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

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(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.
 
(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.
  
__TOC__
+
== Solutions ==
== Solution ==
+
 
 
=== Solution 1 (synthetic) ===
 
=== Solution 1 (synthetic) ===
 +
Without Loss of Generality, assume <math>AB >AC</math>. It is sufficient to prove that <math>\angle OFA = 90^{\circ}</math>, as this would immediately prove that <math>A,P,O,F,N</math> are concyclic.
 +
By applying the Menelaus' Theorem in the Triangle <math>\triangle BFC</math> for the transversal <math>E,M,D</math>, we have (in magnitude)
 +
<cmath> \frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD}
 +
</cmath>
 +
Here, we used that <math>BM=MC</math>, as <math>M</math> is the midpoint of <math>BC</math>. Now, since <math>EC =EA</math> and <math>BD=DA</math>, we have
 +
<cmath> \frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD
 +
</cmath>
 +
Now, note that <math>OE</math> bisects the exterior <math>\angle FED</math> and <math>OD</math> bisects exterior <math>\angle FDE</math>, making <math>O</math> the <math>F</math>-excentre of <math>\triangle FED</math>. This implies that <math>OF</math> bisects interior <math>\angle EFD</math>, making <math>OF \perp AF</math>, as was required.
 +
 +
=== Solution 2 (complex) ===
 +
 +
Let <math>A=1,B=b,C=c</math> where <math>b,c</math> all lie on the unit circle. Then <math>O</math> is 0. As noted earlier, <math>(FOBC)</math> is cyclic. We will find the ghost point <math>F',</math> the second intersection of <math>OBC</math> and <math>ANP</math>.
 +
 +
We know that these two circles already intersect at <math>O</math> so we can reflect over the line between their centers. The center of <math>ANPO</math> is the midpoint of <math>AO</math> namely <math>\frac12</math>. With the tangent formula and then taking the midpoint, we find that the center of <math>OBC</math> is <math>\frac{bc}{b+c}.</math> Then we want to find the reflection of 0 over the line through <math>\frac12</math> and <math>\frac{bc}{b+c}.</math> Then we get
 +
 +
<cmath> \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*} </cmath>
 +
 +
Now it remains to show <math>\angle F'BA=\angle ABM;</math> the other angle equality would follow by symmetry.
 +
 +
Then we get:
 +
<cmath> \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*} </cmath>
 +
Thus <math>\measuredangle F'BA=\measuredangle BAM,</math> so <math>F'=F</math> and we're done.
 +
 +
~cocohearts
 +
 +
=== Solution 3 (synthetic) ===
 
<center><asy>
 
<center><asy>
 
   /* setup and variables */
 
   /* setup and variables */
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[[Without loss of generality]] <math>AB < AC</math>. The intersection of <math>NE</math> and <math>PD</math> is <math>O</math>, the circumcenter of <math>\triangle ABC</math>.  
 
[[Without loss of generality]] <math>AB < AC</math>. The intersection of <math>NE</math> and <math>PD</math> is <math>O</math>, the circumcenter of <math>\triangle ABC</math>.  
  
Let <math>\angle BAM = y</math> and <math>\angle CAM = z</math>. Note <math>D</math> lies on the perpendicular bisector of <math>AB</math>, so <math>AD = BD</math>. So <math>\angle FBC = \angle B - \angle ABD = B - y</math>. Similarly, <math>\angle FCE = C - z</math>, so <math>\angle BFC = 180 - (B + C) + (y + z) = 2A</math>. Notice that <math>\angle BOC</math> intercepts the minor arc <math>BC</math> in the [[circumcircle]] of <math>\triangle ABC</math>, which is double <math>\angle A</math>. Hence <math>\angle BFC = \angle BOC</math>, so <math>FOBC</math> is cyclic.  
+
Let <math>\angle BAM = y</math> and <math>\angle CAM = z</math>. Note <math>D</math> lies on the perpendicular bisector of <math>AB</math>, so <math>AD = BD</math>. So <math>\angle FBC = \angle B - \angle ABD = B - y</math>. Similarly, <math>\angle FCB = C - z</math>, so <math>\angle BFC = 180 - (B + C) + (y + z) = 2A</math>. Notice that <math>\angle BOC</math> intercepts the minor arc <math>BC</math> in the [[circumcircle]] of <math>\triangle ABC</math>, which is double <math>\angle A</math>. Hence <math>\angle BFC = \angle BOC</math>, so <math>BFOC</math> is cyclic.  
  
 +
'''Lemma.''' <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math>
  
''Lemma 1'': <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math>
+
''Proof.''
<cmath>
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<cmath>\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A</cmath>
\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A
+
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>BFOC</math> is cyclic, and <math>OB = OC</math>. Also
</cmath>
+
<cmath>\angle ENM = 90 - \angle MNC = 90 - A</cmath>
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>FOBC</math> is cyclic, and <math>OB = OC</math>. Also
 
<cmath>
 
\angle ENM = 90 - \angle MNC = 90 - A
 
</cmath>
 
 
because <math>NE\perp AC</math>, and <math>MNP</math> is the medial triangle of <math>\triangle ABC</math> so <math>AB \parallel MN</math>. Hence <math>\angle OFE = \angle ENM</math>.
 
because <math>NE\perp AC</math>, and <math>MNP</math> is the medial triangle of <math>\triangle ABC</math> so <math>AB \parallel MN</math>. Hence <math>\angle OFE = \angle ENM</math>.
  
Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp BC</math>. <math>\angle FED = \angle MEC = 2z</math>. Then
+
Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp AC</math>. <math>\angle FED = \angle MEC = 2z</math>. Then
<cmath>
+
<cmath>\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM</cmath>
\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM
 
</cmath>
 
 
Hence <math>\angle FEO = \angle NEM</math>.  
 
Hence <math>\angle FEO = \angle NEM</math>.  
  
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'''
  
 
<center><asy>
 
<center><asy>
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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));
  
/* commented from above asy
+
/* commented in above asy
 
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));  
 
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,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));
Line 76: Line 109:
 
</asy></center>
 
</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 the Lemma, <math>FE: EO = NE: EM\implies FE: EN = OE: EM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence
<cmath>
+
<cmath>\angle EMO = \angle ENF = \angle ONF</cmath>
\angle EMO = \angle ENF = \angle ONF
+
Using essentially the same angle chasing, we can show that <math>\triangle PDM</math> is directly similar to <math>\triangle FDO</math>. It follows that <math>\triangle PDF</math> is directly similar to <math>\triangle MDO</math>. So
</cmath>
+
<cmath>\angle EMO = \angle DMO = \angle DPF = \angle OPF</cmath>
Using essentially the same angle chasing, we can show that <math>\triangle PDM</math> is directly similar to <math>\triangle FMO</math>. It follows that <math>\triangle PDF</math> is directly similar to <math>MDO</math>. So
 
<cmath>
 
\angle EMO = \angle DMO = \angle DPF = \angle OPF
 
</cmath>
 
 
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 (synthetic) ===
+
=== Solution 4 (synthetic) ===
Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle. {{incomplete|solution}}
+
This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because <math>\angle OPA = \angle ONA = 90</math>. Let the circumcircles of triangles <math>APN</math> and <math>BOC</math> intersect at <math>F'</math> and <math>O</math>.
 +
 
 +
'''Lemma.''' If <math>A,B,C</math> are points on circle <math>\omega</math> with center <math>O</math>, and the tangents to <math>\omega</math> at <math>B,C</math> intersect at <math>Q</math>, then <math>AP</math> is the symmedian from <math>A</math> to <math>BC</math>.
  
=== Solution 3 (synthetic) ===
+
''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.
This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because <math>\angle OPA = \angle ONA = 90</math>. Let the circumcircles of triangles <math>APN</math> and <math>BOC</math> intersect at <math>F'</math> and <math>O</math>.
+
 
 +
'''End Lemma'''
  
Lemma. If <math>A,B,C</math> are points on circle <math>\omega</math> with center <math>O</math>, and the tangents to <math>\omega</math> at <math>B,C</math> intersect at <math>Q</math>, then <math>AP</math> is the symmedian from <math>A</math> to <math>BC</math>.
+
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done.
  
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.
 
  
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done. {{incomplete|solution}}
 
  
=== Solution 4 (trigonometric) ===
+
=== Solution 5 (trigonometric) ===
 
By the [[Law of Sines]], <math>\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}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>.
 
By the [[Law of Sines]], <math>\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}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>.
  
 
Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle ACF</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math> (by AA similarity), so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic.
 
Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle ACF</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math> (by AA similarity), so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic.
  
=== Solution 5 (isogonal conjugates) ===
+
=== Solution 6 (isogonal conjugates) ===
 
<center><asy>
 
<center><asy>
 
   /* setup and variables */
 
   /* setup and variables */
Line 136: Line 166:
 
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 6 (symmedians) ===
+
=== Solution 7 (symmedians) ===
 
Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>.
 
Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>.
 
Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian.
 
Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian.
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done. {{incomplete|solution}}
+
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done.
  
=== Solution 7 (inversion) ===
 
{{image}}
 
<center><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));
+
=== Solution 8 (inversion) (Official Solution #2) ===
pair Pa = A + (P-A)/(r*r); D(MP("P'",Pa,NW,s));
+
Invert the figure about a circle centered at <math>A</math>, and let <math>X'</math> denote the image of the point <math>X</math> under this inversion. Find point <math>F_1'</math> so that <math>AB'F_1'C'</math> is a parallelogram and let <math>Z'</math> denote the center of this parallelogram. Note that <math>\triangle BAC\sim\triangle C'AB'</math> and <math>\triangle BAD\sim\triangle D'AB'</math>. Because <math>M</math> is the midpoint of <math>BC</math> and <math>Z'</math> is the midpoint of <math>B'C'</math>, we also have <math>\triangle BAM\sim\triangle C'AZ'</math>. Thus
 +
<cmath>\angle AF_1'B' = \angle F_1'AC' = \angle Z'AC' = \angle MAB = \angle DAB = \angle DBA = \angle AD'B'.</cmath>
 +
Hence quadrilateral <math>AB'D'F_1'</math> is cyclic and, by a similar argument, quadrilateral <math>AC'E'F_1'</math> is also cyclic. Because the images under the inversion of lines <math>BDF</math> and <math>CFE</math> are circles that intersect in <math>A</math> and <math>F'</math>, it follows that <math>F_1' = F'</math>.
 +
 
 +
Next note that <math>B'</math>, <math>Z'</math>, and <math>C'</math> are collinear and are the images of <math>P'</math>, <math>F'</math>, and <math>N'</math>, respectively, under a homothety centered at <math>A</math> and with ratio <math>1/2</math>. It follows that <math>P'</math>, <math>F'</math>, and <math>N'</math> are collinear, and then that the points <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle.
 +
 
 +
<center>[[File:2008usamo2-sol8.png]]</center>
 +
 
 +
=== Solution 9 (Official Solution #1)===
 +
Let <math>O</math> be the circumcenter of triangle <math>ABC</math>. We prove that
 +
<cmath>\angle APO = \angle ANO = \angle AFO = 90^\circ.\qquad\qquad (1)</cmath>
 +
It will then follow that <math>A, P, O, F, N</math> lie on the circle with diameter <math>\overline{AO}</math>. Indeed, the fact that the first two angles in <math>(1)</math> are right is immediate because <math>\overline{OP}</math> and <math>\overline{ON}</math> are the perpendicular bisectors of <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively. Thus we need only prove that <math>\angle AFO = 90^\circ</math>.
 +
 
 +
<center>[[File:2008usamo2-sol9.png]]</center>
 +
 
 +
We may assume, without loss of generality, that <math>AB > AC</math>. This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because <math>\overline{PO}</math> is the perpendicular bisector of <math>\overline{AB}</math>, it follows that triangle <math>ADB</math> is an isosceles triangle with <math>AD = BD</math>. Likewise, triangle <math>AEC</math> is isosceles with <math>AE = CE</math>. Let <math>x = \angle ABD = \angle BAD</math> and <math>y = \angle CAE = \angle ACE</math>, so <math>x + y = \angle BAC</math>.
 +
 
 +
Applying the Law of Sines to triangles <math>ABM</math> and <math>ACM</math> gives
 +
<cmath>\frac{BM}{\sin x} = \frac{AB}{\sin\angle BMA}\quad\text{and}\quad\frac{CM}{\sin y} = \frac{AC}{\sin\angle CMA}.</cmath>
 +
Taking the quotient of the two equations and noting that <math>\sin\angle BMA = \sin\angle CMA</math>, we find
 +
<cmath>\frac{BM}{CM}\frac{\sin y}{\sin x} = \frac{AB}{AC}\frac{\sin\angle CMA}{\sin\angle BMA} = \frac{AB}{AC}.</cmath>
 +
Because <math>BM = MC</math>, we have
 +
<cmath>\frac{\sin x}{\sin y} = \frac{AC}{AB}.\qquad\qquad (2)</cmath>
 +
Applying the Law of Sines to triangles <math>ABF</math> and <math>ACF</math>, we find
 +
<cmath>\frac{AF}{\sin x} = \frac{AB}{\sin\angle AFB}\quad\text{and}\quad\frac{AF}{\sin y} = \frac{AC}{\sin\angle AFC}.</cmath>
 +
Taking the quotient of the two equations yields
 +
<cmath>\frac{\sin x}{\sin y} = \frac{AC}{AB}\frac{\sin\angle AFB}{\sin\angle AFC},</cmath>
 +
so by <math>(2)</math>,
 +
<cmath>\sin\angle AFB = \sin\angle AFC.\qquad\qquad (3)</cmath>
 +
Because <math>\angle ADF</math> is an exterior angle to triangle <math>ADB</math>, we have <math>\angle EDF = 2x</math>. Similarly, <math>\angle DEF = 2y</math>. Hence
 +
<cmath>\angle EFD = 180^\circ - 2x - 2y = 180^\circ - 2\angle BAC.</cmath>
 +
Thus <math>\angle BFC = 2\angle BAC = \angle BOC</math>, so <math>BOFC</math> is cyclic. In addition,
 +
<cmath>\angle AFB + \angle AFC = 360^\circ - 2\angle BAC > 180^\circ,</cmath>
 +
and hence, from <math>(3)</math>, <math>\angle AFB = \angle AFC = 180^\circ - \angle BAC</math>. Because <math>BOFC</math> is cyclic and <math>\triangle BOC</math> is isosceles with vertex angle <math>\angle BOC = 2\angle BAC</math>, we have <math>\angle OFB = \angle OCB = 90^\circ - \angle BAC</math>. Therefore,
 +
<cmath>\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.</cmath>
 +
This completes the proof.
 +
=== Solution 10 (Anti-Steiner point) ===
 +
First, let <math>O</math> be the circumcenter of <math>\triangle{ABC}</math>. Note that since <math>AP \perp PO</math> and <math>AN \perp NO</math>, then <math>A</math>, <math>N</math>, <math>O</math> and <math>P</math>, are concyclic.
 +
 
 +
Notice that <math>NM \parallel AB</math> and <math>AB \perp PO</math>, which means <math>NM \perp PO</math>. It follows analogously that <math>PM \perp NO</math>. This means that <math>M</math> is the orthocenter of triangle <math>{NOP}</math>.
 +
 
 +
Now consider what happens when we reflect line <math>AM</math> over line <math>PO</math>. Since <math>PO</math> is just the perpendicular bisector of <math>AB</math>, point <math>A</math> will map to point <math>B</math>, and point <math>D</math>, which is both line <math>AM</math> and <math>PO</math>, will map to itself. Therefore, we get line <math>BD</math> from this reflection. It follows analogously that by reflecting line <math>AM</math> over line <math>NO</math>, we get <math>CE</math>.
 +
 
 +
Since line <math>AM</math> passes through <math>M</math>, the orthocenter of triangle <math>{NOP}</math>, its reflections over sides <math>PO</math> and <math>NO</math>, which correspond to <math>BD</math> and <math>CE</math> respectively, intersect on <math>\odot{NOP}</math> (this is the Anti-Steiner point application). Therefore, <math>F</math>, the intersection of <math>BD</math> and <math>CE</math>, lies on <math>\odot{NOP}</math>.
 +
 
 +
Since <math>A</math>, <math>N</math>, <math>O</math>, and <math>P</math> are concyclic, then <math>\odot{NOP}</math> is the same as <math>\odot{ANP}</math>, so <math>F</math> lies on <math>\odot{ANP}</math>, which is the same as saying <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> are concyclic, as desired.
  
</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}}
 
  
=== Solution 8 (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}}
  
== Resources ==
+
== See Also ==
 +
* <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>
 +
 
 
{{USAMO newbox|year=2008|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2008|num-b=1|num-a=3}}
 
* <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>
 
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:54, 1 May 2024

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.

2008usamo2-sol8.png

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 $\angle AFO = 90^\circ$.

2008usamo2-sol9.png

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.


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

See Also

  • <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>
2008 USAMO (ProblemsResources)
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