# 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)

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 2 (complex)

Let $A=1,B=b,C=c$ where $b,c$ all lie on the unit circle. Then $O$ is 0. As noted earlier, $(FOBC)$ is cyclic. We will find the ghost point $F',$ the second intersection of $OBC$ and $ANP$.

We know that these two circles already intersect at $O$ so we can reflect over the line between their centers. The center of $ANPO$ is the midpoint of $AO$ namely $\frac12$. With the tangent formula and then taking the midpoint, we find that the center of $OBC$ is $\frac{bc}{b+c}.$ Then we want to find the reflection of 0 over the line through $\frac12$ and $\frac{bc}{b+c}.$ Then we get

\begin{align*} f' &= \frac{\left(\frac{bc}{b+c}-\overline{\frac{bc}{b+c}}\right)\div2}{(1/2)-\overline{\frac{bc}{b+c}}}\\ &= \frac{\frac{bc-1}{2(b+c)}}{\frac{b+c-2}{2(b+c)}}\\ &= \frac{bc-1}{b+c-2}. \end{align*}

Now it remains to show $\angle F'BA=\angle ABM;$ the other angle equality would follow by symmetry.

Then we get: \begin{align*} \frac{f'-b}{b-a}\div\frac{b-a}{a-m} &=\frac{\frac{bc-1}{b+c-2}-b}{b-1}\div\frac{b-1}{1-\frac{b+c}2}\\ &=\frac{\frac{bc-1-b(b+c-2)}{b+c-2}(2-b+c)}{2(b-1)^2}\\ &=\frac{b^2-2b+1}{2(b-1)^2}\\ &=\frac12. \end{align*} Thus $\measuredangle F'BA=\measuredangle BAM,$ so $F'=F$ and we're done.

~cocohearts

### Solution 3 (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 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.

### Solution 10 (Anti-Steiner point)

First, let $O$ be the circumcenter of $\triangle{ABC}$. Note that since $AP \perp PO$ and $AN \perp NO$, then $A$, $N$, $O$ and $P$, are concyclic.

Notice that $NM \parallel AB$ and $AB \perp PO$, which means $NM \perp PO$. It follows analogously that $PM \perp NO$. This means that $M$ is the orthocenter of triangle ${NOP}$.

Now consider what happens when we reflect line $AM$ over line $PO$. Since $PO$ is just the perpendicular bisector of $AB$, point $A$ will map to point $B$, and point $D$, which is both line $AM$ and $PO$, will map to itself. Therefore, we get line $BD$ from this reflection. It follows analogously that by reflecting line $AM$ over line $NO$, we get $CE$.

Since line $AM$ passes through $M$, the orthocenter of triangle ${NOP}$, its reflections over sides $PO$ and $NO$, which correspond to $BD$ and $CE$ respectively, intersect on $\odot{NOP}$ (this is the Anti-Steiner point application). Therefore, $F$, the intersection of $BD$ and $CE$, lies on $\odot{NOP}$.

Since $A$, $N$, $O$, and $P$ are concyclic, then $\odot{NOP}$ is the same as $\odot{ANP}$, so $F$ lies on $\odot{ANP}$, which is the same as saying $A$, $N$, $F$, and $P$ are concyclic, as desired.