Thursday, February 23, 2006

Pair Of Products From P. Topic: Geometry. Level: Olympiad.

Problem: (2000 MOP Rookie Contest, Gabriel Carroll Original) Let $ABCD$ be a square of side $1$ and $P$ be a point in the plane. Prove that $PA \cdot PB + PC \cdot PD \ge 1$.


Solution: First we want to settle the case in which $ P $ is either on the opposite side of $ AB $ as $ CD $ is or vice versa. By symmetry, we can just show it for one side. WLOG, assume $ P $ is on the opposite side of $ AB $ as $ CD $ is (or $ P $ is to the "left" of $ AB $ in the diagram).

But then $ PC \ge BC = 1 $ and $ PD \ge AD = 1 $ so $ PC \cdot PD \ge 1 $ and the inequality clearly holds.

So now we move on to the case in which $ AB $ and $ CD $ are on opposite sides of $ P $, as shown in the diagram. Let $ E $ and $ F $ be the feet of the perpendiculars from $ P $ to $ BC $ and $ AD $, respectively.

We have

$ PA \cdot PB \ge PA \cdot PB \sin{\angle APB} = 2[APB] = [BEFA] $


$ PC \cdot PD \ge PC \cdot PD \sin{\angle CPD} = 2[CPD] = [CEFD] $,

both from the fact that half of the product of two sides in a triangle and the sine of the angle between them gives the area. Hence $ PA \cdot PB + PC \cdot PD \ge [BEFA]+[CEFD] = [ABCD] = 1 $ as desired. QED.


Comment: Using coordinate geometry was pretty tempting in this problem, especially since it turns the problem into an algebraic inequality which are usually easier, but this proof is actually much nicer. Also, knowing all the ways to find the area of a triangle is extremely useful in geometry problems because it gives you much more flexibility in rewriting expressions.


Practice Problem #1: Generalize the problem for a square of side $ k $.

Practice Problem #2: Find and prove $ 5 $ different formulas for the area of a triangle (yes, there are definitely that many).

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