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Difference between revisions of "2015 USAMO Problems/Problem 2"

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===Problem===
 
===Problem===
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Solve in integers the equation
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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.
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<cmath> x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. </cmath>
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===Solution===
 
===Solution===
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We first notice that both sides must be integers, so <math>\frac{x+y}{3}</math> must be an integer.
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WLOG, let the circle be the unit circle centered at the origin, A=(1,0) P=(1-a,b), Q=(1-a,-b), where (1-a)^2+b^2=1. Let angle <XAB=A, which is an acute angle, tanA=t, then X=(1-a,at).
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We can therefore perform the substitution <math>x+y = 3t</math> where <math>t</math> is an integer.
 
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Then:
 
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<math>(3t)^2 - xy = (t+1)^3</math>
 
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<math>9t^2 + x (x - 3t) = t^3 + 3t^2 + 3t + 1</math>
 
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<math>4x^2 - 12xt + 9t^2 = 4t^3 - 15t^2 + 12t + 4</math>
 
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<math>(2x - 3t)^2 = (t - 2)^2(4t + 1)</math>
 
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<math>4t+1</math> is therefore the square of an odd integer and can be replaced with <math>(2n+1)^2 = 4n^2 + 4n +1</math>
 
  
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By substituting using <math>t = n^2 + n</math> we get:
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Angle <BOS=2A, S=(-cos2A,sin2A).
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Let M=(u,v), then T=(2u+cos2A, 2v-sin2A)
  
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<math>(2x - 3n^2 - 3n)^2 = [(n^2 + n - 2)(2n+1)]^2</math>
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The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA.    (E1)
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Use identities (cosA)^2=1/(1+t^2),  cos2A=2(cosA)^2-1= 2/(1+t^2) -1, sin2A=2sinAcosA=2t^2/(1+t^2), we obtain 2vt-at^2=2u+a.  (E1')
  
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<math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math>
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The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2.  (E2)
  
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<math>x = n^3 + 3n^2 - 1</math> or <math>x = n^3 + 3n - 1</math>
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M is the mid-point on the hypotenuse of triangle STX, hence MS=MX, yielding (u+cos2A)^2+(v-sin2A)^2=(u+a-1)^2+(v-at)^2.  (E3)
  
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Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, n^3 + 3n - 1) \cup (n^3 + 3n - 1, n^3 + 3n^2 - 1)</math>
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Expand (E3), using (E2) to replace 2(vsin2A-ucos2A) with 2(u^2+v^2), and using (E1') to replace a(-2vt+at^2) with -a(2u+a), and we obtain
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u^2-u-a+v^2=0, namely (u-1/2)^2+v^2=a+1/4, which is a circle centered at (1/2,0) with radius r=sqrt(a+1/4).

Revision as of 11:11, 12 May 2015

Problem

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ moves along a circle.

Solution

WLOG, let the circle be the unit circle centered at the origin, A=(1,0) P=(1-a,b), Q=(1-a,-b), where (1-a)^2+b^2=1. Let angle <XAB=A, which is an acute angle, tanA=t, then X=(1-a,at).

Angle <BOS=2A, S=(-cos2A,sin2A). Let M=(u,v), then T=(2u+cos2A, 2v-sin2A)

The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA. (E1) Use identities (cosA)^2=1/(1+t^2), cos2A=2(cosA)^2-1= 2/(1+t^2) -1, sin2A=2sinAcosA=2t^2/(1+t^2), we obtain 2vt-at^2=2u+a. (E1')

The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2. (E2)

M is the mid-point on the hypotenuse of triangle STX, hence MS=MX, yielding (u+cos2A)^2+(v-sin2A)^2=(u+a-1)^2+(v-at)^2. (E3)

Expand (E3), using (E2) to replace 2(vsin2A-ucos2A) with 2(u^2+v^2), and using (E1') to replace a(-2vt+at^2) with -a(2u+a), and we obtain u^2-u-a+v^2=0, namely (u-1/2)^2+v^2=a+1/4, which is a circle centered at (1/2,0) with radius r=sqrt(a+1/4).

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