Geometry and Trigonometry 629X(t^2 X−tX−t+ 2 )= 0.The rootX=0 corresponds to the origin. The other rootX=tt 2 −−^2 t gives the desired
parametrization of the hyperbola by rational functions(tt 2 −−^2 t,t
(^2) − 2 t
t^2 −t),treal. So the problem
has little to do with number theory, and we only need to find the leftmost point on the
hyperbola that lies in the half-planeX≥1. Write the equation of the hyperbola as
(
Z−
X
2
) 2
−
(
X
2
− 2
) 2
= 6.
The center is at( 4 , 2 ), and the asymptotes areZ=2 andZ=X−2. Let us first minimize
Xfor the points on the hyperbola and in the half-planeX≥4. We thus minimize the
functionf(X,Z)=Xon the curveg(X, Z)=Z^2 −ZX−Z+ 2 X=0. The Lagrange
multipliers method gives
1 =λ(−Z+ 2 ),
0 =λ( 2 Z−X− 1 ).From the second equation we obtainZ=X+ 21. Substitute ing(X, Z)=0 to obtain
X = 3 ± 2
√
- The further constraintX ≥1 shows thatX = 3 + 2
√
2 gives the
minimum. The same argument shows that the other branch of the hyperbola lies in the
half-planeX<1, and so the answer to the problem ism= 3 + 2
√
2.
(short list of the 42nd International Mathematical Olympiad, 2001)616.We convert to Cartesian coordinates, obtaining the equation of the cardioid
√
x^2 +y^2 = 1 +
x
√
x^2 +y^2
,
or
x^2 +y^2 =√
x^2 +y^2 +x.By implicit differentiation, we obtain
2 x+ 2 y
dy
dx=(x^2 +y^2 )−^1 /^2(
x+y
dy
dx)
+ 1 ,
which yields
dy
dx=
− 2 x+x(x^2 +y^2 )−^1 /^2 + 1
2 y−y(x^2 +y^2 )−^1 /^2.
The points where the tangent is vertical are among those where the denominator cancels.
Solving 2y−y(x^2 +y^2 )−^1 /^2 =0, we obtainy=0orx^2 +y^2 =^14. Combining this