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>. | ||
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>. | Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>. | ||
− | ===Solution 5 | + | ===Solution 5=== |
− | We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D | + | We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A,B, and C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively. |
− | By shoelace the area of <math>CED</math> is Y, and the area of <math>ADB</math> is 3Y. Using division we get that the answer is <math> \mathrm{(C)}</math>. | + | By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>. |
== See also == | == See also == |
Revision as of 08:22, 20 October 2019
Contents
[hide]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 which are , , and respectively. By shoelace the area of is , and the area of is . Using division we get that the answer 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 |
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