AMC10 2021 B

Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is \[2d + d = 3d\] The distance between each of the chords is $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we obtain two different right triangles: - One with base $\dfrac{38}{2} = 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ in the diagram) - Another with base $\dfrac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ in the diagram) By the Pythagorean theorem, we obtain the following system of equations: \begin{align*} 19^2 + d^2 &= r^2, \\ 17^2 + (3d)^2 &= r^2. \end{align*} That is, \begin{align*} 361 + d^2 &= r^2, \\ 289 + 9d^2 &= r^2. \end{align*} Set the right-hand sides equal: \[ 361 + d^2 = 289 + 9d^2 \] \[ 361 - 289 = 9d^2 - d^2 \] \[ 72 = 8d^2 \] \[ d^2 = 9 \implies d = 3. \] Thus, \[ 2d = 6. \]