Difference between revisions of "1993 IMO Problems/Problem 2"
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<cmath>\measuredangle ADX + \measuredangle XDY + \measuredangle YDB = \measuredangle ACB + \measuredangle XDY = \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> | + | <cmath>\implies \measuredangle XDY = 90^{\circ}</cmath> which means circles <math>(ADC)</math> and <math>(ADB)</math> are orthogonal. <math>\square</math> |
==See Also== | ==See Also== | ||
{{IMO box|year=1993|num-b=1|num-a=3}} | {{IMO box|year=1993|num-b=1|num-a=3}} |
Revision as of 04:50, 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 |