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

 
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== Problem ==
 
== Problem ==
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Show that for any quadrilateral inscribed in a circle of radius <math>\displaystyle 1,</math> the length of the shortest side is less than or equal to <math>\displaystyle \sqrt{2}</math>.
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Show that for any quadrilateral inscribed in a [[circle]] of [[radius]] <math>1,</math> the length of the shortest side is less than or equal to <math>\sqrt{2}</math>.
  
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== Solution ==
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== Solution 1 ==
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Let <math>\displaystyle a,b,c,d</math> be the sides and <math>\displaystyle e,f</math> be the diagonals. By Ptolemy's theorem, <math>\displaystyle ab+cd = ef</math>. However, the diameter is the longest possible diagonal, so <math>\displaystyle e,f \le 2</math> and <math>\displaystyle ab+cd \le 4</math>.  
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Let <math>a,b,c,d</math> be the [[edge]]-[[length]]s and <math>e,f</math> be the lengths of the [[diagonal]]s of the [[quadrilateral]]. By [[Ptolemy's Theorem]], <math>ab+cd = ef</math>. However, each diagonal is a [[chord]] of the circle and so must be shorter than the [[diameter]]: <math>e,f \le 2</math> and thus <math>ab+cd \le 4</math>.  
  
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If <math>\displaystyle a,b,c,d > \sqrt{2}</math>, then <math>\displaystyle ab+cd > 4,</math> which is impossible. Proof by contradiction.
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If <math>a,b,c,d > \sqrt{2}</math>, then <math>ab+cd > 4,</math> which is impossible. Thus, at least one of the sides must have length less than <math>\sqrt 2</math>, so certainly the shortest side must.
  
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== Solution 2 ==
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* [[1969 Canadian MO Problems/Problem 8|Previous Problem]]
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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>.
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* [[1969 Canadian MO Problems/Problem 10|Next Problem]]
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* [[1969 Canadian MO Problems|Back to Exam]]
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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>.
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{{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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