Geometry and Trigonometry 627P=
k+ 2 r
3 kA+
2 k− 2 r
3 kD
=
k−r−3 cosθ
3 kB+
2 k+r+3 cosθ
3 kE
=
k−r+3 cosθ
3 kC+
2 k+r−3 cosθ
3 kF.
An algebraic computation shows thatAD=BE=CF =k,soPis an equicevian
point, andPDAP =(^2 (kk+− 22 r)r)is independent ofA.
To find the other equicevian point note that if we replacekby−kandθby−θ, then
Aremains the same. In this new parametrization, we have the points
D′=((r−k)cosθ, 0 ),E′=(
( 2 k^2 −rk− 3 )cosθ−k−r
r− 2 k+3 cosθ,
k( 2 r−k)sinθ
r− 2 k+3 cosθ)
,
F′=
(
( 2 k^2 −rk− 3 )cosθ+k+r
r− 2 k−3 cosθ,
k( 2 r−k)sinθ
r− 2 k−3 cosθ)
,
P′=
(
r− 2 k
3cosθ,
k− 2 r
3sinθ)
.
Of course,P′is again an equicevian point, andAP′
P′D′=( 2 k+ 2 r)
(k− 2 r), which is also independent
ofA. Whenr =0, the pointsPandP′are distinct, since sinθ =0. Whenr=0, the
two pointsPandP′coincide whenA=S, a case ruled out by the hypothesis. Asθ
varies,PandP′trace an ellipse. Moreover, since
(
r± 2 k
3) 2
−
(
k± 2 r
3) 2
= 1 ,
this ellipse has foci atBandC.
(American Mathematical Monthly, proposed by C.R. Pranesachar)
614.The interesting case occurs of course whenbandcare not both equal to zero. Set
d =√
b^2 +c^2 and define the angleαby the conditions cosα=√b
b^2 +c^2and sinα=
√c
b^2 +c^2. The integral takes the form
∫
dx
a+dcos(x−α)
,
which, with the substitutionu=x−α, becomes the simpler
∫
du
a+dcosu