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>. | ||
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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> | ||
+ | ~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 20:33, 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. ~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 |