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=== |
− | + | <asy> | |
+ | size(8cm); | ||
+ | pair A=(1,0); | ||
+ | pair B=(-1,0); | ||
+ | pair P=dir(70); | ||
+ | pair Q=dir(-70); | ||
+ | pair O=(0,0); | ||
− | + | pair X=0.3*P + 0.7*Q; | |
− | + | pair Y=5*X-4*A; | |
+ | pair S=intersectionpoints(A--Y,circle(O,1))[1]; | ||
+ | pair Z=(A-X)*dir(-90) + X; | ||
+ | pair T=intersectionpoint(X--Z,circle(O,1)); | ||
+ | pair M=(S+T)/2; | ||
− | + | draw(circle(O,1)); | |
− | + | draw(B--A--P--B--Q--A--S--T--X); | |
+ | draw(P--Q); | ||
+ | dot("$A$",A,dir(A)); | ||
+ | dot("$B$",B,dir(B)); | ||
+ | dot("$P$",P,dir(P)); | ||
+ | dot("$Q$",Q,dir(Q)); | ||
+ | dot("$X$", X, SE); | ||
+ | dot("$S$",S,dir(S)); | ||
+ | dot("$T$",T,dir(T)); | ||
+ | dot("$M$",M,dir(M)); | ||
+ | dot((0,0)); | ||
+ | </asy> | ||
− | |||
− | + | 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>. | ||
+ | |||
+ | The condition <math>TX \perp AX</math> yields: <math>(2v-\sin(2A)-at)/(2u+\cos(2A)+a-1)=\cot A. </math> (E1) | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==More Solutions== | ||
+ | https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_2 |
Latest revision as of 16:45, 29 April 2020
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.
More Solutions
https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_2