Difference between revisions of "1980 AHSME Problems/Problem 19"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | Let the center of the circle be on the origin with equation <math>x^2 + y^2 = r^2</math>. |
| + | As the chords are bisected by the x-axis their y-coordinates are <math>10, 8, 4</math> respectively. Let the chord of length <math>10</math> have x-coordinate <math>a</math>. Let <math>d</math> be the common distance between chords. Thus, the coordinates of the top of the chords will be <math>(a, 10), (a+d, 8), (a+2d, 4)</math> for the chords of length <math>20, 16</math>, and <math>8</math> respectively. | ||
| + | As these points fall of the circle, we get three equations: | ||
| + | <math>\\a^2 + 100 = r^2</math> | ||
| + | <math>\\(a+d)^2 + 64 = r^2</math> | ||
| + | <math>\\(a+2d)^2 + 16 = r^2</math> | ||
| + | Subtracting the first equation from the second we get: | ||
| + | <math>\\(a+d)^2 - a^2 - 36 = 0</math> | ||
| + | <math>\\d(2a+d) = 36</math> | ||
| + | Similarly, by subtracting the first equation from the third we get: | ||
| + | <math>\\(a+2d)^2 - a^2 - 84 = 0</math> | ||
| + | <math>\\2d(2a+2d) = 84</math> | ||
| + | <math>\\d(a+d) = 21</math> | ||
| + | Subtracting these two equations gives us <math>ad = 15</math>. Expanding the second equation now gives us | ||
| + | <math>\\a^2 + 2ad + d^2 + 64 = r^2</math> | ||
| + | <math>\\a^2 + d^2 + 94 = r^2</math> | ||
| + | Subtracting the first equation from this yields: | ||
| + | <math>\\d^2 - 6 = 0</math> | ||
| + | <math>\\d = \sqrt{6}</math> | ||
| + | Combining this with <math>ad = 15</math> we get | ||
| + | <math>\\\sqrt{6}a = 15</math> | ||
| + | <math>\\a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}</math> | ||
| + | Plugging this into the first equation finally us | ||
| + | <math>\\(\frac{5\sqrt{6}}{2})^2 + 100 = r^2</math> | ||
| + | <math>\frac{150}{4} + 100 = \frac{550}{4} = r^2</math> | ||
| + | <math>\\r = \sqrt{\frac{550}{4}} = \frac{5\sqrt{22}}{2}</math> | ||
| + | <math>\fbox{D}</math> | ||
== See also == | == See also == | ||
Latest revision as of 16:41, 12 September 2021
Problem
Let 
and 
be three parallel chords of a circle on the same side of the center.
The distance between 
and 
is the same as the distance between and .
The lengths of the chords are 
, and 
. The radius of the circle is
Solution
Let the center of the circle be on the origin with equation 
.
As the chords are bisected by the x-axis their y-coordinates are 
respectively. Let the chord of length 
have x-coordinate 
. Let 
be the common distance between chords. Thus, the coordinates of the top of the chords will be 
for the chords of length , and respectively.
As these points fall of the circle, we get three equations:
Subtracting the first equation from the second we get:
Similarly, by subtracting the first equation from the third we get:
Subtracting these two equations gives us 
. Expanding the second equation now gives us
Subtracting the first equation from this yields:
Combining this with we get
Plugging this into the first equation finally us
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
