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

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== Solution ==
 
== Solution ==
 
=== Solution 1 (synthetic) ===
 
=== Solution 1 (synthetic) ===
=== Solution 4 (synthetic) ===
 
 
<center><asy>
 
<center><asy>
 
   /* setup and variables */
 
   /* setup and variables */
Line 14: Line 13:
 
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);
 
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(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,N,s)));
+
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s)));
 
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));
 
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));
 
D(C--D(MP("F",F,NW,s)));  
 
D(C--D(MP("F",F,NW,s)));  
 
D(B--O--C,linetype("4 4")+linewidth(0.7));
 
D(B--O--C,linetype("4 4")+linewidth(0.7));
 +
D(M--N,linetype("4 4")+linewidth(0.7));
 
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));
 
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));
 
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));
 
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));
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Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.''
 
Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar. ''End Lemma 1.''
  
 +
<center><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=(1,4); /* 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(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s)));
 +
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));
 +
D(C--D(MP("F",F,NW,s)));
 +
D(B--O--C,linetype("4 4")+linewidth(0.7));
 +
D(F--N); D(O--M);
 +
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));
 +
 +
/* commented from above asy
 +
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));
 +
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));
 +
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));
 +
picture p = new picture;
 +
draw(p,circumcircle(B,O,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></center>
  
 
By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence
 
By the similarity in Lemma 1, <math>FE: EO = NE: EM\implies FE: EN = OE: NM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence
Line 62: Line 86:
 
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired.
 
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired.
  
=== Solution 2 (isogonal conjugates) ===
+
=== Solution 2 (synthetic) ===
 +
Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}}
 +
 
 +
=== Solution 3 (isogonal conjugates) ===
 
<center><asy>
 
<center><asy>
 
   /* setup and variables */
 
   /* setup and variables */
Line 95: Line 122:
 
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.
 
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.
  
=== Solution 3 (inversion) ===
+
=== Solution 4 (inversion) ===
 
{{image}}
 
{{image}}
 
<center><asy>
 
<center><asy>
Line 101: Line 128:
 
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);
 
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 */
 
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */
 +
real r = 1.2; /* inversion radius */
 +
 
   /* construction and drawing */
 
   /* 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);
 
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);
Line 106: Line 135:
 
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,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(D(MP("D",D,SE,s))--MP("P",P,W,s));
D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7));
+
D(B--D(MP("F",F,s)));  
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));
 
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));
  
/* removal of code from original
+
D(CR(A,r));
 +
pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s));
  
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></center>
 
</asy></center>
 
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}}
 
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}}
  
=== Solution 4 (trigonometric) ===
+
=== Solution 5 (trigonometric) ===
 
Use the [[Law of Sines]] to show that <math>\angle BFA = \angle AFC</math>. It follows that <math>\triangle BFA \sim \triangle AFC</math>. Noting the [[spiral similarity]] from <math>P</math> to <math>N</math>, {{incomplete|solution}}
 
Use the [[Law of Sines]] to show that <math>\angle BFA = \angle AFC</math>. It follows that <math>\triangle BFA \sim \triangle AFC</math>. Noting the [[spiral similarity]] from <math>P</math> to <math>N</math>, {{incomplete|solution}}
  
=== Solution 5 (analytical) ===
+
=== Solution 6 (analytical) ===
 +
We let <math>A</math> be at the [[origin]], <math>B</math> be at the point <math>(a,0)</math>, and <math>C</math> be at the point <math>(b,c):\ b<a</math>. Then the equation of the perpendicular bisector of <math>\overline{AB}</math> is <math>x = a/2</math>, and {{incomplete|solution}}
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 16:53, 2 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 (synthetic)

[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=(1,4); /* 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(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s)));  D(B--O--C,linetype("4 4")+linewidth(0.7)); D(M--N,linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,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);  /* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */ [/asy]

Without loss of generality $AB < AC$. The intersection of $NE$ and $PD$ is $O$, the circumcenter of $\triangle ABC$.

Let $\angle BAM = y$ and $\angle CAM = z$. Note $D$ lies on the perpendicular bisector of $AB$, so $AD = BD$. So $\angle FBC = \angle B - \angle ABD = B - y$. Similarly, $\angle FCE = C - z$, so $\angle BFC = 180 - (B + C) + (y + z) = 2A$. Notice that $\angle BOC$ intercepts the minor arc $BC$ in the circumcircle of $\triangle ABC$, which is double $\angle A$. Hence $\angle BFC = \angle BOC$, so $FOBC$ is cyclic.


Lemma 1: $\triangle FEO$ is directly similar to $\triangle NEM$ \[\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A\] since $F$, $E$, $C$ are collinear, $FOBC$ is cyclic, and $OB = OC$. Also \[\angle ENM = 90 - \angle MNC = 90 - A\] because $NE\perp AC$, and $MNP$ is the medial triangle of $\triangle ABC$ so $AB \parallel MN$. Hence $\angle OFE = \angle ENM$.

Notice that $\angle AEN = 90 - z = \angle CEN$ since $NE\perp BC$. $\angle FED = \angle MEC = 2z$. Then \[\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM\] Hence $\angle FEO = \angle NEM$.

Hence $\triangle FEO$ is similar to $\triangle NEM$ by AA similarity. It is easy to see that they are oriented such that they are directly similar. End Lemma 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=(1,4); /* 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(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s)));  D(B--O--C,linetype("4 4")+linewidth(0.7)); D(F--N); D(O--M); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));  /* commented from above asy D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));  D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,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]

By the similarity in Lemma 1, $FE: EO = NE: EM\implies FE: EN = OE: NM$. $\angle FEN = \angle OEM$ so $\triangle FEN\sim\triangle OEM$ by SAS similarity. Hence \[\angle EMO = \angle ENF = \angle ONF\] Using essentially the same angle chasing, we can show that $\triangle PDM$ is directly similar to $\triangle FMO$. It follows that $\triangle PDF$ is directly similar to $MDO$. So \[\angle EMO = \angle DMO = \angle DPF = \angle OPF\] Hence $\angle OPF = \angle ONF$, so $FONP$ is cyclic. In other words, $F$ lies on the circumcircle of $\triangle PON$. Note that $\angle ONA = \angle OPA = 90$, so $APON$ is cyclic. In other words, $A$ lies on the circumcircle of $\triangle PON$. $A$, $P$, $N$, $O$, and $F$ all lie on the circumcircle of $\triangle PON$. Hence $A$, $P$, $F$, and $N$ lie on a circle, as desired.

Solution 2 (synthetic)

Hint: consider $CF$ intersection with $PM$; show that the resulting intersection lies on the desired circle. Template:Incomplete

Solution 3 (isogonal conjugates)

[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--A--F,linetype("4 4")+linewidth(0.7)); D(MP("O'",circumcenter(A,P,N),NW,s)); 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 4 (inversion)


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


[asy] 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 */ real r = 1.2; /* inversion radius */    /* 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(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));  D(CR(A,r)); pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s));  [/asy]

We consider an inversion by an arbitrary radius about $A$. We want to show that $P', F',$ and $N'$ are collinear. Notice that $D', A,$ and $P'$ lie on a circle with center $B'$, and similarly for the other side. We also have that $B', D', F', A$ form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that $A B' F' C'$ is a parallelogram, indicating that $F'$ is the midpoint of $P'N'$. Template:Incomplete

Solution 5 (trigonometric)

Use the Law of Sines to show that $\angle BFA = \angle AFC$. It follows that $\triangle BFA \sim \triangle AFC$. Noting the spiral similarity from $P$ to $N$, Template:Incomplete

Solution 6 (analytical)

We let $A$ be at the origin, $B$ be at the point $(a,0)$, and $C$ be at the point $(b,c):\ b<a$. Then the equation of the perpendicular bisector of $\overline{AB}$ is $x = a/2$, and Template:Incomplete

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
  • <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>