# Ib AMO 1992 Problem 3

Problem 3 of the IberoAmerican Mathematical Olympiad 1992

C is the incircle of an equilateral triangle of side length 2.

(a) Show that for all points P on C the sum of the squares of the distances to the vertices A, B and C is 5.
(b) Show that for all points P on C it is possible to construct a triangle with sides equal to the segments AP, BP and CP, and the area \frac{\sqrt{3}}{4}.

Solution by Steve Dinh (dedicated to the Indian ladies at Portnov school)

(a) We have the existing formula relating the distances from a point to the vertices of a triangle a² + b² + c² = 3(d² +e² + f² -3m²) where a, b and c are the three sides of the triangle and d, e and f are distances from the point to the triangle’s vertices. Applying the formula to this case with m = GP = r = 1/ \sqrt{3}, 4 = AP² + BP² + CP² - 1 or AP² + BP² + CP² = 5. (See also Leo Moser's trick.)

(b) Rotate ΔABC 60° clockwise around point C. We have A → A’, B → A, P → P’ and AP’ = BP, PC = PP’ = P’C and the triangle AP’P has its side lengths equal to the segments AP, BP and CP.

Now draw the circumcircle ABC, and let Q be the intersection of AP’ with the circumcircle. Since after the rotation, ΔABP = ΔA’AP’, ∠ABP = ∠A’AP’, the three points B, P and Q are collinear. Let H be the foot of P to AP’, the area of the ΔAPP’ = ½PH × AP’. Now extend AP to intercept the two circles at I and J, respectively. We have PQ ×PB = AP ×PJ, but, since the two circles are concentric, AP = IJ and PQ ×PB = AP ×AI = AM² = 1.

However, PB = AP’ and ∠AQB = ∠BQC = 60° and PH = ½ PQ \sqrt{3}, the area of the triangle APP’ = ½ PH ×AP’ = (¼ PQ \sqrt{3}) ×PB = ¼ \sqrt{3} PQ ×BP = ¼ \sqrt{3}.

Extensions of problem: The following are drawn from this problem:

1. Sum of distances from point Q to vertices of triangle ABC is AQ² + BQ² + CQ² = 8 (*) since GQ = 2r = 2/ \sqrt{3} or BQ² = 8 - AQ² - CQ².

2. Using the law of the cosine function AC² = AQ² + CQ² - 2 AQ ×CQ cos 120° or AQ² + CQ² + AQ ×CQ = 4 (**) or (AQ + CQ)² = AQ ×CQ + 4 (***) Now subtract (**) from (*), we have BQ² = AQ ×CQ + 4 Combine with (***), we have BQ = AQ + CQ. (Incidently, this is van Schooten's theorem.)

3. The area of triangle with length segments AP, BP and CP is always constant as long as P is on the inner circle. One can derive another problem to find the locus of the points P in the plane of an equilateral triangle ABC for which the triangle formed with PA, PB and PC has constant area.

4. The problem can be reversed: If for every point P in the interior of a triangle, one can construct a triangle having sides equal to PA, PB and PC then the triangle is equilateral.

5. In triangle ABC, AB is the longest side. Prove that for any point P in the interior of the triangle, PA + PB > PC.

(Relevant here is Pompeiu's theorem.)