**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}}