Difference between revisions of "2015 USAMO Problems/Problem 2"
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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> | ||
+ | == Solution 3(synthetic) == | ||
+ | To begin with, we connect <math>\overline{AT}</math> and we construct the nine-point circle of <math>\triangle AST</math> centered at <math>N_9</math>. | ||
+ | Lemma <math>1</math>: <math>AX \cdot AS = AP^2</math> | ||
+ | We proceed on a directed angle chase. We get <math>\measuredangle ASP = \measuredangle AQP = \measuredangle QPA</math>, so <math>\triangle PAX \sim \triangle XAP</math> and the desired result follows by side length ratios. | ||
+ | |||
+ | Lemma <math>2</math>: The locus of <math>N_9</math> as <math>X</math> moves along <math>\overline{PQ}</math> is a circle centered about <math>A</math>. | ||
+ | We add the midpoint of <math>\overline{AS}</math>, <math>N</math>, and let the circumradius of <math>\triangle AST</math> be <math>R</math>. Taking the power of <math>A</math> with respect to <math>(N_9)</math>, we get <cmath>AN_9^2 - \left(\frac{1}{2} R\right)^2 = \text{Pow}_{(N_9)} A = AX \cdot AN = \frac{1}{2} AX \cdot AS = \frac{1}{2} AP^2.</cmath> | ||
+ | Hence, <math>AN_9 = \sqrt{\frac{1}{4}R^2 + \frac{1}{2}AP^2}</math>, which remains constant as <math>X</math> moves. | ||
+ | |||
+ | Next, consider the homothety of scale factor <math>\frac{2}{3}</math> about <math>O</math> mapping <math>N_9</math> to <math>G</math>. This means that the locus of <math>G</math> is a circle as well. | ||
+ | |||
+ | Finally, we take a homothety of scale factor <math>\frac{3}{2}</math> about <math>A</math> mapping <math>G</math> to <math>M</math>. Hence, the locus of <math>M</math> is a circle, as desired. | ||
===Fake Solution=== | ===Fake Solution=== | ||
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 |
Revision as of 18:37, 28 July 2019
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(synthetic)
To begin with, we connect and we construct the nine-point circle of
centered at
.
Lemma
:
We proceed on a directed angle chase. We get
, so
and the desired result follows by side length ratios.
Lemma : The locus of
as
moves along
is a circle centered about
.
We add the midpoint of
,
, and let the circumradius of
be
. Taking the power of
with respect to
, we get
Hence,
, which remains constant as
moves.
Next, consider the homothety of scale factor about
mapping
to
. This means that the locus of
is a circle as well.
Finally, we take a homothety of scale factor about
mapping
to
. Hence, the locus of
is a circle, as desired.
Fake Solution
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