Difference between revisions of "2005 AMC 12A Problems/Problem 15"
(Solution 3 (using power of a point) added) |
(→Solution 1) |
||
Line 30: | Line 30: | ||
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 <math>\frac{CD}{CF}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>). | 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 <math>\frac{CD}{CF}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>). | ||
− | Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{3}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. | + | Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac({1}{3}r)^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. |
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]: | Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]: | ||
Line 39: | Line 39: | ||
Our answer is <math>\frac{CD}{CF} = \frac{\frac{2\sqrt{2}}{3}r}{\frac{2\sqrt{2}}{9}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>. | Our answer is <math>\frac{CD}{CF} = \frac{\frac{2\sqrt{2}}{3}r}{\frac{2\sqrt{2}}{9}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>. | ||
− | |||
===Solution 2=== | ===Solution 2=== |
Revision as of 04:54, 15 August 2016
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 centre 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
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.