Difference between revisions of "1991 USAMO Problems/Problem 5"
Eaz shadow (talk | contribs) |
|||
(One intermediate revision by the same user not shown) | |||
Line 117: | Line 117: | ||
</cmath> | </cmath> | ||
is a constant equal to <math>r := \tfrac{AC+BC-AB}2</math>, and so <math>C</math> lies on the circle with center <math>C</math> and radius <math>r</math>. | is a constant equal to <math>r := \tfrac{AC+BC-AB}2</math>, and so <math>C</math> lies on the circle with center <math>C</math> and radius <math>r</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=G4UVUZ7UemY | ||
+ | |||
+ | |||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 19:53, 27 October 2024
Contents
[hide]Problem
Let be an arbitrary point on side of a given triangle and let be the interior point where intersects the external common tangent to the incircles of triangles and . As assumes all positions between and , prove that the point traces the arc of a circle.
Solution 1
Let the incircle of and the incircle of touch line at points , respectively; let these circles touch at , , respectively; and let them touch their common external tangent containing at , respectively, as shown in the diagram below.
We note that On the other hand, since and are tangents from the same point to a common circle, , and similarly , so On the other hand, the segments and evidently have the same length, and , so . Thus If we let be the semiperimeter of triangle , then , and , so Similarly, so that Thus lies on the arc of the circle with center and radius intercepted by segments and . If we choose an arbitrary point on this arc and let be the intersection of lines and , then becomes point in the diagram, so every point on this arc is in the locus of .
Solution 2
Define the same points as in the first solution. First extend to intersect at a point ; without loss of generality let lie in between and . Then the incircle of is also the incircle of , while the incircle of is the -excircle of . It follows that ; denote this equality by .
Now remark that Hence is a constant equal to , and so lies on the circle with center and radius .
Video Solution
https://www.youtube.com/watch?v=G4UVUZ7UemY
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1991 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.