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Difference between revisions of "1998 USAMO Problems/Problem 2"

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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,
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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 AED</math>, so <math>\angle ACF = \angle AED</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>
 
<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>
  
 
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html
 
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html
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== Solution 2 ==
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We will use signed lengths. WLOG let <math>AC = 4</math>, <math>AM = x</math>, and <math>AE = y > 0</math>. Then <math>AB = 2</math> and <math>AD = 1</math>. Therefore, <math>DM = x - 1</math> and <math>MC = 4 - x</math>. By Power of a Point, <math>AE \cdot AF = AB^2 = 4</math>, so <math>AF = \frac{4}{AE} = \frac{4}{y}</math>, and <math>EF = \frac{4}{y} - y = \frac{4 - y^2}{y}</math>. Since <math>M</math> is on the perpendicular bisectors of <math>DE</math> and <math>CF</math>, we have <math>|DM| = |EM| = |x - 1|</math> and <math>|CM| = |FM| = |4 - x|</math>.
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By Stewart's Theorem on <math>\triangle MFA</math> and cevian <math>DM</math>, <cmath>\frac{4}{y}(y(\frac{4-y^2}{y}) + (x-1)^2) = y(4-x)^2 + \frac{4-y^2}{y} \cdot x^2;</cmath>
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<cmath>4(4 - y^2 + (x-1)^2) = y^2(4-x)^2 + (4-y^2)x^2;</cmath>
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<cmath>16 - 4y^2 + 4x^2 - 8x + 4 = x^2y^2 - 8xy^2 + 16y^2 + 4x^2 - x^2y^2;</cmath>
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<cmath>8xy^2 - 8x- 20y^2 + 20 = (8x - 20)(y^2 - 1) = 0.</cmath>
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So either <math>8x - 20 = 0</math> or <math>y^2 - 1 = 0</math>. If <math>y^2 - 1 = 0</math>, then <math>AE = y = 1 = AD</math> and <math>AF = AC = 4</math>, so the perpendicular bisectors of <math>CF</math> and <math>DE</math> are the same line, and they do not intersect at a point. Therefore, <math>AM = x = \frac{5}{2}</math> and <math>MC = \frac{3}{2}</math>, so <math>\frac{AM}{MC} = \boxed{\frac{5}{3}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 16:26, 20 June 2019

Problem

Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.

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]

First, $AD=\frac{AB}{2}=\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC$. Therefore, $\triangle ACF \sim \triangle AED$, so $\angle ACF = \angle AED$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, which implies that $MC=\frac{CD}{2}$. Putting it all together,

$\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}$

Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html

Solution 2

We will use signed lengths. WLOG let $AC = 4$, $AM = x$, and $AE = y > 0$. Then $AB = 2$ and $AD = 1$. Therefore, $DM = x - 1$ and $MC = 4 - x$. By Power of a Point, $AE \cdot AF = AB^2 = 4$, so $AF = \frac{4}{AE} = \frac{4}{y}$, and $EF = \frac{4}{y} - y = \frac{4 - y^2}{y}$. Since $M$ is on the perpendicular bisectors of $DE$ and $CF$, we have $|DM| = |EM| = |x - 1|$ and $|CM| = |FM| = |4 - x|$.

By Stewart's Theorem on $\triangle MFA$ and cevian $DM$, \[\frac{4}{y}(y(\frac{4-y^2}{y}) + (x-1)^2) = y(4-x)^2 + \frac{4-y^2}{y} \cdot x^2;\] \[4(4 - y^2 + (x-1)^2) = y^2(4-x)^2 + (4-y^2)x^2;\] \[16 - 4y^2 + 4x^2 - 8x + 4 = x^2y^2 - 8xy^2 + 16y^2 + 4x^2 - x^2y^2;\] \[8xy^2 - 8x- 20y^2 + 20 = (8x - 20)(y^2 - 1) = 0.\]

So either $8x - 20 = 0$ or $y^2 - 1 = 0$. If $y^2 - 1 = 0$, then $AE = y = 1 = AD$ and $AF = AC = 4$, so the perpendicular bisectors of $CF$ and $DE$ are the same line, and they do not intersect at a point. Therefore, $AM = x = \frac{5}{2}$ and $MC = \frac{3}{2}$, so $\frac{AM}{MC} = \boxed{\frac{5}{3}}$.

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

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