Difference between revisions of "2008 USAMO Problems/Problem 2"
I like pie (talk | contribs) (Fixed link) |
|||
(37 intermediate revisions by 18 users not shown) | |||
Line 1: | Line 1: | ||
== 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 | + | (''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. | ||
− | |||
− | |||
− | |||
− | |||
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == 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}} | ||
− | |||
− | |||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:26, 30 August 2020
Contents
Problem
(Zuming Feng) Let be an acute, scalene triangle, and let , , and be the midpoints of , , and , respectively. Let the perpendicular bisectors of and intersect ray in points and respectively, and let lines and intersect in point , inside of triangle . Prove that points , , , and all lie on one circle.
Solutions
Solution 1 (synthetic)
Without loss of generality . The intersection of and is , the circumcenter of .
Let and . Note lies on the perpendicular bisector of , so . So . Similarly, , so . Notice that intercepts the minor arc in the circumcircle of , which is double . Hence , so is cyclic.
Lemma. is directly similar to
Proof. since , , are collinear, is cyclic, and . Also because , and is the medial triangle of so . Hence .
Notice that since . . Then Hence .
Hence is similar to by AA similarity. It is easy to see that they are oriented such that they are directly similar.
End Lemma
By the similarity in the Lemma, . so by SAS similarity. Hence Using essentially the same angle chasing, we can show that is directly similar to . It follows that is directly similar to . So Hence , so is cyclic. In other words, lies on the circumcircle of . Note that , so is cyclic. In other words, lies on the circumcircle of . , , , , and all lie on the circumcircle of . Hence , , , and lie on a circle, as desired.
Solution 2 (synthetic)
Without Loss of Generality, assume . It is sufficient to prove that , as this would immediately prove that are concyclic. By applying the Menelaus' Theorem in the Triangle for the transversal , we have (in magnitude) Here, we used that , as is the midpoint of . Now, since and , we have Now, note that bisects the exterior and bisects exterior , making the -excentre of . This implies that bisects interior , making , as was required.
Solution 3 (synthetic)
Hint: consider intersection with ; 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 . Let the circumcircles of triangles and intersect at and .
Lemma. If are points on circle with center , and the tangents to at intersect at , then is the symmedian from to .
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 (the intersection of ray and the circumcircle of ) is colinear with and , and because line is the diameter of that circle, , so is the point in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that satisfies the original construction for , showing ; we are done.
Solution 5 (trigonometric)
By the Law of Sines, . Since and similarly , we cancel to get . Obviously, so .
Then and . Subtracting these two equations, so . Therefore, (by AA similarity), so a spiral similarity centered at takes to and to . Therefore, it takes the midpoint of to the midpoint of , or to . So and is cyclic.
Solution 6 (isogonal conjugates)
Construct on such that . Then . Then , so , or . Then , so . Then we have
and . So and are isogonally conjugate. Thus . Then
.
If is the circumcenter of then so is cyclic. Then .
Then . Then is a right triangle.
Now by the homothety centered at with ratio , is taken to and is taken to . Thus is taken to the circumcenter of and is the midpoint of , which is also the circumcenter of , so all lie on a circle.
Solution 7 (symmedians)
Median of a triangle implies . Trig ceva for shows that is a symmedian. Then is a median, use the lemma again to show that , and similarly , so you're done.
Solution 8 (inversion) (Official Solution #2)
Invert the figure about a circle centered at , and let denote the image of the point under this inversion. Find point so that is a parallelogram and let denote the center of this parallelogram. Note that and . Because is the midpoint of and is the midpoint of , we also have . Thus Hence quadrilateral is cyclic and, by a similar argument, quadrilateral is also cyclic. Because the images under the inversion of lines and are circles that intersect in and , it follows that .
Next note that , , and are collinear and are the images of , , and , respectively, under a homothety centered at and with ratio . It follows that , , and are collinear, and then that the points , , , and lie on a circle.
Solution 9 (Official Solution #1)
Let be the circumcenter of triangle . We prove that It will then follow that lie on the circle with diameter . Indeed, the fact that the first two angles in are right is immediate because and are the perpendicular bisectors of and , respectively. Thus we need only prove that .
We may assume, without loss of generality, that . This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because is the perpendicular bisector of , it follows that triangle is an isosceles triangle with . Likewise, triangle is isosceles with . Let and , so .
Applying the Law of Sines to triangles and gives Taking the quotient of the two equations and noting that , we find Because , we have Applying the Law of Sines to triangles and , we find Taking the quotient of the two equations yields so by , Because is an exterior angle to triangle , we have . Similarly, . Hence Thus , so is cyclic. In addition, and hence, from , . Because is cyclic and is isosceles with vertex angle , we have . Therefore, 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 (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.