Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 IMO Problems/Problem 2"
(→‎Solution)
 
(9 intermediate revisions by 3 users not shown)
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>.
 
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{AB\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 ==
 
== Solution ==
−
{{solution}}
+
[[File:IMO 1993 A2.jpg]]
 +
Let us construct a point <math>B'</math> satisfying the following conditions: <math>B', B</math> are on the same side of AC, <math>BC = B'C</math> and <math>\angle BCB' = 90^{\circ}</math>.
 +
 
 +
Hence <math>\triangle ADB \sim \triangle ACB'</math>.
 +
 
 +
<cmath>\implies \frac{AB}{BD} = \frac{AB'}{B'C} </cmath>
 +
 
 +
Also considering directed angles mod <math>180^{\circ}</math>,
 +
 
 +
<cmath>\measuredangle CAB' = \measuredangle DAB \implies \measuredangle CAD = \measuredangle BAB' </cmath>.
 +
 
 +
Also, <math>\frac{AB'}{AB} = \frac{B'C}{BD} = \frac{BC}{BD} = \frac{AC}{AD} </math>.
 +
 
 +
<math>\implies \triangle ABB' \sim \triangle ADC</math>.
 +
 
 +
Hence, <math>\frac{CD}{AC} = \frac{BB'}{AB'}</math>.
 +
 
 +
Finally, we get <math>\frac{AB \cdot CD}{AC \cdot BD} = \frac{BB'}{CB'} = \boxed{\sqrt{2}} </math>.
 +
 
 +
For the second part, let the tangent to the circle <math>(ADC)</math> be <math>DX</math> and the tangent to the circle <math>(ADB)</math> be <math>DY</math>.
 +
 
 +
<math>\measuredangle ADX = \measuredangle ACD</math> due to the tangent-chord theorem.
 +
 
 +
<math>\measuredangle YDB = \measuredangle DCB</math> for the same reason.
 +
 
 +
Hence, <cmath>\measuredangle ADX + \measuredangle YDB = \measuredangle ACB</cmath>
 +
 
 +
We also have <cmath>\measuredangle ADB = \measuredangle ACB + 90^{\circ}</cmath>
 +
 
 +
<cmath>\measuredangle ADX + \measuredangle XDY + \measuredangle YDB = \measuredangle ACB + \measuredangle XDY = \measuredangle ACB + 90^{\circ}</cmath>.
 +
 
 +
<cmath>\implies \measuredangle XDY = 90^{\circ}</cmath> which means circles <math>(ADC)</math> and <math>(ADB)</math> are orthogonal.                <math>\square</math>
 +
~reyaansh_agrawal
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1993|num-b=1|num-a=3}}
 
{{IMO box|year=1993|num-b=1|num-a=3}}

Latest revision as of 21:33, 25 August 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 $\frac{AB\cdot CD}{AC\cdot BD}$.

(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 $\frac{AB \cdot CD}{AC \cdot BD} = \frac{BB'}{CB'} = \boxed{\sqrt{2}}$.

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. $\square$ ~reyaansh_agrawal

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_IMO_Problems/Problem_2&oldid=225788"