Difference between revisions of "2015 USAJMO Problems/Problem 3"
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===Solution 3=== | ===Solution 3=== | ||
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Note that each point <math>X</math> on <math>PQ</math> corresponds to exactly one point on arc <math>PBQ</math>. Also notice that since <math>AB</math> is the diameter of <math>\omega</math>, <math>\angle ASB</math> is always a right angle; therefore, point <math>T</math> is always <math>B</math>. WLOG, assume that <math>\omega</math> is on the coordinate plane, and <math>B</math> corresponds to the origin. The locus of <math>M</math>, since the locus of <math>S</math> is arc <math>PBQ</math>, is the arc that is produced when arc <math>PBQ</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin, which resides on the circle <math>\psi</math>, which is produced when <math>\omega</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin. By MSmathlete1018 | Note that each point <math>X</math> on <math>PQ</math> corresponds to exactly one point on arc <math>PBQ</math>. Also notice that since <math>AB</math> is the diameter of <math>\omega</math>, <math>\angle ASB</math> is always a right angle; therefore, point <math>T</math> is always <math>B</math>. WLOG, assume that <math>\omega</math> is on the coordinate plane, and <math>B</math> corresponds to the origin. The locus of <math>M</math>, since the locus of <math>S</math> is arc <math>PBQ</math>, is the arc that is produced when arc <math>PBQ</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin, which resides on the circle <math>\psi</math>, which is produced when <math>\omega</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin. By MSmathlete1018 | ||
The above solution is false; it gives the wrong locus. The step where they mess up is in stating that <math>T=B</math>, however this is not true because <math>\angle TXA</math> is right, not <math>\angle TSA</math>. | The above solution is false; it gives the wrong locus. The step where they mess up is in stating that <math>T=B</math>, however this is not true because <math>\angle TXA</math> is right, not <math>\angle TSA</math>. |
Revision as of 09:53, 16 April 2017
Contents
Problem
Quadrilateral is inscribed in circle with and . Let be a variable point on segment . Line meets again at (other than ). Point lies on arc of such that is perpendicular to . Let denote the midpoint of chord . As varies on segment , show that moves along a circle.
Solution 1
We will use coordinate geometry.
Without loss of generality, let the circle be the unit circle centered at the origin, , where .
Let angle , which is an acute angle, , then .
Angle , . Let , then .
The condition yields: (E1)
Use identities , , , we obtain . (E1')
The condition that is on the circle yields , namely . (E2)
is the mid-point on the hypotenuse of triangle , hence , yielding . (E3)
Expand (E3), using (E2) to replace with , and using (E1') to replace with , and we obtain , namely , which is a circle centered at with radius .
Solution 2
Let the midpoint of be . We claim that moves along a circle with radius .
We will show that , which implies that , and as is fixed, this implies the claim.
by the median formula on .
by the median formula on .
.
As , from right triangle .
By , .
Since is the circumcenter of , and is the circumradius, the expression is the power of point with respect to . However, as is also the power of point with respect to , this implies that .
By ,
Finally, by AA similarity ( and ), so .
By , , so , as desired.
Solution 3
The below solution is false.
Note that each point on corresponds to exactly one point on arc . Also notice that since is the diameter of , is always a right angle; therefore, point is always . WLOG, assume that is on the coordinate plane, and corresponds to the origin. The locus of , since the locus of is arc , is the arc that is produced when arc is dilated by with respect to the origin, which resides on the circle , which is produced when is dilated by with respect to the origin. By MSmathlete1018
The above solution is false; it gives the wrong locus. The step where they mess up is in stating that , however this is not true because is right, not .