Difference between revisions of "2005 AMC 10A Problems/Problem 23"
(→Solution 5) |
Dairyqueenxd (talk | contribs) (→Problem) |
||
Line 22: | Line 22: | ||
draw(rightanglemark(D,C,B,2));</asy> | draw(rightanglemark(D,C,B,2));</asy> | ||
− | <math> \ | + | <math> \textbf{(A) } \frac{1}{6}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3} </math> |
==Solution 1== | ==Solution 1== |
Revision as of 12:56, 14 December 2021
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
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
.
Solution 5 (Video)
Video solution: https://youtu.be/i6eooSSJF64
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.