Vol. 11 No 1 - Pi Mu Epsilon

. '
62 CLAYTON W. DODGE
D5 = (312 + x - y) 2 + (512 + y - x) 2 , and
D~ = (7 12 +X - y) 2 + ((112) - x + y) 2 .
Using our Observation, we see that the optimum distance occurs when D 1 = D 2 , when
x + Y = ~. Thus M and N will lie on the diagonals through A and C' respectively.
D A A' D'
M4
y
X
w z
c B B'
Nt
C'
A y A'
M3 X
D c
C'
B
A
c
B
~ w
z N2
D2
y X /1
/'
M2
D4 // D3
D
D'
//
..... -::-4-::-
/ z
//
~/
wNJ
z
w
--I
X M1
Dl
y
A A' B'
FIG. 6. (by Rex Wu)
Substitute x = ~ - y into the expression for D 3 to get D1 = 8y2 + 8.
Again use our Observation to set D 1 = D 3 to get 2y 2 + 2y - 1 = 0, so that
y = ( v'3 - 1) 12 and x = 1 - v'312.
Finally,
Di = (1 - 2y) 2 + 3 2 = (2 + VJ) 2 + 3 2 = 16 - 4VJ,
so that D1 = J16 - 4J3 ~ 3.011942358.
We can use this same technique in a 1 x 1 x n box to get that the final coordinates
for NI and N are
(x ) = (n - Jn2=l 1 - n + Jn2=1)
,y 2 ' 2
Also solved by Rex H. Wu, part (a) , Brooklyn, NY, and the Proposer. Two incorrect
solutions were also received. The proposer has written an unpublished paper entitled "Kotani's Ant
Problem," which generalizes the problem to a 1 x ax b box with rather interesting results.
r
D'
B'
A
C'
PROBLEM DEPARTMENT 63
952. [Fal11998] Proposed by Peter A. Lindstrom, Batavia, New York.
Let A, B, C denote the measures of the angles and a, b, c the lengths of the
opposite sides of a triangle. Show that
sin A sin B + sin B sin C + sin C sin A
(a + b + c) ( b + c - a) ( c + a - b) (a + b - c) (be + ca + ab)
4a 2 b 2 c 2
I. Solution by Kevin P. Wagner, student, University of South Florida, Saint
Petersburg, Florida.
By the laws of sines and cosines, we have (sin A) I a = (sin B) I b = (sin C) I c and
cosC = (a 2 + b 2 - c 2 )12ab, so that
. 2c 1 a+ -c
Slll = - b
(
2 b2 2)2
2a
(a+ b + c)(a + b- c)(a + c- b)(b + c- a)
4a 2 b 2
Substituting this value into the right side RHS of the desired equation yields
RHS = ab + ac +be sin2 C = absin 2 C + asin 2 C + bsin 2 C = LHS
c2 c2 c c
by using the law of sines.
II. Solution by Grand Valley State University Problem Group, Grand Valley
State University, Allendale, Michigan.
The area K of the triangle is given by K = (ll2)bcsinA, so that
. A 2K
sm =bc
and two similar expressions for sin B and sin C. Recall Heron's formula,
Now we have
K 2 = (a + b + c) (a + b - c) ( b + c - a) ( c + a - b)
16
sin A sin B + sin B sin C + sin C sin A
4K 2 4K 2 4K 2
= a2bc + ab2c + abc2
4K 2 (ab +be+ ac)
a2b2c2
(a + b + c) (a + b - c) ( b + c - a) ( b + c - a) ( ab + be + ac)
4a 2 b 2 c 2
Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain, Scott H. Brown,
Auburn University at Montgomery, AL, Paul S. Bruckman, Berkeley, CA, William Chau,
AT&T Laboratories, Middletown, NJ, Chantel Cleghorn, Hardin Dunham, and Daniel
H ermann, Angelo State University, San Angelo, TX, Erin Cooper, Alma College, MI, Russell
Euler and Jawad Sadek, Northwest Missouri State University, Maryville, George P. Evanovich,
Saint Peter's College, Jersey City, NJ, Mark Evans, Louisville, KY, Richard I. Hess, Rancho
I ,alas Verdes, CA, Joe Howard, New Mexico Highlands University, Las Vegas, Edward John
K oslowska, Southwest Texas Junior College, Eagle Pass, Henry S. Lieberman, Waban, MA,
David E. Manes, SUNY College at Oneonta, NY, Yoshinobu Murayoshi, Okinawa, Japan,
William H. Peirce, Rangeley, ME, Shiva K. Saksena, University of North Carolina at Wilmiugton,
H.-J. Seiffert, Berlin, Germany, Rex H. Wu, Brooklyn, NY, Monte J. Zerger, Adams
Sl.atc College, Alamosa, CO, and the Proposer.
Late solution to Problem 934 by Andrew Ostergaard, student, Hopatcong, NJ.