Solutions - Georg Mohr-Konkurrencen

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Let C be the midpoint of the segment AB . By hypothesis, we have
tan α + tan β
|OC| =
2
= tan γ , hence OP C = γ and |P C| = 1
cos γ .
Let Q be the point symmetric to P with respect to C . The quadrilateral
P AQB is a parallelogram, and therefore |AQ| = |P B| = 1
ally,
. Eventu-
cos β
2 1 1
= +
cos δ cos α cos β
and hence δ > γ .
Another solution. Set x =
and
= |P A| + |AQ| > |P Q| = 2 · |P C| = 2
cos γ ,
α + β
2
and y =
α − β
2
, then α = x+y , β = x−y
cos α cos β = 1
(cos 2x + cos 2y) =
2
= 1
2 (1 − 2 sin2 x) + 1
2 (2 cos2 y − 1) = cos 2 y − sin 2 x .
By the conditions of the problem,
and
tan γ = 1
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sin α sin β
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+ =
2 cos α cos β
1 sin(α + β)
·
2 cos α cos β
1
cos δ
1
�
1 1
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= + =
2 cos α cos β
1 cos α + cos β
·
2 cos α cos β
Using (3) we hence obtain
tan 2 δ − tan 2 γ =
showing that δ > γ .
= sin x cos x
cos α cos β
= cos x cos y
cos α cos β .
1
cos2 δ − 1 − tan2γ = cos2x cos2y − sin 2 x cos2x cos2α cos2 − 1 =
β
= cos2 x(cos 2 y − sin 2 x)
(cos 2 y − sin 2 x) 2
= cos2 x − cos 2 y + sin 2 x
cos 2 y − sin 2 x
9
− 1 =
=
cos2x cos2y − sin 2 − 1 =
x
sin 2 y
cos α cos β
> 0 ,
(3)