Difference between revisions of "2014 IMO Problems/Problem 3"
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For <math>\triangle{LAK}</math>, <math>\angle{ALK}=90-\angle{DML}=90-\angle{CMK}=\angle{MCH}=y+z</math>, and similarly, <math>\angle{AKL}=w+x</math>; hence, | For <math>\triangle{LAK}</math>, <math>\angle{ALK}=90-\angle{DML}=90-\angle{CMK}=\angle{MCH}=y+z</math>, and similarly, <math>\angle{AKL}=w+x</math>; hence, | ||
<cmath>\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3) </cmath> | <cmath>\frac{\sin{(w+x)}}{\sin{(y+z)}}=\frac{AL}{AK} \qquad \qquad (3) </cmath> | ||
− | + | Combining <math>(1), (2), (3)</math>, we have | |
<cmath>\frac{AL}{AK}=\frac{LT}{KS}=\frac{AL-AT}{AK-AS}=\frac{AT}{AS}</cmath> | <cmath>\frac{AL}{AK}=\frac{LT}{KS}=\frac{AL-AT}{AK-AS}=\frac{AT}{AS}</cmath> | ||
Therefore, <math>TS \parallel KL</math>, and <math>\angle{ATS} = \angle{ALK}=y+z</math>. | Therefore, <math>TS \parallel KL</math>, and <math>\angle{ATS} = \angle{ALK}=y+z</math>. |
Revision as of 03:31, 8 September 2018
Problem
Convex quadrilateral has . Point is the foot of the perpendicular from to . Points and lie on sides and , respectively, such that lies inside and
Prove that line is tangent to the circumcircle of
Solution
Denote , , , , , , , . Since and , we have , .
Since , the tangent of the circumcircle of at point is perpendicular to ; therefore, the circumcenter of (point ) is on . Similarly, the circumcenter of (point ) is on . In addition, is the perpendicular bisector of .
Extend to meet circumcircle of at , and extend to meet circumcircle of at . Then, since , and are the perpendicular bisector of and , respectively; hence is the circumcenter of . Since and are midpoints on and , ; also, , so . Since is the circumcenter, is also the perpendicular bisector of . Hence,
We have Hence, , or Since quadrilaterals and are cyclic, we have , ; so, Hence, Similarly,
Now we apply law of Sines repeatedly on pairs of triangles. For and , , , , ; hence, For , , ; hence, For , , and similarly, ; hence, Combining , we have Therefore, , and . Let the circumcircle of meets at . We have, And, This proves is the diameter of the circle and the center of the circle is on AH.
Solution by .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2014 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |