| Solution by Dag Jonsson, Uppsala, Sweden.
[The solution contains a drawing I will attempt to describe: Draw a
circle with point A at the top and chord BC horizontal at the bottom.
Draw triangle ABC. Draw point G at the midpoint of BC. Draw points D
and E approximately two diameters (not to scale) to the right of points
A and G, respectively. Draw rectangle ADEG. Label BG and GC with a,
label AD with a+b, label CD with a+b, and label CE with b. -- edp]
Let G be the midpoint of BC and form the rectangle AGED as in the
figure. Let |BG|=|GC|=a and |CE|=b. Then |CD|=|AD|=a+b and
|BD|^2 = |BE|^2 + (|CD|^2 - |CE|^2)
= (2a+b)^2 + (a+b)^2 - b^2 = 5a^2 + 6ab + b^2.
Now
angle DBE <= 30 degrees <=> |BE| >= sqrt(3)/2 * |BD| <=>
4 |BE|^2 >= 3 |BD|^2 <=>
16 a^2 + 16ab + 4b^2 >= 15a^2 +18ab + 3b^2 <=>
a^2+b^2 >= 2ab,
which of course is true.
[There's a second solution by Waldemare Pompe, student, University of
Warsaw, Poland, that is more geometric and less algebraic.]
|