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

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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.  
 
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.  
  
Let <math>A</math>, <math>B</math>, and <math>C</math> be the upper ends of the <math>C_1</math>, <math>C_2</math>, and <math>C_3</math>, respectively.
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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.
  
 
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>.
 
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>.

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AHSME Problems and Solutions

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