Difference between revisions of "2005 AMC 12A Problems/Problem 15"
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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</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}}{ | + | 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}}{6x^2\sqrt{2}}</math> or <math>\frac{1}{3}</math> = <math>\boxed{(C)}</math>. |
~Math_Genius_164 | ~Math_Genius_164 |
Revision as of 12:32, 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:
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 = .
~Math_Genius_164
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.