Difference between revisions of "2015 USAJMO Problems/Problem 5"
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Let <math>ABCD</math> be a cyclic quadrilateral. Prove that there exists a point <math>X</math> on segment <math>\overline{BD}</math> such that <math>\angle BAC=\angle XAD</math> and <math>\angle BCA=\angle XCD</math> if and only if there exists a point <math>Y</math> on segment <math>\overline{AC}</math> such that <math>\angle CBD=\angle YBA</math> and <math>\angle CDB=\angle YDA</math>. | Let <math>ABCD</math> be a cyclic quadrilateral. Prove that there exists a point <math>X</math> on segment <math>\overline{BD}</math> such that <math>\angle BAC=\angle XAD</math> and <math>\angle BCA=\angle XCD</math> if and only if there exists a point <math>Y</math> on segment <math>\overline{AC}</math> such that <math>\angle CBD=\angle YBA</math> and <math>\angle CDB=\angle YDA</math>. | ||
− | == Solution == | + | == Solution 1 == |
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 14:25, 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 1
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
All angles are directed. Note that lines are isogonal in and are isogonal in . From the law of sines it follows that
Therefore, the ratio equals
Now let be a point of such that . We apply the above identities for to get that . So , the converse follows since all our steps are reversible.
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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 |