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. | ||
== 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 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 |