Difference between revisions of "1998 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | {{ | + | |
| + | <center> | ||
| + | <asy> | ||
| + | |||
| + | pair O,A,B,C,D,E,F,DEb,CFb,Fo,M; | ||
| + | O=(0,0); | ||
| + | A=(1.732,1); | ||
| + | B=(0,1); | ||
| + | C=(-1.732,1); | ||
| + | D=(0.866,1); | ||
| + | Fo=(-1,-0.5); | ||
| + | |||
| + | path AC,AF,DE,CF,DEbM,CFbM,C1,C2; | ||
| + | C1=circle(O,2); | ||
| + | C2=circle(O,1); | ||
| + | |||
| + | E=intersectionpoints(A--Fo,C2)[0]; | ||
| + | F=intersectionpoints(A--Fo,C2)[1]; | ||
| + | DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0); | ||
| + | CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0); | ||
| + | M=(-0.433,1); | ||
| + | |||
| + | path AC=A--C; | ||
| + | path AF=A--F; | ||
| + | path DEbM=DEb--M; | ||
| + | path CFbM=CFb--M; | ||
| + | path DE=D--E; | ||
| + | path CF=C--F; | ||
| + | |||
| + | draw(AC); | ||
| + | draw(AF); | ||
| + | draw(DE); | ||
| + | draw(CF); | ||
| + | draw(DEbM); | ||
| + | draw(CFbM); | ||
| + | draw(C1); | ||
| + | draw(C2); | ||
| + | |||
| + | label("\(A\)",A,NE); | ||
| + | label("\(B\)",B,N); | ||
| + | label("\(C\)",C,NW); | ||
| + | label("\(D\)",D,N); | ||
| + | label("\(E\)",E,SE); | ||
| + | label("\(F\)",F,SW); | ||
| + | label("\(M\)",M,N); | ||
| + | |||
| + | </asy> | ||
| + | </center> | ||
| + | |||
| + | First, <math>AD=\frac{AB}{2}=\frac{AC}{4}</math>. Because <math>E</math>,<math>F</math> and <math>B</math> all lie on a circle, <math>AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC</math>. Therefore, <math>\triangle ACF \sim \triangle AEB</math>, so <math>\angle ACF = \angle AEB</math>. Thus, quadrilateral <math>CFED</math> is cyclic, and <math>M</math> must be the center of the circumcircle of <math>CFED</math>, which implies that <math>MC=\frac{CD}{2}</math>. Putting it all together, | ||
| + | |||
| + | <math>\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}</math> | ||
== See Also == | == See Also == | ||
Revision as of 16:58, 24 June 2014
Problem
Let 
and 
be concentric circles, with in the interior of . From a point 
on one draws the tangent 
to (
). Let 
be the second point of intersection of and , and let 
be the midpoint of . A line passing through intersects at 
and 
in such a way that the perpendicular bisectors of 
and 
intersect at a point 
on . Find, with proof, the ratio 
.
Solution
![[asy] pair O,A,B,C,D,E,F,DEb,CFb,Fo,M; O=(0,0); A=(1.732,1); B=(0,1); C=(-1.732,1); D=(0.866,1); Fo=(-1,-0.5); path AC,AF,DE,CF,DEbM,CFbM,C1,C2; C1=circle(O,2); C2=circle(O,1); E=intersectionpoints(A--Fo,C2)[0]; F=intersectionpoints(A--Fo,C2)[1]; DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0); CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0); M=(-0.433,1); path AC=A--C; path AF=A--F; path DEbM=DEb--M; path CFbM=CFb--M; path DE=D--E; path CF=C--F; draw(AC); draw(AF); draw(DE); draw(CF); draw(DEbM); draw(CFbM); draw(C1); draw(C2); label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,NW); label("\(D\)",D,N); label("\(E\)",E,SE); label("\(F\)",F,SW); label("\(M\)",M,N); [/asy]](http://latex.artofproblemsolving.com/7/1/c/71c6a9a039f6d75f381250dfec4ada49afd43cd9.png)
First, 
. Because , and 
all lie on a circle, 
. Therefore, 
, so 
. Thus, quadrilateral 
is cyclic, and must be the center of the circumcircle of , which implies that 
. Putting it all together,
See Also
| 1998 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
