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Difference between revisions of "1961 IMO Problems/Problem 5"

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==Solution==
 
==Solution==
  
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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.
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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.
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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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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}}
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[[Category:Olympiad Geometry Problems]]
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[[Category:Geometric Construction Problems]]

Latest revision as of 03:12, 7 June 2021

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

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
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