1961 IMO Problems/Problem 5


Construct a triangle ABC if the following elements are given: $AC = b, AB = c$, and $\angle AMB = \omega \left(\omega < 90^{\circ}\right)$ where M is the midpoint of BC. Prove that the construction has a solution if and only if

$b \tan{\frac{\omega}{2}} \le c < b$

In what case does equality hold?


Prolong BA to a point D such that $BD = 2AB$. Take circle through B and D such that the minor arc BD is equal to $2*\omega$ so that for points P on the major arc BD we have $\angle BPD = \omega$. Draw a circle with center A and radius AC, and the point of intersection of this circle and the major arc BD will be C. In general there are two possibilities for C.

Let X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require $AX \geqslant AC > AB$. But $\frac{AB}{AX} =  \tan{\frac{\omega}{2}}$, so we get the condition in the question

See Also

1961 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions