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

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== Problem ==
 
== Problem ==
(''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 bisectors 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.
 +
 
 +
== Solutions ==
 +
 
 +
=== Solution 1 (synthetic) ===
 +
<center><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></center>
 +
 
 +
[[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 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>
 +
 
 +
''Proof.''
 +
<cmath>\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A</cmath>
 +
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>BFOC</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>.
 +
 
 +
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>\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM</cmath>
 +
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'''
 +
 
 +
<center><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></center>
 +
 
 +
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>\angle EMO = \angle ENF = \angle ONF</cmath>
 +
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>\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.
 +
 
 +
=== Solution 2 (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 3 (synthetic) ===
 +
Hint: consider <math>CF</math> intersection with <math>PM</math>; 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 <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>.
 +
 
 +
''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 <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.
 +
 
 +
 
 +
 
 +
=== 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>.
 +
 
 +
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 6 (isogonal conjugates) ===
 +
<center><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></center>
 +
 
 +
Construct <math>T</math> on <math>AM</math> such that <math>\angle BCT = \angle ACF</math>. Then <math>\angle BCT = \angle CAM</math>. Then <math>\triangle AMC\sim\triangle CMT</math>, so <math>\frac {AM}{CM} = \frac {CM}{TM}</math>, or <math>\frac {AM}{BM} = \frac {BM}{TM}</math>. Then <math>\triangle AMB\sim\triangle BMT</math>, so <math>\angle CBT = \angle BAM = \angle FBA</math>. Then we have
 +
 
 +
<math>\angle CBT = \angle ABF</math> and <math>\angle BCT = \angle ACF</math>. So <math>T</math> and <math>F</math> are [[isogonal conjugate|isogonally conjugate]]. Thus <math>\angle BAF = \angle CAM</math>. Then
 +
 
 +
<math>\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC</math>.
 +
 
 +
If <math>O</math> is the [[circumcenter]] of <math>\triangle ABC</math> then <math>\angle BFC = 2\angle BAC = \angle BOC</math> so <math>BFOC</math> is cyclic. Then <math>\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC</math>.
 +
 
 +
Then <math>\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90</math>. Then <math>\triangle AFO</math> is a right triangle.
 +
 
 +
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 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>.
 +
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.
 +
 
 +
 
 +
 
 +
=== Solution 8 (inversion) (Official Solution #2) ===
 +
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.
  
__TOC__
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 2 ===
 
  
 
{{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 00:26, 30 August 2020

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.

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.


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)
Preceded by
Problem 1
Followed by
Problem 3
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