652 Geometry and Trigonometry
l− 0. 002 ≤PQ,PR,RP≤l+ 0. 002.On the other hand, since there is no equilateral triangle whose vertices have integer
coordinates, we may assume thatPQ =QR. Therefore,
|PQ^2 −QR^2 |=(P Q+QR)|PQ−QR|
≤((l+ 0. 002 )+(l+ 0. 002 ))((l+ 0. 002 )−(l− 0. 002 ))
≤ 2 × 96. 002 × 0. 004 < 1.However,PQ^2 −QR^2 is an integer. This contradiction proves the claim.
(short list of the 44th International Mathematical Olympiad, 2003)
655.Imagine instead that the figure is fixed and the points move on the cylinder, all
rigidly linked to each other. LetPbe one of thenpoints; when another point tracesS,
Pitself will trace a figure congruent toS. So after all the points have tracedS,Palone
has traced a surfaceFof area strictly less thann.
On the other hand, if we rotateParound the cylinder or translate it back and forth by
n
4 πr, we trace a surface of area exactly equal ton. Choose on this surface a pointP
′thatdoes not lie inF, and consider the transformation that mapsPtoP′. The fact thatP′
is not inFmeans that at this moment none of the points lies inS. This transformation,
therefore, satisfies the required condition.
(M. Pimsner, S. Popa,Probleme de geometrie elementara ̆(Problems in elementary
geometry), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1979) ̆
656.The left-hand side is equal to
cos 20◦sin 40◦−sin 10◦cos 10◦=2 sin 20◦cos^220 ◦−sin 20◦
2=1
2
(3 sin 20◦−4 sin^320 ◦)=1
2
sin 60◦=√
3
4
.
(Romanian Mathematical Olympiad, 1967, proposed by C. Ionescu- ̧Tiu)657.Because−π 2 <− 1 ≤sinx≤ 1 <π 2 , cos(sinx) >0. Hence sin(cosx) >0, and
so cosx>0. So the only possible solutions can lie in the interval(−π 2 ,π 2 ). Note that
ifxis a solution, then−xis also a solution; thus we can restrict our attention to the first
quadrant. Rewrite the equation as
sin(cosx)=sin(π
2
−sinx)
.
Then cosx =π 2 −sinx, and so sinx+cosx=π 2. This equality cannot hold, since
the range of the functionf(x)=sinx+cosx=
√
2 cos(π 4 −x)is[−√
2 ,
√
2 ], and
π
2 >