# How do I find the diameters of the circles in this geometry puzzle?

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.

• Related (pretty-much a duplicate, although it asks for the area of the crescent): math.stackexchange.com/questions/35898/… .
– Blue
Commented Feb 12, 2017 at 1:18
• 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.
– dxiv
Commented Feb 12, 2017 at 1:19
• 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. Commented Feb 12, 2017 at 1:48
• I can't resist pointing out that a crescent moon actually never looks like that. Commented Feb 12, 2017 at 17:07
• Hint: DCE and ECA are similar triangles. Commented Feb 13, 2017 at 3:53

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:

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 $50-9 =\fbox{41}$

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.

• @JohnHughes It was indeed wrong. Thanks for pointing it out. Commented Feb 12, 2017 at 5:31
• So the answer is 25cm and 20.5cm? Commented Feb 12, 2017 at 8:12
• @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. Commented Feb 12, 2017 at 15:12

$\triangle DCE$, $\triangle ECA$ 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}