Difference between revisions of "1993 IMO Problems/Problem 2"
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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>. | ||
− | (a) Calculate the ratio <math>\frac{ | + | (a) Calculate the ratio <math>\frac{AB\cdot CD}{AC\cdot BD}</math>. |
(b) Prove that the tangents at <math>C</math> to the circumcircles of <math>\Delta ACD</math> and <math>\Delta 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. |
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 |