**Problem**: (1998 India National Olympiad - #4) Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.

**Solution**: This reeks of Ptolemy... anything involving cyclic quadrilaterals, products of sides, inequalities and you definitely want to try Ptolemy. If you don't know this, just use your Byakugan and everything should be ok.

In any case, let's try using Ptolemy, which states that

$ AB \cdot CD + AD \cdot BC = AC \cdot BD $

if $ ABCD $ is cyclic. Well, we know $ AC \cdot BD \le 2 \cdot 2 = 4 $ since they can only be as long as the diameter. So

$ AB \cdot CD + AD \cdot BC \le 4 $.

How can we relate this to $ AB \cdot BC \cdot CD \cdot DA $, though? How about... AM-GM! This gives us

$ AB \cdot BC \cdot CD \cdot DA \le \left(\frac{AB \cdot CD + AD \cdot BC}{2}\right)^2 \le 2^2 = 4 $,

which is good. But wait, we have the product is $ \ge 4 $ and is $ \le 4 $, so it in fact must equal $ 4 $. This means equality holds at all the steps, or $ AC $ and $ BD $ are both diameters and $ AB \cdot CD = AD \cdot BC $.

Well, if you think about this, if $ AC $ and $ BD $ are both diameters, then $ ABCD $ is a rectangle so $ AB = CD $ and $ AD = BC $. Combining this with $ AB \cdot CD = AD \cdot BC $, we see that $ AB = BC = CD = DA $ or $ ABCD $ is a square. QED.

--------------------

Comment: After realizing that Ptolemy would probably help (through experience or Byakugan), it wasn't a huge leap to find AM-GM and develop the complete solution. It's good practice to be on the lookout for Ptolemy (in its equality and inequality forms) because it is a very useful result in problems such as this one.

--------------------

Practice Problem: (1991 AIME - #14) A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A$.

## No comments:

## Post a Comment