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Difference between revisions of "1980 AHSME Problems/Problem 19"

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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Place this circle in the coordinate plane such that the center is at the origin all three chords are to the right of, and parallel to, the <math>y</math>-axis.  The equation of this circle is <math>x^2 + y^2 = r^2</math>, where <math>r > 0</math> is the radius of the circle.
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Let <math>A</math>, <math>B</math>, and <math>C</math> be the upper ends of chords <math>C_1</math>, <math>C_2</math>, and <math>C_3</math>, respectively.
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Since the <math>x</math>-axis is perpendicular to the three chords, it must bisect each chord.  Let <math>a</math> be the <math>x</math>-coordinate <math>A</math>, and <math>d</math> be the distance between <math>C_1</math> and <math>C_2</math>, which is also the distance between <math>C_2</math> and <math>C_3</math>.  Then <math>A</math> is at <math>(a, 10)</math>, <math>B</math> is at <math>(a+d, 8)</math>, and <math>C</math> is at <math>(a+2d, 4)</math>.
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Since <math>A</math>, <math>B</math>, and <math>C</math> all lie on this circle, the following equations must hold:
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<cmath>\\a^2 + 100 = r^2\ (\textrm{i})</cmath>
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<cmath>\\(a+d)^2 + 64 = r^2\ (\textrm{ii})</cmath>
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<cmath>\\(a+2d)^2 + 16 = r^2\ (\textrm{iii})</cmath>
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Subtracting equation <math>(\textrm{i})</math> from equation <math>(\textrm{ii})</math> yields:
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<cmath>\\(a+d)^2 - a^2 - 36 = 0</cmath>
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<cmath>\\2ad + d^2 = 36\ (\textrm{iv})</cmath>
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Subtracting equation <math>(\textrm{i})</math> from equation <math>(\textrm{iii})</math> yields:
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<cmath>\\(a+2d)^2 - a^2 - 84 = 0</cmath>
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<cmath>\\4ad + 4d^2 = 84</cmath>
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<cmath>\\ad + d^2 = 21\ (\textrm{v})</cmath>
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Subtracting <math>(\textrm{v})</math> from <math>(\textrm{iv})</math> yields:
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<cmath>\\ad = 15\ (\textrm{vi})</cmath>
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Subtracting <math>(\textrm{vi})</math> from <math>(\textrm{v})</math> yields <math>d^2 - 6 = 0</math>, so <math>d = \sqrt{6}</math>.
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Substituting <math>d = \sqrt{6}</math> into <math>(\textrm{vi})</math> yields <math>\sqrt{6}a = 15</math>, so <math>a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}</math>.
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Substituting <math>a = \frac{5\sqrt{6}}{2}</math> yields  <math>r^2 = \left(\frac{5\sqrt{6}}{2}\right)^2 + 100 = \frac{275}{2}</math>, so <math>r = \sqrt{\frac{275}{2}} = \boxed{(\textbf{D})\ \frac{5\sqrt{22}}{2}}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 16:59, 15 August 2025

Problem

Let $C_1, C_2$ and $C_3$ be three parallel chords of a circle on the same side of the center. The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3$. The lengths of the chords are $20, 16$, and $8$. The radius of the circle is

$\text{(A)} \ 12 \qquad \text{(B)} \ 4\sqrt{7} \qquad \text{(C)} \ \frac{5\sqrt{65}}{3} \qquad \text{(D)}\ \frac{5\sqrt{22}}{2}\qquad \text{(E)}\ \text{not uniquely determined}$


Solution

Place this circle in the coordinate plane such that the center is at the origin all three chords are to the right of, and parallel to, the $y$-axis. The equation of this circle is $x^2 + y^2 = r^2$, where $r > 0$ is the radius of the circle.

Let $A$, $B$, and $C$ be the upper ends of chords $C_1$, $C_2$, and $C_3$, respectively.

Since the $x$-axis is perpendicular to the three chords, it must bisect each chord. Let $a$ be the $x$-coordinate $A$, and $d$ be the distance between $C_1$ and $C_2$, which is also the distance between $C_2$ and $C_3$. Then $A$ is at $(a, 10)$, $B$ is at $(a+d, 8)$, and $C$ is at $(a+2d, 4)$.

Since $A$, $B$, and $C$ all lie on this circle, the following equations must hold: \[\\a^2 + 100 = r^2\ (\textrm{i})\] \[\\(a+d)^2 + 64 = r^2\ (\textrm{ii})\] \[\\(a+2d)^2 + 16 = r^2\ (\textrm{iii})\] Subtracting equation $(\textrm{i})$ from equation $(\textrm{ii})$ yields: \[\\(a+d)^2 - a^2 - 36 = 0\] \[\\2ad + d^2 = 36\ (\textrm{iv})\] Subtracting equation $(\textrm{i})$ from equation $(\textrm{iii})$ yields: \[\\(a+2d)^2 - a^2 - 84 = 0\] \[\\4ad + 4d^2 = 84\] \[\\ad + d^2 = 21\ (\textrm{v})\] Subtracting $(\textrm{v})$ from $(\textrm{iv})$ yields: \[\\ad = 15\ (\textrm{vi})\] Subtracting $(\textrm{vi})$ from $(\textrm{v})$ yields $d^2 - 6 = 0$, so $d = \sqrt{6}$.

Substituting $d = \sqrt{6}$ into $(\textrm{vi})$ yields $\sqrt{6}a = 15$, so $a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}$.

Substituting $a = \frac{5\sqrt{6}}{2}$ yields $r^2 = \left(\frac{5\sqrt{6}}{2}\right)^2 + 100 = \frac{275}{2}$, so $r = \sqrt{\frac{275}{2}} = \boxed{(\textbf{D})\ \frac{5\sqrt{22}}{2}}$

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

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