USAMO 76/2 solution

5th USAMO 1976

Problem 2

AB is a fixed chord of a circle, not a diameter. CD is a variable diameter. Find the locus of the intersection of AC and BD.

Solution

Let the lines meet at X and suppose X lies outside the circle. ∠AXB = ∠AXD (same angle) = 180 deg - ∠XAD - ∠XDA = 90^o - ∠XDA (CD is a diameter, so angle CAD = 90^o) = 90^o - ∠BDA (same angle). But ∠BDA is constant, so ∠AXB is constant and hence X lies on a circle through A and B.

Let O be the center of the circle ABC and O' the center of the circle ABX. We have ∠AOB = 2 ∠ADB, ∠AO'B = 2 ∠AXB, so ∠AOB + ∠AO'B = 180^o. Hence ∠OAO' + ∠OBO' = 180^o. But ∠OAO' = ∠OBO', so ∠OAO' = 90^o. In other words, the circles are orthogonal.

If X lies inside the circle center O, then ∠AXB = ∠XAD + ∠XDA = ∠CAD + ∠ADB (same angles) = 90^o + ∠ADB (CD diameter). So X lies on the same circle.

Conversely, suppose X lies on the circle O'. Extend XA, XB to meet the circle center O at C and D respectively. If X lies outside the circle center O, assume C does not lie inside the circle center O' (if not use D). Then ∠CAD = ∠AXD + ∠ADX = ∠AXB + ∠ADB (same angles) = (∠AO'B + ∠AOB)/2 = 90^o. Hence CD is a diameter.

If X lies inside the circle O', then ∠ABX = 90^o + ∠ADB. But ∠AXB = ∠ADB + ∠XBD. So ∠CBD = ∠XBD = 90^o. Hence CD is a diameter.

5th USAMO 1976