Difference between revisions of "1991 USAMO Problems/Problem 5"
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<cmath> CC_b - DD_b = BC - DB, </cmath> | <cmath> CC_b - DD_b = BC - DB, </cmath> | ||
so that | so that | ||
− | <cmath> CE = \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} = \frac{AC + BC - AB}{2} . </cmath> | + | <cmath> \begin{align*} |
+ | CE &= \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} \ | ||
+ | &= \frac{AC + BC - AB}{2} . | ||
+ | \end{align*} </cmath> | ||
Thus <math>E</math> lies on the arc of the circle with center <math>C</math> and radius <math>(AB+BC-AB)/2</math> intercepted by segments <math>CA</math> and <math>CB</math>. If we choose an arbitrary point <math>X</math> on this arc and let <math>D</math> be the intersection of lines <math>CX</math> and <math>AB</math>, then <math>X</math> becomes point <math>E</math> in the diagram, so every point on this arc is in the locus of <math>E</math>. <math>\blacksquare</math> | Thus <math>E</math> lies on the arc of the circle with center <math>C</math> and radius <math>(AB+BC-AB)/2</math> intercepted by segments <math>CA</math> and <math>CB</math>. If we choose an arbitrary point <math>X</math> on this arc and let <math>D</math> be the intersection of lines <math>CX</math> and <math>AB</math>, then <math>X</math> becomes point <math>E</math> in the diagram, so every point on this arc is in the locus of <math>E</math>. <math>\blacksquare</math> | ||
Revision as of 13:01, 13 January 2008
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
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 .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1991 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |