1993 IMO Problems/Problem 2

Problem

Let $D$ be a point inside acute triangle $ABC$ such that $\angle ADB = \angle ACB+\frac{\pi}{2}$ and $AC\cdot BD=AD\cdot BC$ .

(a) Calculate the ratio $\frac{AC\cdot CD}{AB\cdot BD}$ .

(b) Prove that the tangents at $C$ to the circumcircles of $\Delta ACD$ and $\Delta BCD$ are perpendicular.

Let us construct a point $B'$ satisfying the following conditions: $B', B$ are on the same side of AC, $BC = B'C$ and $\angle BCB' = 90^{\circ}$ .

Hence $\triangle ADB \sim \triangle ACB'$ .

$\implies \frac{AB}{BD} = \frac{AB'}{B'C}$

Also considering directed angles mod $180^{\circ}$ ,

$\measuredangle CAB' = \measuredangle DAB \implies \measuredangle CAD = \measuredangle BAB'$ .

Also, $\frac{AB'}{AB} = \frac{B'C}{BD} = \frac{BC}{BD} = \frac{AC}{AD}$ .

$\implies \triangle ABB' \sim \triangle ADC$ .

Hence, $\frac{CD}{AC} = \frac{BB'}{AB'}$ .

Finally, we get $\frac{AB \dot CD}{AC \dot BD} = \frac{BB'}{CB'} = \boxed{\sqrt{2}}$ .

For the second part, let the tangent to the circle $(ADC)$ be $DX$ and the tangent to the circle $(ADB)$ be $DY$ .

$\measuredangle ADX = \measuredangle ACD$ due to the tangent-chord theorem.

$\measuredangle YDB = \measuredangle DCB$ for the same reason.

Hence, $\measuredangle ADX + \measuredangle YDB = \measuredangle ACB$

We also have $\measuredangle ADB = \measuredangle ACB + 90^{\circ}$

$\measuredangle ADX + \measuredangle XDY + \measuredangle YDB = \measuredangle ACB + \measuredangle XDY = \measuredangle ACB + 90^{\circ}$ .

$\implies \measuredangle XDY = 90^{\circ}$ , which means circles $(ADC)$ and $(ADB)$ are orthogonal. $\blacksquare$

1993 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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