Difference between revisions of "2005 AMC 12A Problems/Problem 15"
Line 181: | Line 181: | ||
Let the center of the circle be <math>O</math>. | Let the center of the circle be <math>O</math>. | ||
− | The area of <math>ADB</math> is <math>\frac{1}{2} \cdot 6x \cdot 2x\sqrt{2}</math> = <math>6x^2\sqrt{2}</math>. The area of <math>DCE</math> is <math>\frac{1}{2} \cdot DC \cdot DE \cdot sin CDE</math>. We find <math> | + | The area of <math>ADB</math> is <math>\frac{1}{2} \cdot 6x \cdot 2x\sqrt{2}</math> = <math>6x^2\sqrt{2}</math>. The area of <math>DCE</math> is <math>\frac{1}{2} \cdot DC \cdot DE \cdot</math> sin <math>CDE</math>. We find sin <math>CDE</math> is <math>\frac{OC}{OD}</math> = <math>\frac{1}{3}</math>. Substituting <math>DC = 2x\sqrt{2}</math> and <math>DE = 6x</math>, we get <math>[DCE]</math> = <math>\frac{1}{2} \cdot 2x\sqrt{2} \cdot 6x \cdot \frac{1}{3}</math> = <math>2x^2\sqrt{2}</math>. Hence, the ratio between the areas of <math>DCE</math> and <math>ADB</math> is equal to <math>\frac{2x^2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> = <math>\boxed{(C)}</math>. |
== See also == | == See also == |
Revision as of 04:06, 23 December 2019
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
due to angles being equal and CPCTC. Thus
and
. Know we know the area of
or
. We also know that the area of
or
. Thus the area of
or
. We also can calculate the area of
to be
or
. Thus
is equal to
+
or
or
. The ratio between
and
is equal to
or
.
Solution 5
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for , and notice how
is a 180 degree rotation of
, using the rotation matrix formula we get
. WLOG say that this circle has radius
. We can now find points
,
, and
which are
,
, and
respectively.
By shoelace the area of
is
, and the area of
is
. Using division we get that the answer is
.
Solution 6 (Mass Points)
We set point as a mass of 2. This means that point
has a mass of
since
. This implies that point
has a mass of
and the center of the circle has a mass of
. After this, we notice that points
and
both must have a mass of
since
and they are both radii of the circle.
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply which is
(the reciprocal of 3)
-Brudder
Solution 7 (Slight Trigonometry)
Let the center of the circle be .
The area of
is
=
. The area of
is
sin
. We find sin
is
=
. Substituting
and
, we get
=
=
. Hence, the ratio between the areas of
and
is equal to
or
=
.
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.