Difference between revisions of "2015 USAJMO Problems/Problem 5"
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== Solution == | == Solution == | ||
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+ | Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math>, so an inversion centered at <math>A</math> with power <math>r^2=AB\cdot AD</math> composed with a reflection about the angle bisector of <math>\angle DAB</math> swaps the pairs <math>(D,B)</math> and <math>(C,X)</math>. Thus, <cmath>\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1</cmath>so that <math>ACBD</math> is a harmonic quadrilateral. By symmetry, if <math>Y</math> exists, then <math>(B,D;A,C)=-1</math>. We have shown the two conditions are equivalent, whence both directions follow<math>.\:\blacksquare\:</math> | ||
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+ | == Solution 2== | ||
Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math>, so an inversion centered at <math>A</math> with power <math>r^2=AB\cdot AD</math> composed with a reflection about the angle bisector of <math>\angle DAB</math> swaps the pairs <math>(D,B)</math> and <math>(C,X)</math>. Thus, <cmath>\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1</cmath>so that <math>ACBD</math> is a harmonic quadrilateral. By symmetry, if <math>Y</math> exists, then <math>(B,D;A,C)=-1</math>. We have shown the two conditions are equivalent, whence both directions follow<math>.\:\blacksquare\:</math> | Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math>, so an inversion centered at <math>A</math> with power <math>r^2=AB\cdot AD</math> composed with a reflection about the angle bisector of <math>\angle DAB</math> swaps the pairs <math>(D,B)</math> and <math>(C,X)</math>. Thus, <cmath>\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1</cmath>so that <math>ACBD</math> is a harmonic quadrilateral. By symmetry, if <math>Y</math> exists, then <math>(B,D;A,C)=-1</math>. We have shown the two conditions are equivalent, whence both directions follow<math>.\:\blacksquare\:</math> |
Revision as of 14:16, 13 April 2017
Contents
[hide]Problem
Let be a cyclic quadrilateral. Prove that there exists a point on segment such that and if and only if there exists a point on segment such that and .
Solution
Note that lines are isogonal in , so an inversion centered at with power composed with a reflection about the angle bisector of swaps the pairs and . Thus, so that is a harmonic quadrilateral. By symmetry, if exists, then . We have shown the two conditions are equivalent, whence both directions follow
Solution 2
Note that lines are isogonal in , so an inversion centered at with power composed with a reflection about the angle bisector of swaps the pairs and . Thus, so that is a harmonic quadrilateral. By symmetry, if exists, then . We have shown the two conditions are equivalent, whence both directions follow
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2015 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |