Solutions to Problem Set 11, Math 461, Spring 2010
A
B
C
X
Problem 1 (2E.1) Suppose that the orthocenter and incenter of △ABC are the same
point. Prove that the triangle is equilateral.
Solution Let X be the point where the angle bisector of 6 A intersects BC. Since AX
contains the incenter, it also contains the orthocenter. Thus AX ⊥ BC. Therefore
△AXB ∼
= △AXC by ASA. So AB = AC. Similarly, we get BC = BA(= AC). So the
triangle is equilateral.
Problem 2 (2E.3) Show that in a right triangle, the inradius, circumradius, and semiperimeter are related
by the formula s = r + 2R.
Solution Let c by the length of the hypotenuse, and a and b the lengths
of the sides. Then R = c/2,
√
s = (a + b + c)/2, and r = ab/(a + b + c) from Problem 2.26. Since c = a2 + b2 , we get
r=
y
B
D
c
x
A
b
a
C
ab
a+b−c
ab(a + b − c)
=
=
= s − c = s − 2R.
(a + b + c)
(a + b)2 − c2
2
Problem 3 (2E.4) Let CD be an altitude of △ABC and assume that 6 C =
90◦ . Let r1 and r2 be the inradii of △CAD and △CBD, respectively, and
show that r + r1 + r2 = CD, where r is the inradius of △ABC.
Solution Let a = BC, b = AC, c = AB, x = AD, and y = BD. From the
solution of (2E.3) above, we get the following:
a+b−c
,
2
x + CD − b
,
r1 =
2
y + CD − a
.
r2 =
2
r=
Adding these three equations together and observing that c = x + y gives us the result.
1