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

Revision as of 10:29, 21 November 2023

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 $\frac{AC\cdot CD}{AC\cdot BD}$ .

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

@@ Line 2: / Line 2: @@
 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}.}
-\begin{enumerate}
+(a) Calculate the ratio <math>\frac{AC\cdot CD}{AC\cdot BD}</math>.
-\item Calculate the ratio <math>\frac{AC\cdot CD}{AC\cdot BD}</math>
-\item Prove that the tangents at <math>C</math> to the circumcircles of <math>\triangle ACD</math> and <math>\triangle BCD</math> are perpendicular.
+(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}
 == Solution ==

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