1969 Canadian MO Problems/Problem 9
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 |