Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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==Solution 4== | ==Solution 4== | ||
+ | <asy> | ||
+ | unitsize(2.5cm); | ||
+ | defaultpen(fontsize(10pt)+linewidth(.8pt)); | ||
+ | dotfactor=3; | ||
+ | pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); | ||
+ | pair D=dir(aCos(C.x)), E=(-D.x,-D.y); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(D--E--C); | ||
+ | draw(unitcircle,white); | ||
+ | drawline(D,C); | ||
+ | dot(O); | ||
+ | clip(unitcircle); | ||
+ | draw(unitcircle); | ||
+ | label("$E$",E,SSE); | ||
+ | label("$B$",B,E); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
+ | |||
+ | We know that <math>BC=2*AC</math> and <math>AB</math> is a diameter. Without loss of generality, assume the diameter has length 6, which means that <math>AC=2</math> and <math>BC=4</math>. | ||
+ | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}} |
Revision as of 00:43, 16 January 2018
Problem
Let be a diameter of a circle and let
be a point on
with
. Let
and
be points on the circle such that
and
is a second diameter. What is the ratio of the area of
to the area of
?
Solution 1
Let us assume that the diameter is of length .
is
of diameter and
is
.
is the radius of the circle, so using the Pythagorean theorem height
of
is
. This is also the height of the
.
Area of the is
=
.
The height of can be found using the area of
and
as base.
Hence the height of is
=
.
The diameter is the base for both the triangles and
.
Hence, the ratio of the area of to the area of
is
=
Solution 2
Since and
share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from
to
.
.
Since , then
. So the ratio of the two altitudes is
Solution 3
Say the center of the circle is point ;
Without loss of generality, assume
, so
and the diameter and radius are
and
, respectively. Therefore,
, and
.
The area of
can be expressed as
happens to be the area of
. Furthermore,
or
Therefore, the ratio is
Solution 4
We know that and
is a diameter. Without loss of generality, assume the diameter has length 6, which means that
and
.
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.