Difference between revisions of "2015 USAJMO Problems/Problem 3"
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===Solution 1=== | ===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. | We will use coordinate geometry. | ||
Line 52: | Line 81: | ||
By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, as desired. <math>QED</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 | |
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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