Vol. 11 No 1 - Pi Mu Epsilon

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54 CLAYTON W. DODGE
For x E [0, oo ), we have xe- x < 1, whence
Also solved by Paul S. Bruckman, Berkeley, CA, Kenneth B. Davenport, Frackville, PA,
Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Robert C.
Gebhardt, Hopatcong, NJ, Richard I. Hess (who gave a 42 decimal place answer), Rancho Palos
Verdes, CA, Joe Howard, New Mexico Highlands University, Las Vegas, and the Proposer.
*945. [Fa111ggs] Proposed by the late Jack Garfunkel, Flushing, New York.
Let A, B, C be the angles of a triangle and A' ,B' , C' those of another triangle
with A 2: B 2: C, A > C, A' 2: B' 2: C' , and A' > C' . Prove or disprove that if
A- C 2: 3(A' - C'), then E cos(A/2) ~ E sin( A').
Solution by Paul S. Bruckman, Berkeley, California.
We shall show that the coefficient 3 is unnecessary, that the stated conclusion is
true whenever A- C 2: A'- C' and the rest of the hypothesis is true. Let U, V, W
be arbitrary numbers and define R(U, V, W) = cos(U /2) + cos(V /2) + cos(W/2) and
S(U, V, W) = sin(U) + sin(V) + sin(W).
We prove the following identity by using trigonometric formulas on its product
term to replace the product of two sine factors by a sum and then the resulting two
products of a sine and a cosine by sums,
(1) S(U, V, W) = sin(U + V + W) + 4sin(U; V )sin( V ~ W )sin( W; U ).
PROBLEM DEPARTMENT 55
Clearly, Fvv < 0 in its permissible domain. For any fixed B , since FD is a decreasing
function of D, it suffices to show that limD-+O+ Fv(D, B) ::; 0. We have
lim Fv(D, B)= (3- B + 2B- 3 - B) cos(7r/3)) = 0.
D-+O+
This implies that Fv(D, B) < 0 for all permissible D and B. Therefore, S::; S' .
946. [Fall 1ggs] Proposed by Ayoub B. Ayoub, Pennsylvania State University,
Abington College, Abington, Pennsylvania.
Let M be a point inside (outside) triangle ABC if LA is acute (obtuse) and let
mLM BA + mLMCA =goo .
a) Prove that (BC · AM) 2 = (AB · CM? + (CA · BM) 2 •
b) Show that the Pythagorean theorem is a special case of the formula of part
(a).
Solution by the Proposer.
We consider the case where LA is acute. From point M drop perpendiculars M X,
MY, and M Z onto sides BC, AB, and C A respectively, as shown in the figure. Now
M, X, B, andY lie on a circle with diameter d = BM. Then
(2) XY = dsinB = BMsinB
and similarly,
(3) ZX = CMsinC and YZ = AMsinA.
c
Note that
R(A, B, C) = cos ('' - ~ - C) + cos ('' - ~ - A) + cos ('' - ~ - B)
=S(B+C C+A A+B)
2 ' 2 ' 2
= sin( A+ B +C) +
= sin,- +4sin (";A) sin (";B) sin (";C) = S(A, B, C)
by identity 1.
LetS = S(A, B, C) and S' = S'(A', B', C'). We need to show that S ~ S' under
the revised hypotheses. Set A-C = 6D, so that 0 < D < 1r /6. Then there is a B with
- 1 ~ B ~ 1 such that A= 7r/3 + (3- B)D, B = 7r/3 + 2BD, and C = 7r/3- (3 + B)D.
Let F(D, B) = S(A, B , C). It suffices to show that Fv(D, B) < 0 for all permissible
D and B. Since F is analytic in either variable, we have
and
FD(D, B) = (3- B) cos( A) + 2B cos(B) - (3 +B) cos( C)
Fvv(D, B) = -(3 - B) 2 sin(A) - 4B 2 sin(B) - (3 + B) 2 sin( C).
A
Recall that the law of sines, as applied to triangle ABC, states
(4)
sin A
BC
sinB
CA
y
sinC
AB
Since LMBY = LMXY and LMCZ = LMXZ, as shown in the figure by the spotted
angles and the triangle-marked angles, then LYXZ = LMBY +LMCZ = goo . Hence
(YZ)2 = (XY) 2 + (ZX) 2 , into which we substitute equations 2 and 3, and then 4
and simplify to get
(AMsinA) 2 = (BMsinB) 2 + (CMsinC) 2
B