1969 Canadian MO Problems/Problem 9

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 $\sqrt{2 - 2$ (Error compiling LaTeX. Unknown error_msg)cos \theta $}$ (Error compiling LaTeX. Unknown error_msg). 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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1969 Canadian MO Problems/Problem 9

Problem

Solution 1

Solution 2