1993 IMO Problems/Problem 2

Revision as of 03:50, 25 August 2024 by Reyaansh agrawal (talk | contribs) (Solution)

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.

Solution

IMO 1993 A2.jpg 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. $\square$

See Also

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