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

(I'll construct T later -_-)
(<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 ==
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=== Solution 1 ===
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=== Solution 1 (isogonal conjugates) ===
 
<center><asy>
 
<center><asy>
 
   /* setup and variables */
 
   /* setup and variables */
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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 2 ===
+
=== Solution 2 (inversion) ===
 +
Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) {{incomplete|solution}}
 +
 
 +
=== Solution 3 (trigonometric) ===
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=== Solution 4 (synthetic) ===
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 19:15, 1 May 2008

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.

Solution

Solution 1 (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--F--A,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)); 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 2 (inversion)

Invert by an arbitrary radius about A. We want to show that P', F', and N' are collinear. Notice that D', A, and P' lie on a circle with center B', and similarly for the other side. We also have that B', D', F', A form a cyclic quad and similar for the other side. We can then use some angle chasing to prove that A B' F' C' is a paralellogram, meaning that F' the midpoint of P'N'. (On the actual test, I showed that A B' F' C' was a parallelogram, and didn't realize that that implied F' was on P'N'.) Template:Incomplete

Solution 3 (trigonometric)

Solution 4 (synthetic)

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)
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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>
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