Difference between revisions of "2005 AMC 12A Problems/Problem 15"
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Revision as of 22:24, 13 June 2018
Contents
Problem
Let be a diameter of a circle and
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
Solution 1
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or (
is the foot of the perpendicular from
to
).
Call the radius . Then
,
. Using the Pythagorean Theorem in
, we get
.
Now we have to find . Notice
, so we can write the proportion:
![$\frac{OF}{OC} = \frac{OC}{OD}$](http://latex.artofproblemsolving.com/0/2/2/02291691270e57eea057ea561416318b8de9fc01.png)
![$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$](http://latex.artofproblemsolving.com/6/e/1/6e1dcd67ba34e71f9f880510df1fd482941f6e37.png)
![$OF = \frac 19r$](http://latex.artofproblemsolving.com/5/a/4/5a4dcf36a069d28f1265cf8c1b825c9ae46c0cbd.png)
By the Pythagorean Theorem in , we have
.
Our answer is .
Solution 2
Let the center of the circle be .
Note that .
is midpoint of
.
is midpoint of
Area of
Area of
Area of
Area of
.
Solution 3
Let be the radius of the circle. Note that
so
.
By Power of a Point Theorem, , and thus
Then the area of is
. Similarly, the area of
is
, so the desired ratio is
Solution 4
Let the center of the circle be .
Without loss of generality, let the radius of the circle be equal to
. Thus,
and
. As a consequence of
,
and
. Also, we know that
and
are both equal to
due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to
or
. Now we know that the area of
is equal to
or
. Know we need to find the area of
. By simple inspection
/cong
due to angles being equal and CPCTC. Thus
and
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.