2008 USAMO Problems/Problem 2

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 (isogonal conjugates)

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

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 (inversion)

Invert 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 quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) Template:Incomplete

Solution 3 (trigonometric)

Solution 4 (synthetic)

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>

Page

Toolbox

Search