Difference between revisions of "1991 USAMO Problems/Problem 5"
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− | == Solution == | + | == Solution 1 == |
Let the incircle of <math>ACD</math> and the incircle of <math>BCD</math> touch line <math>AB</math> at points <math>D_a,D_b</math>, respectively; let these circles touch <math>CD</math> at <math>C_a</math>, <math>C_b</math>, respectively; and let them touch their common external tangent containing <math>E</math> at <math>T_a,T_b</math>, respectively, as shown in the diagram below. | Let the incircle of <math>ACD</math> and the incircle of <math>BCD</math> touch line <math>AB</math> at points <math>D_a,D_b</math>, respectively; let these circles touch <math>CD</math> at <math>C_a</math>, <math>C_b</math>, respectively; and let them touch their common external tangent containing <math>E</math> at <math>T_a,T_b</math>, respectively, as shown in the diagram below. | ||
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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> | ||
+ | == Solution 2 == | ||
+ | Define the same points as in the first solution. First extend <math>T_aT_b</math> to intersect <math>AB</math> at a point <math>P</math>; without loss of generality let <math>A</math> lie in between <math>B</math> and <math>P</math>. Then the incircle of <math>\triangle ACD</math> is also the incircle of <math>\triangle PED</math>, while the incircle of <math>\triangle BCD</math> is the <math>P</math>-excircle of <math>\triangle PED</math>. It follows that <math>EC_a = C_bD</math>; denote this equality by <math>(*)</math>. | ||
+ | |||
+ | Now remark that | ||
+ | <cmath> | ||
+ | CC_a + CC_b = \frac{AC+AD-CD}2 + \frac{BC+DC-BD}2 = \frac{AC+BC-AB}2 + CD. | ||
+ | </cmath> | ||
+ | Hence | ||
+ | <cmath> | ||
+ | CC_a + CC_b - CD = CC_a - C_bD \stackrel{(*)}= CC_a - C_aE = CE | ||
+ | </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>. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 11:18, 4 September 2018
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 .
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.