Solutions - Georg Mohr-Konkurrencen

strictly convex on (0, ∞) . Consequently,
1
cos γ =
�
1 + tan2 � �
tan α + tan β
γ = f(tan γ) = f
<
2
< f(tan α) + f(tan β)
=
2
1
� �
1 1
+ =
2 cos α cos β
1
cos δ ,
and hence γ < δ .
Remark. The use of calculus can be avoided. We only need the midpointconvexity
of f , i.e., the inequality
�
1 + 1
4 (u + v)2 < 1�
1�
1 + u2 + 1 + v2 2
2
for u, v > 0 and u �= v , which is equivalent (via squaring) to
1 + uv < � (1 + u 2 )(1 + v 2 ) .
The latter inequality reduces (again by squaring) to 2uv < u 2 + v 2 , holding
trivially.
Alternative solution. Draw a unit segment OP in the plane and take points
A and B on the same side of line OP so that � P OA = � P OB = 90 ◦ ,
� OP A = α and � OP B = β (see Figure 1). Then we have |OA| = tan α,
|OB| = tan β , |P A| = 1
cos α
Q
B
C
and |P B| = 1
cos β .
Figure 1
8
β
◆
α
A O
P