Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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− | == Solution | + | == Solution 5 == |
− | Video solution | + | |
+ | Assume the diameter is <math>3</math>. | ||
+ | |||
+ | <math>AC = 1</math> | ||
+ | |||
+ | Get the height <math>CD = \sqrt{(AC)(BC)} = \sqrt2</math> via power of a point. | ||
+ | |||
+ | <math>CO = AO - AC = 1/2</math>. | ||
+ | |||
+ | By altitude of right triangle <math>\triangle CDO</math>: Altitude from <math>C</math> to <math>DE</math> is same as altitude from <math>C</math> to <math>DO</math> is <math>\frac{(CO)(CD)}{DO} = \frac{(1/2)(\sqrt2)}{\frac 3 2}</math>. | ||
+ | |||
+ | <math>\triangle DCE</math> and <math>\triangle ABD</math> have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is | ||
+ | <math>\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}} {\sqrt2} = \boxed{1/3}</math>. | ||
+ | |||
+ | ~oinava | ||
+ | |||
+ | == Video solution == | ||
+ | https://youtu.be/i6eooSSJF64 | ||
==See Also== | ==See Also== |
Latest revision as of 20:57, 9 July 2024
Contents
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
WLOG, Let us assume that the diameter is of length .
The length of is
and
is
.
is the radius of the circle, which is
, so using the Pythagorean Theorem the height
of
is
. This is also the height of the
.
The area of 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
Let the point G be the reflection of point across
. (Point G is on the circle).
Let , then
. The diameter is
. To find
, there are two ways (presented here):
1. Since is the diameter,
. Using power of points,
2. Use the geometric mean theorem,
(These are the same equations but obtained through different formulae)
Therefore . Since
is a diameter,
is right. By the Pythagorean theorem,
As established before, is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures
) so
is the altitude of
, and
is the base. Therefore
is the base of
and
is the height.
The required ratio is
The answer is
.
~JH. L
Solution 5
Assume the diameter is .
Get the height via power of a point.
.
By altitude of right triangle : Altitude from
to
is same as altitude from
to
is
.
and
have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is
.
~oinava
Video solution
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.