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 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> | ||
+ | |||
+ | Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | <cmath>4(4 - y^2 + (x-1)^2) = y^2(4-x)^2 + (4-y^2)x^2;</cmath> | ||
+ | <cmath>16 - 4y^2 + 4x^2 - 8x + 4 = x^2y^2 - 8xy^2 + 16y^2 + 4x^2 - x^2y^2;</cmath> | ||
+ | <cmath>8xy^2 - 8x- 20y^2 + 20 = (8x - 20)(y^2 - 1) = 0.</cmath> | ||
+ | |||
+ | 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 == | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:26, 20 June 2019
Contents
[hide]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,
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html
Solution 2
We will use signed lengths. WLOG let ,
, and
. Then
and
. Therefore,
and
. By Power of a Point,
, so
, and
. Since
is on the perpendicular bisectors of
and
, we have
and
.
By Stewart's Theorem on and cevian
,
So either or
. If
, then
and
, so the perpendicular bisectors of
and
are the same line, and they do not intersect at a point. Therefore,
and
, so
.
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.