circumcircle at points X, Y , and Z, respectively. Show that lines
AX, BY , and CZ bisect the three angles of △XY Z.
Solution Let Z ′ be the point on CZ that intersects AB, and let Y ′ be the point on
BY that intersects AC. Then △ABY ′ and △ACZ ′ are right triangles. Now we see
C
B
6 CAZ ′ ∼
= 6 BAY ′ as they are the same angle. Then
Z
6
CAZ ′ + 6 ACZ ′ = 6 BAY ′ + 6 ABY ′ =
π
,
2
X
so 6 ABY = 6 ABY ′ = 6 ACZ ′ = 6 ACZ. Since 6 AXZ and 6 ABZ subtend the same arcs, we get 6 AXZ =
6 ABZ, and similarly, 6 ACY = 6 AXY . Then
6
AXZ = 6 ABZ = 6 ACY = 6 AXY,
so AX bisects 6 Y XZ. Similarly, CZ bisects 6 XZY and BY bisects 6 XY Z.
H
Problem 3 (2A.5) In the figure at left, altitude AD of △ABC has been extended to
A
meet the circumcircle at point X. Point H is chosen on line AD so that HD = DX.
P
Show that the reflection of H in line AC lies on the circle. The solution of 2A.4 is
contained in the solution below.
D
B
C
Solution First we do the proof for when A is an obtuse angle, as shown by the figure
in the text and at left. Construct the segment BX. Then DH = HX and 6 BDX
is a right angle, so by SAS, △BDX ∼
= △BDH. Thus 6 HBC ∼
= 6 XBC. Let P
∼
d since they are arcs
d
XC,
be
the
point
where
BH
crosses
the
circle.
Then
P
AC
=
X
subtended by equal angles. Then 6 HBD ∼
= 6 CAX, as they subtend congruent arcs
of the same triangle. Let A′ be the intersection of AC and BH. Since △ACD is a right triangle, we get
6
BA′ C = π − 6 A′ CB − 6 CBA′ = π − 6 ACD − 6 CAD = π/2.
So 6 BA′ C is a right angle. Since △HBD is a right triangle, we further conclude by that 6 BHD ∼
= 6 ACD.
Since AP BD is an inscribed quadrilateral, we conclude 6 AP B and 6 ACB are supplementary angles. Since
6 AP B and 6 HP A are supplementary, we get
6
P HA ∼
= 6 ACB ∼
= 6 HP A.
By the converse of pons asinorum, △AP H is isosceles with base HP . We thus conclude P is the reflection
of H in the line AC.
A
P
In the case where 6 A is acute, we get a figure as in the diagram at right. The
d
proof above goes through mostly verbatim, but we replace “Pd
AC” with “P
C”; we
get 6 AHP ∼
= 6 BHD by vertical angles, and 6 AP B ∼
= 6 ACB as they subtend the
H
same arc; we conclude
D
B
C
∼ 6 ACB =
∼ 6 BHD =
∼ 6 AHP,
6 AP H =
so △AHP is again isosceles. Thus P is again the reflection of H in AC. If 6 A is a
right angle, then A = H = P , so the conclusion is trivial.
1
X