Draw a Circle Size of Dime

Vii Circles Iii

Task

Seven circles of the same size can be placed in the design shown below: The vi outer circles bear on the one in the middle and each circle on the outside also touches its two neighbors in the outside ring.

1. If twelve circles are placed effectually a central circle, as pictured below, what is the relationship betwixt the diameters of the outer circles and the diameter of the inner circle? Explain.
2. If four circles are placed around a central circle, as pictured below, what is the relationship between the diameter of the outer circles and the diameter of the inner circle? Explain.

IM Commentary

This job is intended for instructional purposes just. It provides an opportunity to model a concrete situation with mathematics. In one case a representative movie of the situation described in the problem is drawn (the teacher may provide guidance here equally necessary), the solution of the task requires an understanding of the definition of the sine part. When the chore is consummate, new insight is shed on the ''Seven Circles I'' problem which initiated this investigation equally is noted at the end of the solution.

This problem leads to what could make for a fun in form activeness, namely measuring the diameters of different size coins (pennies, nickels, dimes, quarters, and half dollars) and using this information to gauge how many of each sized coin would fit around the circumference of a given coin. This information will be recorded in a dissever chore, ''Coins in a circular pattern.''

In the solution to both parts of this problem, the trigonometric function $\sin{x}$ needs to be evaluated, in one case for the benchmark angle of $45$ degrees and the other time for an angle of $fifteen$ degrees. While not a benchmark angle, the value for $\sin(15)$ can exist plant using a double bending formula. In the 2d solution, the values for $\sin{10}$ are evaluated explicitly while in the first solution a calculator is used to notice an approximate value.

Solutions

Solution: i

1. Allow $O$ be the center of the central circle and $P$, $Q$ centers of adjacent circles in the band and $R$ the midpoint of segment $PQ$ equally pictured beneath:

   

   Allow $r$ be the radius of the inner circle and $due south$ the radii of the twelve outer circles. Segment $PO$ is congruent to segement $QO$ every bit both have a length of $s + r$. Segment $PR$ is congruent to segment $QR$ because $R$ is the midpoint of $PQ$. Finally segment $OR$ is coinciding to segment $OR$. By $SSS$ nosotros conclude that triangle $POR$ is congurent to triangle $QOR$. Since angles $ORP$ and $ORQ$ are congruent and, taken together, add together upwardly to $180$ degrees they are both right angles. Because there are twelve equal circles in the ring, the measure of angle $POQ$ is one 12th of $360$ degrees or $30$ degrees. Angles $POR$ and $QOR$ are congruent and so they much each be $15$ degrees.

   The sine of angle $POR$ is the length of the side opposite the angle, $|PR|$ divided by the length of the hypotenuse $|OP|$. So we take \begin{eqnarray} \sin(fifteen^\circ) &=& \frac{|PR|}{|OP|} \\ &=& \frac{south}{r+s}. \end{eqnarray} Multiplying both sides by $r+south$ gives $r\sin{15^\circ} + s\sin{15^\circ} = s$. Solving for $south$ in terms of $r$ we find $$ due south = \left(\frac{ \sin{15^\circ}}{one-\sin{15^\circ}}\right) r. $$ For students who know double angle formulas, $\sin{fifteen}$ tin can be calculated directly as in the second solution below. Evaluating on a estimator, we find that $southward$ is about $0.349r$: as the movie shows, the circles in the ring are substantially smaller than the inner circumvolve. For function (b) below, nosotros should find, if the picture is accurate, that $s$ is substantially larger than $r$. The example where $s = r$ was seen in ''Seven Circles I.''

2. The method used in part (a) works more than generally every bit we will verify hither for the case where at that place are iv circles in the band. We go along to denote the radius of the inner circle past $r$ merely use $t$ this time for the radius of the outer circumvolve. Choosing $P$ and $Q$ to be centers of adjacent circles in the outer ring and $R$ the midpoint of segment $PQ$ we find the following picture:

   

   The remainder of the argument from role (a) now applies, the only modify beingness that the measure of angle $POR$ is now $45$ degrees instead of $15$ degrees. The sine of angle $POR$ is the length of the side opposite the bending, $|PR|$ divided by the length of the hypotenuse $|OP|$. So we accept \brainstorm{eqnarray} \sin(45^\circ) &=& \frac{|PR|}{|OP|} \\ &=& \frac{t}{r+t}. \stop{eqnarray} Multiplying both sides past $r+t$ gives $r\sin{45^\circ} +t\sin{45^\circ} = t$. Solving for $t$ in terms of $r$ we find $$ t = \left(\frac{ \sin{45^\circ}}{1-\sin{45^\circ}}\right) r. $$ Here students may use the fact that $\sin{45} = \frac{\sqrt{ii}}{two}$ as is washed in the 2d solution below. Evaluating on a calculator, we notice that $t$ is about $2.41r$, validating the pic which shows that the circles in the outer band are substantially larger than the fundamental circumvolve.

   Now that all of this work has been done nosotros can check, every bit was asked in ''Seven Circles I'' that if six circles of radius $r^\prime number$ are places around a circle of radius $r$ in the pattern shown at the beginning of the problem, then $$ r^\prime = \left(\frac{ \sin{30^\circ}}{one-\sin{30^\circ}}\right) r = r $$ so that all seven circles in this pattern take the same size.

Solution: 2 Double bending formula

1. In order to calculate $\sin{15^\circ}$ exactly we may use the double bending formula for the cosine which is $$ \cos{2x} = 1 - 2\sin^ii{x}. $$ Plugging in $ten = fifteen^\circ$ we find $$ 1 - 2\sin^2{15^\circ} = \cos{30^\circ} = \frac{\sqrt{3}}{2}. $$ Solving for $\sin{15^\circ}$ we find

   \brainstorm{marshal} \sin{15^\circ} &= \sqrt{\frac{one}{2} - \frac{\sqrt{3}}{iv}} \\ &= \frac{\sqrt{2 - \sqrt{iii}}}{ii}. \end{marshal}

   This exact expression can then be used to find the verbal relationship betwixt $due south$ and $r$ since we know from the start solution that $$ due south = \left( \frac{\sin{fifteen^\circ}}{1-\sin{xv^\circ}}\right)r. $$ Simplifying this expression any further is rather tedious, but one ends with, amid other possible expressions, the answer $$ south = \left(\frac{\sqrt{3}-ane}{1+2\sqrt{two}-\sqrt{3}}\right) r. $$
2. For function $b$ finding and exact value for $t$ is simpler than for office (a) because $\sin{45^\circ} = \frac{\sqrt{ii}}{two}$. Using this we detect

   \begin{align} t &= \left( \frac{\sin{45^\circ}}{1-\sin{45^\circ}}\right)r \\ &= \left( \frac{\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{ii}}{2}}\right)r \\ &= (\sqrt{two} +1) r. \end{align}

Seven Circles III

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Source: https://tasks.illustrativemathematics.org/content-standards/tasks/710

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