Difference between revisions of "2008 USAMO Problems/Problem 2"

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(I'll construct T later -_-)
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== Problem ==
 
== Problem ==
(''Zuming Feng'') Let <math>ABC</math> be an acute, scalene triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the perpendicular bisectors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.
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(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.
  
 
__TOC__
 
__TOC__
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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<center><asy>
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  /* setup and variables */
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size(280);
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pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);
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pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */
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  /* construction and drawing */
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pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);
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D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));
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D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));
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D(D(MP("D",D,SE,s))--MP("P",P,W,s));
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D(B--D(MP("F",F,s))); D(O--F--A,linetype("4 4")+linewidth(0.7));
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D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));
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D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));
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picture p = new picture;
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draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));
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draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7));
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clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);
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</asy></center>
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Construct <math>T</math> on <math>AM</math> such that <math>\angle BCT = \angle ACF</math>. Then <math>\angle BCT = \angle CAM</math>. Then <math>\triangle AMC\sim\triangle CMT</math>, so <math>\frac {AM}{CM} = \frac {CM}{TM}</math>, or <math>\frac {AM}{BM} = \frac {BM}{TM}</math>. Then <math>\triangle AMB\sim\triangle BMT</math>, so <math>\angle CBT = \angle BAM = \angle FBA</math>. Then we have
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<math>\angle CBT = \angle ABF</math> and <math>\angle BCT = \angle ACF</math>. So <math>T</math> and <math>F</math> are [[isogonal conjugate|isogonally conjugate]]. Thus <math>\angle BAF = \angle CAM</math>. Then
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<math>\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC</math>.
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If <math>O</math> is the [[circumcenter]] of <math>\triangle ABC</math> then <math>\angle BFC = 2\angle BAC = \angle BOC</math> so <math>BFOC</math> is cyclic. Then <math>\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC</math>.
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Then <math>\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90</math>. Then <math>\triangle AFO</math> is a right triangle.
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Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle.
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=== Solution 2 ===
 
=== Solution 2 ===
  

Revision as of 19:12, 1 May 2008

Problem

(Zuming Feng) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$, inside of triangle $ABC$. Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.

Solution

Solution 1

[asy]   /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */   /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(O--F--A,linetype("4 4")+linewidth(0.7)); D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); [/asy]

Construct $T$ on $AM$ such that $\angle BCT = \angle ACF$. Then $\angle BCT = \angle CAM$. Then $\triangle AMC\sim\triangle CMT$, so $\frac {AM}{CM} = \frac {CM}{TM}$, or $\frac {AM}{BM} = \frac {BM}{TM}$. Then $\triangle AMB\sim\triangle BMT$, so $\angle CBT = \angle BAM = \angle FBA$. Then we have

$\angle CBT = \angle ABF$ and $\angle BCT = \angle ACF$. So $T$ and $F$ are isogonally conjugate. Thus $\angle BAF = \angle CAM$. Then

$\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC$.

If $O$ is the circumcenter of $\triangle ABC$ then $\angle BFC = 2\angle BAC = \angle BOC$ so $BFOC$ is cyclic. Then $\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC$.

Then $\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90$. Then $\triangle AFO$ is a right triangle.

Now by the homothety centered at $A$ with ratio $\frac {1}{2}$, $B$ is taken to $P$ and $C$ is taken to $N$. Thus $O$ is taken to the circumcenter of $\triangle APN$ and is the midpoint of $AO$, which is also the circumcenter of $\triangle AFO$, so $A,P,N,F,O$ all lie on a circle.

Solution 2

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

2008 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions
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