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

Revision as of 17:53, 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 (Problems • Resources)
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>

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