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.

Solutions

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, $\angle FCB = C - z$ , 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 $BFOC$ is cyclic.

Lemma. $\triangle FEO$ is directly similar to $\triangle NEM$

Proof. $\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A$ since $F$ , $E$ , $C$ are collinear, $BFOC$ 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 $NE\perp AC$ . $\angle FED = \angle MEC = 2z$ . 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

$[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s))); D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s)); D(C--D(MP("F",F,NW,s))); D(B--O--C,linetype("4 4")+linewidth(0.7)); D(F--N); D(O--M); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); /* commented in above asy D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6)); D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); */ [/asy]$

By the similarity in the Lemma, $FE: EO = NE: EM\implies FE: EN = OE: EM$ . $\angle FEN = \angle OEM$ 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 $\triangle FDO$ . It follows that $\triangle PDF$ is directly similar to $\triangle MDO$ . 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)

Without Loss of Generality, assume $AB >AC$ . It is sufficient to prove that $\angle OFA = 90^{\circ}$ , as this would immediately prove that $A,P,O,F,N$ are concyclic. By applying the Menelaus' Theorem in the Triangle $\triangle BFC$ for the transversal $E,M,D$ , we have (in magnitude) $\frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD}$ Here, we used that $BM=MC$ , as $M$ is the midpoint of $BC$ . Now, since $EC =EA$ and $BD=DA$ , we have $\frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD$ Now, note that $OE$ bisects the exterior $\angle FED$ and $OD$ bisects exterior $\angle FDE$ , making $O$ the $F$ -excentre of $\triangle FED$ . This implies that $OF$ bisects interior $\angle EFD$ , making $OF \perp AF$ , as was required.

Solution 3 (synthetic)

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

Solution 4 (synthetic)

This solution utilizes the phantom point method. Clearly, APON are cyclic because $\angle OPA = \angle ONA = 90$ . Let the circumcircles of triangles $APN$ and $BOC$ intersect at $F'$ and $O$ .

Lemma. If $A,B,C$ are points on circle $\omega$ with center $O$ , and the tangents to $\omega$ at $B,C$ intersect at $Q$ , then $AP$ is the symmedian from $A$ to $BC$ .

Proof. This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.

End Lemma

It is easy to see $Q$ (the intersection of ray $OM$ and the circumcircle of $\triangle BOC$ ) is colinear with $A$ and $F'$ , and because line $OM$ is the diameter of that circle, $\angle QBO = \angle QCO = 90$ , so $Q$ is the point $Q$ in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that $F'$ satisfies the original construction for $F$ , showing $F=F'$ ; we are done.

Solution 5 (trigonometric)

By the Law of Sines, $\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}$ . Since $\angle ABF = \angle ABD = \angle BAD = \angle BAM$ and similarly $\angle ACF = \angle CAM$ , we cancel to get $\sin\angle AFC = \sin\angle AFB$ . Obviously, $\angle AFB + \angle AFC > 180^\circ$ so $\angle AFC = \angle AFB$ .

Then $\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF$ and $\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC$ . Subtracting these two equations, $\angle FAB - \angle FCA = \angle FCA - \angle FAB$ so $\angle BAF = \angle ACF$ . Therefore, $\triangle ABF\sim\triangle CAF$ (by AA similarity), so a spiral similarity centered at $F$ takes $B$ to $A$ and $A$ to $C$ . Therefore, it takes the midpoint of $\overline{BA}$ to the midpoint of $\overline{AC}$ , or $P$ to $N$ . So $\angle APF = \angle CNF = 180^\circ - \angle ANF$ and $APFN$ is cyclic.

Solution 6 (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 7 (symmedians)

Median $AM$ of a triangle $ABC$ implies $\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}$ . Trig ceva for $F$ shows that $AF$ is a symmedian. Then $FP$ is a median, use the lemma again to show that $AFP = C$ , and similarly $AFN = B$ , so you're done.

Solution 8 (inversion) (Official Solution #2)

Invert the figure about a circle centered at $A$ , and let $X'$ denote the image of the point $X$ under this inversion. Find point $F_1'$ so that $AB'F_1'C'$ is a parallelogram and let $Z'$ denote the center of this parallelogram. Note that $\triangle BAC\sim\triangle C'AB'$ and $\triangle BAD\sim\triangle D'AB'$ . Because $M$ is the midpoint of $BC$ and $Z'$ is the midpoint of $B'C'$ , we also have $\triangle BAM\sim\triangle C'AZ'$ . Thus $\angle AF_1'B' = \angle F_1'AC' = \angle Z'AC' = \angle MAB = \angle DAB = \angle DBA = \angle AD'B'.$ Hence quadrilateral $AB'D'F_1'$ is cyclic and, by a similar argument, quadrilateral $AC'E'F_1'$ is also cyclic. Because the images under the inversion of lines $BDF$ and $CFE$ are circles that intersect in $A$ and $F'$ , it follows that $F_1' = F'$ .

Next note that $B'$ , $Z'$ , and $C'$ are collinear and are the images of $P'$ , $F'$ , and $N'$ , respectively, under a homothety centered at $A$ and with ratio $1/2$ . It follows that $P'$ , $F'$ , and $N'$ are collinear, and then that the points $A$ , $P$ , $F$ , and $N$ lie on a circle.

Solution 9 (Official Solution #1)

Let $O$ be the circumcenter of triangle $ABC$ . We prove that $\angle APO = \angle ANO = \angle AFO = 90^\circ.\qquad\qquad (1)$ It will then follow that $A, P, O, F, N$ lie on the circle with diameter $\overline{AO}$ . Indeed, the fact that the first two angles in $(1)$ are right is immediate because $\overline{OP}$ and $\overline{ON}$ are the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ , respectively. Thus we need only prove that $\angle AFO = 90^\circ$ .

We may assume, without loss of generality, that $AB > AC$ . This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because $\overline{PO}$ is the perpendicular bisector of $\overline{AB}$ , it follows that triangle $ADB$ is an isosceles triangle with $AD = BD$ . Likewise, triangle $AEC$ is isosceles with $AE = CE$ . Let $x = \angle ABD = \angle BAD$ and $y = \angle CAE = \angle ACE$ , so $x + y = \angle BAC$ .

Applying the Law of Sines to triangles $ABM$ and $ACM$ gives $\frac{BM}{\sin x} = \frac{AB}{\sin\angle BMA}\quad\text{and}\quad\frac{CM}{\sin y} = \frac{AC}{\sin\angle CMA}.$ Taking the quotient of the two equations and noting that $\sin\angle BMA = \sin\angle CMA$ , we find $\frac{BM}{CM}\frac{\sin y}{\sin x} = \frac{AB}{AC}\frac{\sin\angle CMA}{\sin\angle BMA} = \frac{AB}{AC}.$ Because $BM = MC$ , we have $\frac{\sin x}{\sin y} = \frac{AC}{AB}.\qquad\qquad (2)$ Applying the Law of Sines to triangles $ABF$ and $ACF$ , we find $\frac{AF}{\sin x} = \frac{AB}{\sin\angle AFB}\quad\text{and}\quad\frac{AF}{\sin y} = \frac{AC}{\sin\angle AFC}.$ Taking the quotient of the two equations yields $\frac{\sin x}{\sin y} = \frac{AC}{AB}\frac{\sin\angle AFB}{\sin\angle AFC},$ so by $(2)$ , $\sin\angle AFB = \sin\angle AFC.\qquad\qquad (3)$ Because $\angle ADF$ is an exterior angle to triangle $ADB$ , we have $\angle EDF = 2x$ . Similarly, $\angle DEF = 2y$ . Hence $\angle EFD = 180^\circ - 2x - 2y = 180^\circ - 2\angle BAC.$ Thus $\angle BFC = 2\angle BAC = \angle BOC$ , so $BOFC$ is cyclic. In addition, $\angle AFB + \angle AFC = 360^\circ - 2\angle BAC > 180^\circ,$ and hence, from $(3)$ , $\angle AFB = \angle AFC = 180^\circ - \angle BAC$ . Because $BOFC$ is cyclic and $\triangle BOC$ is isosceles with vertex angle $\angle BOC = 2\angle BAC$ , we have $\angle OFB = \angle OCB = 90^\circ - \angle BAC$ . Therefore, $\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.$ This completes the proof.

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

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

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