Difference between revisions of "2015 USAMO Problems/Problem 2"
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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>. | Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>. | ||
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− | Solution 2 | + | ===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>. | 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>. | ||
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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. | 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>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>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle APO</math>. | ||
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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 PAS \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. - Spacesam |
Latest revision as of 09:58, 20 August 2021
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. - Spacesam