# APMO 1995 Problem 4

Problem 4 of Asian Pacific Mathematical Olympiad 1995

Let w be a circle with radius R and centre O, and S a fixed point in the interior of w. Let AA' and BB' be perpendicular chords through S. Consider the rectangles SAMB, SBNA, SAMB, and SBNA. Find the set of all points M, N, M, and N when A moves around the whole circle.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (tặng bạn Nguyễn Tấn) Let a = SB and b = SB. Also let MA and MO intercept the circle at P and Q, respectively.

Now let MQ = c.

Since MN || BB', we have BP = BA.

∠PBB = ∠ABB or ∠PBM = ∠ABN

ΔPBM = ΔABN; therefore, MP = NA = SB = b

From point M outside the circle, we have:

MP × MA = MQ × (MQ + 2R)

or ab = c(c + 2R)

or c² + 2Rc  ab = 0

we have c = - R + \sqrt{R^2 + {ab}}

Therefore, OM = c + R = \sqrt{R^2 + {ab}}

The same proof can be used for other points N', M' and N; we have

OM = ON' = OM' = ON = \sqrt{R^2 + {ab}}

Since S is a fixed point inside circle w, the product ab is fixed. From there we conclude that the set of all points M, N, M, and N when A moves around the whole circle is a larger circle that has the same center with circle w and has the radius r = OM = \sqrt{R^2 + {ab}}