Difference between revisions of "1969 Canadian MO Problems/Problem 9"
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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>. | 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>. | ||
| − | == Solution == | + | == Solution 1 == |
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>. | 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>. | ||
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. | 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 == | ||
| + | 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 cos x} </math>. Since <math>cos x </math> is positive in the interval <math>(0, 90^{circ}) </math>, <math>2- 2 cos x \leq 2 </math> and so <math>\sqrt{2 - 2 cos x} \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}} | ||
Revision as of 11:40, 28 May 2024
Problem
Show that for any quadrilateral inscribed in a circle of radius 
the length of the shortest side is less than or equal to 
.
Solution 1
Let 
be the edge-lengths and 
be the lengths of the diagonals of the quadrilateral. By Ptolemy's Theorem, 
. However, each diagonal is a chord of the circle and so must be shorter than the diameter: 
and thus 
.
If 
, then 
which is impossible. Thus, at least one of the sides must have length less than 
, 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 
, because had all 4 angles been greater than , the sum of the subtended angles would have been greater than 
, which is impossible as the sum should be exactly . Hence, atleast one (but not all 4) of the angles would be less than or equal to and so the smallest one most certainly is also 
.
Now assuming the angle between the two radii at the ends of the smallest side is 
, the length of the smallest side is 
. Since 
is positive in the interval 
, 
and so 
.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 10 |
