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My family and I like to do a daily quiz but this particular question has had us baffled for weeks. Please help. We only have basic mathematical knowledge.

Quiz Question

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    $\begingroup$ Related (pretty-much a duplicate, although it asks for the area of the crescent): math.stackexchange.com/questions/35898/… . $\endgroup$ – Blue Feb 12 '17 at 1:18
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    $\begingroup$ Hint: first note that the difference between the diameters is $9$. Then let $G$ be the center of the small circle, and look closely at $\triangle GCE$ to find a second relation between the radii. Nice family habit, btw. $\endgroup$ – dxiv Feb 12 '17 at 1:19
  • $\begingroup$ Thank you for your reply. I did see that link but even though it had the same graphic it discussed the area. We are trying now to solve it with the clues you have given us. At least its something in the right direction. $\endgroup$ – Birdman2000 Feb 12 '17 at 1:48
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    $\begingroup$ I can't resist pointing out that a crescent moon actually never looks like that. $\endgroup$ – Ilmari Karonen Feb 12 '17 at 17:07
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    $\begingroup$ Hint: DCE and ECA are similar triangles. $\endgroup$ – Akiva Weinberger Feb 13 '17 at 3:53
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The diagram you need to draw, with $r$ as the radius of the larger circle (giving $r{-}\frac 92$ as the radius of the smaller) is:

enter image description here

where $G$ is the centre of the smaller circle. From here you should be able to use Pythagoras to solve.


Since suitable time has now elapsed, the completion to a solution should look something like:

$$\require{cancel}\begin{align} \left( r-\frac 92 \right)^2 &= (r-5)^2+\left( \frac 92 \right)^2 \\ r^2 - 9r +\left( \frac 92 \right)^2 &= r^2 - 10r + 25 +\left( \frac 92 \right)^2 \\ \cancel{r^2} - 9r +\cancel{\left( \frac 92 \right)^2} &= \cancel{r^2} - 10r + 25 +\cancel{\left( \frac 92 \right)^2} \\ 10r-9r&=25\\ r &= 25 \end{align}$$

So the diameter of the large circle is $2\cdot 25 = \fbox{50}$ and of the smaller circle $50-9 =\fbox{41}$

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Let $d$ be the length of the segment $DC$, and $r$ the radius of the larger circle. Thanks to the geometric mean theorem of elementary geometry we can write:

$$ \begin{align} d \cdot r &= (r-5)^2 \\ d+r &= 2r - 9 \enspace. \end{align}$$ This simplifies to $$ \begin{align} d \cdot r &= r^2 - 10r + 25 \\ d &= r - 9 \enspace. \end{align}$$ Therefore the radius of the larger circle is $25$ cm and the radius of the smaller circle is $(2\cdot 25 - 9) / 2 = 41/2 = 20.5$ cm.

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  • $\begingroup$ @JohnHughes It was indeed wrong. Thanks for pointing it out. $\endgroup$ – Fabio Somenzi Feb 12 '17 at 5:31
  • $\begingroup$ So the answer is 25cm and 20.5cm? $\endgroup$ – Birdman2000 Feb 12 '17 at 8:12
  • $\begingroup$ @Birdman2000 The problem asks for the diameters, which are 50 cm and 41 cm. The length of the segment $DC$ is 16 cm, and the altitude on the hypotenuse of the right triangle $\triangle DEA$ is 20 cm, which is the geometric mean of 25 and 16. $\endgroup$ – Fabio Somenzi Feb 12 '17 at 15:12
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$\triangle DCE$, $\triangle ECA$ and $\triangle AED$ are right-angled triangles. We therefore have

$$ \begin{align} AD^2 &= AE^2 + DE^2 \\ &= AC^2 + CE^2 + EC^2 + CD^2 \\ &= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2 \end{align}$$

But we also have $AD = 2AC - 9$, so we can solve for $AD$:

$$ \begin{align} (2AC - 9)^2 &= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2 \\ 4AC^2 - 36 AC + 81 &= AC^2 + 2AC^2 - 20AC + 50 + AC^2 - 18AC + 81 \\ 2AC &= 50 \\ AD &= 41 \end{align}$$

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