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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | |||
+ | </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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.