Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. | <math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. | ||
− | <math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle AOC</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>. | + | <math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle AOC</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>. |
Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. | Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. | ||
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The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. | The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. | ||
− | Hence the height of <math>\triangle DCE</math> is <math>\ | + | Hence the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>. |
The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>. | The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>. | ||
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | ||
− | <math>\ | + | <math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\dfrac{1}{3} \Rightarrow C</math> |
==Solution 2== | ==Solution 2== |
Revision as of 12:33, 26 June 2018
Contents
[hide]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
.
Solution 5
WLOG, let ,
, so radius of the circle is
and
. As in solution 1, By same altitude, the ratio
, where
is the point where
extended meets circle
. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is
.
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 |
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