Difference between revisions of "1961 IMO Problems/Problem 5"
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Prolong BA to a point D such that <math>BD = 2AB</math>. Take circle through B and D such that the minor arc BD is equal to <math>2*\omega</math> so that for points P on the major arc BD we have <math>\angle BPD = \omega</math>. 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. | Prolong BA to a point D such that <math>BD = 2AB</math>. Take circle through B and D such that the minor arc BD is equal to <math>2*\omega</math> so that for points P on the major arc BD we have <math>\angle BPD = \omega</math>. 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 <math>AX \geqslant AC > AB</math>. But <math>\frac{AB}{AX} = \tan{\frac{\omega}{2}}</math>, so we get the condition in the question | + | 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 <math>AX \geqslant AC > AB</math>. But <math>\frac{AB}{AX} = \tan{\frac{\omega}{2}}</math>, so we get the condition in the question |
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| + | ==See Also== | ||
{{IMO box|year=1961|num-b=4|num-a=6}} | {{IMO box|year=1961|num-b=4|num-a=6}} | ||
| + | [[Category:Olympiad Geometry Problems]] | ||
| + | [[Category:Geometric Construction Problems]] | ||
Latest revision as of 03:12, 7 June 2021
Problem
Construct a triangle ABC if the following elements are given: 
, and 
where M is the midpoint of BC. Prove that the construction has a solution if and only if
In what case does equality hold?
Solution
Prolong BA to a point D such that 
. Take circle through B and D such that the minor arc BD is equal to 
so that for points P on the major arc BD we have 
. 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 
. But 
, 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 | ||
