Difference between revisions of "2005 AMC 10A Problems/Problem 23"
m (→Solution 1) |
(→Solution 5) |
||
(4 intermediate revisions by 2 users not shown) | |||
Line 31: | Line 31: | ||
The length of <math>AC</math> is <math>\frac{1}{3}</math> and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. | The length of <math>AC</math> is <math>\frac{1}{3}</math> and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. | ||
− | <math>OD</math> is the radius of the circle, which is <math>\frac{1}{2}</math>, so using the Pythagorean | + | <math>OD</math> is the radius of the circle, which is <math>\frac{1}{2}</math>, so using the Pythagorean Theorem the height <math>CD</math> of <math>\triangle DCO</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>. |
− | + | The area of <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. | |
− | The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. | + | The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. |
− | Hence the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>. | + | Hence, the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>. |
− | The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math> | + | The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>, |
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | ||
Line 72: | Line 72: | ||
WLOG, let <math>AC=1</math>, <math>BC=2</math>, so radius of the circle is <math>\frac{3}{2}</math> and <math>OC=\frac{1}{2}</math>. As in solution 1, By same altitude, the ratio <math>[DCE]/[ABD]=PE/AB</math>, where <math>P</math> is the point where <math>DC</math> extended meets circle <math>O</math>. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is <math>\frac{1}{3}</math>. | WLOG, let <math>AC=1</math>, <math>BC=2</math>, so radius of the circle is <math>\frac{3}{2}</math> and <math>OC=\frac{1}{2}</math>. As in solution 1, By same altitude, the ratio <math>[DCE]/[ABD]=PE/AB</math>, where <math>P</math> is the point where <math>DC</math> extended meets circle <math>O</math>. Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is <math>\frac{1}{3}</math>. | ||
+ | |||
+ | == Solution 5 (Video) == | ||
+ | Video solution: https://youtu.be/i6eooSSJF64 | ||
==See Also== | ==See Also== |
Latest revision as of 15:16, 26 November 2020
Contents
Problem
Let be a diameter of a circle and let 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 1
WLOG, Let us assume that the diameter is of length .
The length of is and is .
is the radius of the circle, which is , so using the Pythagorean Theorem the height of is . This is also the height of the .
The area of is = .
The height of can be found using the area of and as base.
Hence, the height of is = .
The diameter is the base for both the triangles and ,
Hence, the ratio of the area of to the area of is =
Solution 2
Since and share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from to .
.
Since , then . So the ratio of the two altitudes is
Solution 3
Say the center of the circle is point ; Without loss of generality, assume , so and the diameter and radius are and , respectively. Therefore, , and . The area of can be expressed as happens to be the area of . Furthermore, or Therefore, the ratio is
Solution 4
WLOG, let , , so radius of the circle is and . As in solution 1, By same altitude, the ratio , where is the point where extended meets circle . Note that angle P = 90 deg, so DCO ~ DPE with ratio 1:2, so PE = 1. Thus, our ratio is .
Solution 5 (Video)
Video solution: https://youtu.be/i6eooSSJF64
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.