1961 IMO Problems/Problem 5

Problem

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?

Solution

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.

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

Page

Toolbox

Search

1961 IMO Problems/Problem 5

Problem

Solution

See Also