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}.}
 
\renewcommand{\labelenumi}{\alph{enumi}.}
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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.
 
\item Prove that the tangents at <math>C</math> to the circumcircles of <math>\triangle ACD</math> and <math>\triangle BCD</math> are perpendicular.
 
\end{enumerate}
 
\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}}

Revision as of 11:25, 21 November 2023

Problem

Let $D$ be a point inside acute triangle $ABC$ such that $\angle ADB = \angle ACB+\frac{\pi}{2}$ and $AC\cdot BD=AD\cdot BC$. \renewcommand{\labelenumi}{\alph{enumi}.} \begin{enumerate} \item Calculate the ratio $\frac{AC\cdot CD}{AC\cdot BD}$ \item Prove that the tangents at $C$ to the circumcircles of $\triangle ACD$ and $\triangle BCD$ are perpendicular. \end{enumerate}

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