Difference between revisions of "2008 USAMO Problems/Problem 2"
(<blank edit> Solution 1 credit to tjhance, asy by myself, Solution 2 outline credit to calc rulz, solutions 3,4 were outlined on forums but still have to write up) |
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__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | === Solution 1 (isogonal conjugates) === | + | === Solution 1 (synthetic) === |
+ | === Solution 4 (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,N,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(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 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. | ||
+ | |||
+ | |||
+ | ''Lemma 1'': <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math> | ||
+ | <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>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>. | ||
+ | |||
+ | Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp BC</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 1.'' | ||
+ | |||
+ | |||
+ | 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 | ||
+ | <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 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. | ||
+ | |||
+ | === Solution 2 (isogonal conjugates) === | ||
<center><asy> | <center><asy> | ||
/* setup and variables */ | /* setup and variables */ | ||
Line 15: | Line 73: | ||
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,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(D(MP("D",D,SE,s))--MP("P",P,W,s)); | ||
− | D(B--D(MP("F",F,s))); D(O-- | + | 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(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); | ||
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)); | ||
Line 36: | Line 95: | ||
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 | + | === Solution 3 (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 */ | ||
+ | /* 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(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); | ||
+ | D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); | ||
+ | |||
+ | /* removal of code from original | ||
+ | |||
+ | 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); | ||
− | === Solution | + | */ |
− | === Solution | + | </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 4 (trigonometric) === | ||
+ | Use the [[Law of Sines]] to show that <math>\angle BFA = \angle AFC</math>. It follows that <math>\triangle BFA \sim \triangle AFC</math>. Noting the [[spiral similarity]] from <math>P</math> to <math>N</math>, {{incomplete|solution}} | ||
+ | |||
+ | === Solution 5 (analytical) === | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 16:28, 2 May 2008
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.
Contents
[hide]Solution
Solution 1 (synthetic)
Solution 4 (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 1: is directly similar to
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 1.
By the similarity in Lemma 1, . 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 (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 3 (inversion)
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
We consider an inversion by an arbitrary radius about . We want to show that and are collinear. Notice that and lie on a circle with center , and similarly for the other side. We also have that form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that is a parallelogram, indicating that is the midpoint of . Template:Incomplete
Solution 4 (trigonometric)
Use the Law of Sines to show that . It follows that . Noting the spiral similarity from to , Template:Incomplete
Solution 5 (analytical)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
2008 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>