Difference between revisions of "2015 USAJMO Problems/Problem 3"
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Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle. | Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle. | ||
− | ===Solution | + | ===Solution 1=== |
− | |||
− | + | We will use coordinate geometry. | |
− | |||
− | + | Without loss of generality, | |
− | + | let the circle be the unit circle centered at the origin, | |
+ | <cmath>A=(1,0) P=(1-a,b), Q=(1-a,-b)</cmath>, | ||
+ | where <math>(1-a)^2+b^2=1</math>. | ||
− | + | Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>. | |
− | + | Angle <math>\angle BOS=2A</math>, <math>S=(-\cos(2A),\sin(2A))</math>. | |
+ | Let <math>M=(u,v)</math>, then <math>T=(2u+\cos(2A), 2v-\sin(2A))</math>. | ||
− | Expand (E3), using (E2) to replace 2( | + | The condition <math>TX \perp AX</math> yields: <math>(2v-\sin(2A)-at)/(2u+\cos(2A)+a-1)=\cot A. </math> (E1) |
− | u^2-u-a+v^2=0, namely (u-1 | + | |
+ | Use identities <math>(\cos A)^2=1/(1+t^2)</math>, <math>\cos(2A)=2(\cos A)^2-1= 2/(1+t^2) -1</math>, <math>\sin(2A)=2\sin A\cos A=2t^2/(1+t^2)</math>, we obtain <math>2vt-at^2=2u+a</math>. (E1') | ||
+ | |||
+ | The condition that <math>T</math> is on the circle yields <math>(2u+\cos(2A))^2+ (2v-\sin(2A))^2=1</math>, namely <math>v\sin(2A)-u\cos(2A)=u^2+v^2</math>. (E2) | ||
+ | |||
+ | <math>M</math> is the mid-point on the hypotenuse of triangle <math>STX</math>, hence <math>MS=MX</math>, yielding <math>(u+\cos(2A))^2+(v-\sin(2A))^2=(u+a-1)^2+(v-at)^2</math>. (E3) | ||
+ | |||
+ | Expand (E3), using (E2) to replace <math>2(v\sin(2A)-u\cos(2A))</math> with <math>2(u^2+v^2)</math>, and using (E1') to replace <math>a(-2vt+at^2)</math> with <math>-a(2u+a)</math>, and we obtain | ||
+ | <math>u^2-u-a+v^2=0</math>, namely <math>(u-\frac{1}{2})^2+v^2=a+\frac{1}{4}</math>, which is a circle centered at <math>(\frac{1}{2},0)</math> with radius <math>r=\sqrt{a+\frac{1}{4}}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | |||
+ | Let the midpoint of <math>AO</math> be <math>K</math>. We claim that <math>M</math> moves along a circle with radius <math>KP</math>. | ||
+ | |||
+ | We will show that <math>KM^2 = KP^2</math>, which implies that <math>KM = KP</math>, and as <math>KP</math> is fixed, this implies the claim. | ||
+ | |||
+ | <math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle AMO</math>. | ||
+ | |||
+ | <math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle APO</math>. | ||
+ | |||
+ | <math>KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)</math>. | ||
+ | |||
+ | As <math>OP = OT</math>, <math>OP^2-OM^2 = MT^2</math> from right triangle <math>OMT</math>. <math>(1)</math> | ||
+ | |||
+ | By <math>(1)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^2)</math>. | ||
+ | |||
+ | Since <math>M</math> is the circumcenter of <math>\triangle XTS</math>, and <math>MT</math> is the circumradius, the expression <math>AM^2-MT^2</math> is the power of point <math>A</math> with respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies that <math>AM^2-MT^2=AX*AS</math>. <math>(2)</math> | ||
+ | |||
+ | By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</math> | ||
+ | |||
+ | Finally, <math>\triangle APX \sim \triangle ASP</math> by AA similarity (<math>\angle XAP = \angle SAP</math> and <math>\angle APX = \angle AQP = \angle ASP</math>), so <math>AX*AS = AP^2</math>. <math>(3)</math> | ||
+ | |||
+ | By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, as desired. <math>QED</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | 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 |
Revision as of 14:15, 6 April 2016
Contents
[hide]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
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