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

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Hence, <math>\frac{CD}{AC} = \frac{BB'}{AB'}</math>.
 
Hence, <math>\frac{CD}{AC} = \frac{BB'}{AB'}</math>.
  
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Finally, we get <math>\frac{AB \dot CD}{AC \dot BD} = \frac{BB'}{CB'} = \boxed{\sqrt{2}} </math>.
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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>.
 
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>.
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<cmath>\implies \measuredangle XDY = 90^{\circ}</cmath> which means circles <math>(ADC)</math> and <math>(ADB)</math> are orthogonal.                <math>\square</math>
 
<cmath>\implies \measuredangle XDY = 90^{\circ}</cmath> which means circles <math>(ADC)</math> and <math>(ADB)</math> are orthogonal.                <math>\square</math>
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~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
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