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Difference between revisions of "1969 Canadian MO Problems/Problem 9"

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Note that the shortest side would subtend the smallest angle at the center of the circle. It is easily provable that this angle is <math>90^{\circ} </math>, because had all 4 angles been greater than <math>90^{\circ} </math>, the sum of the subtended angles would have been greater than <math>360^{\circ} </math>, which is impossible as the sum should be exactly <math>360^{\circ} </math>. Hence, atleast one (but not all 4) of the angles would be less than or equal to <math>90^{\circ} </math> and so the smallest one most certainly is also <math>\leq 90^{\circ} </math>.  
 
Note that the shortest side would subtend the smallest angle at the center of the circle. It is easily provable that this angle is <math>90^{\circ} </math>, because had all 4 angles been greater than <math>90^{\circ} </math>, the sum of the subtended angles would have been greater than <math>360^{\circ} </math>, which is impossible as the sum should be exactly <math>360^{\circ} </math>. Hence, atleast one (but not all 4) of the angles would be less than or equal to <math>90^{\circ} </math> and so the smallest one most certainly is also <math>\leq 90^{\circ} </math>.  
  
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Now assuming the angle between the two radii at the ends of the smallest side is <math>\theta </math>, the length of the smallest side is <math>\sqrt{2 - 2 </math>cos \theta <math>} </math>. Since <math>cos \theta </math> is positive in the interval <math>(0, 90^{\circ}) </math>, <math>2- 2 </math>cos \theta \leq 2 <math> and so </math>\sqrt{2 - 2 <math>cos \theta </math>} \leq \sqrt{2} $.  
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Now assuming the angle between the two radii at the ends of the smallest side is <math>\theta </math>, the length of the smallest side is <math>\sqrt{2 - 2 cos \theta} </math>. Since <math>cos \theta </math> is positive in the interval <math>(0, 90^{\circ}) </math>, <math>2- 2 cos \theta \leq 2 </math> and so <math>\sqrt{2 - 2 cos \theta } \leq \sqrt{2} </math>.  
  
 
{{Old CanadaMO box|num-b=8|num-a=10|year=1969}}
 
{{Old CanadaMO box|num-b=8|num-a=10|year=1969}}

Latest revision as of 10:53, 28 May 2024

Problem

Show that for any quadrilateral inscribed in a circle of radius $1,$ the length of the shortest side is less than or equal to $\sqrt{2}$.

Solution 1

Let $a,b,c,d$ be the edge-lengths and $e,f$ be the lengths of the diagonals of the quadrilateral. By Ptolemy's Theorem, $ab+cd = ef$. However, each diagonal is a chord of the circle and so must be shorter than the diameter: $e,f \le 2$ and thus $ab+cd \le 4$.

If $a,b,c,d > \sqrt{2}$, then $ab+cd > 4,$ which is impossible. Thus, at least one of the sides must have length less than $\sqrt 2$, so certainly the shortest side must.

Solution 2

Note that the shortest side would subtend the smallest angle at the center of the circle. It is easily provable that this angle is $90^{\circ}$, because had all 4 angles been greater than $90^{\circ}$, the sum of the subtended angles would have been greater than $360^{\circ}$, which is impossible as the sum should be exactly $360^{\circ}$. Hence, atleast one (but not all 4) of the angles would be less than or equal to $90^{\circ}$ and so the smallest one most certainly is also $\leq 90^{\circ}$.

Now assuming the angle between the two radii at the ends of the smallest side is $\theta$, the length of the smallest side is $\sqrt{2 - 2 cos \theta}$. Since $cos \theta$ is positive in the interval $(0, 90^{\circ})$, $2- 2 cos \theta \leq 2$ and so $\sqrt{2 - 2 cos \theta } \leq \sqrt{2}$.

1969 Canadian MO (Problems)
Preceded by
Problem 8 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 10


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