Difference between revisions of "1993 IMO Problems/Problem 2"

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==Problem==
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Let <math>D</math> be a point inside acute triangle <math>ABC</math> such that <math>\angle ADB = \angle ACB+\frac{\pi}{2}</math> and <math>AC\cdot BD=AD\cdot BC</math>.
 
Let <math>D</math> be a point inside acute triangle <math>ABC</math> such that <math>\angle ADB = \angle ACB+\frac{\pi}{2}</math> and <math>AC\cdot BD=AD\cdot BC</math>.
\renewcommand{\labelenumi}{\alph{enumi}.}
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\begin{enumerate}
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(a) Calculate the ratio <math>\frac{AC\cdot CD}{AB\cdot BD}</math>.
\item Calculate the ratio <math>\frac{AC\cdot CD}{AC\cdot BD}</math>
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\item Prove that the tangents at <math>C</math> to the circumcircles of <math>\triangle ACD</math> and <math>\triangle BCD</math> are perpendicular.
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(b) Prove that the tangents at <math>C</math> to the circumcircles of <math>\Delta ACD</math> and <math>\Delta BCD</math> are perpendicular.
\end{enumerate}
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== Solution ==
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{{solution}}
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==See Also==
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{{IMO box|year=1993|num-b=1|num-a=3}}

Latest revision as of 23:02, 21 January 2024

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

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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