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>. | ||
− | Finally, we get <math>\frac{AB \ | + | 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>. |
Latest revision as of 04:53, 25 August 2024
Problem
Let be a point inside acute triangle
such that
and
.
(a) Calculate the ratio .
(b) Prove that the tangents at to the circumcircles of
and
are perpendicular.
Solution
Let us construct a point
satisfying the following conditions:
are on the same side of AC,
and
.
Hence .
Also considering directed angles mod ,
.
Also, .
.
Hence, .
Finally, we get .
For the second part, let the tangent to the circle be
and the tangent to the circle
be
.
due to the tangent-chord theorem.
for the same reason.
Hence,
We also have
.
which means circles
and
are orthogonal.
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 |