Difference between revisions of "1969 Canadian MO Problems/Problem 9"
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== Problem == | == Problem == | ||
− | Show that for any quadrilateral inscribed in a circle of radius <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> | + | 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> | + | 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. |
− | ---- | + | == 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>. | |
− | + | ||
− | + | 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 11:53, 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 |