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We have three circles where their centers are on the same line and PR is tangent to both small circles as shown below.

two circles lined up within a larger blue one which minimally encloses them, with shared tangent PR drawn in

If |PR|=12 unit, what is the area of blue part of the circle?

Reference: A Turkish Journal: Bilim Teknik

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    $\begingroup$ Are the circles in a ratio of size to each other? $\endgroup$ Commented Nov 13, 2017 at 17:23
  • $\begingroup$ as in is the smallest circle half the size of the second largest etc... $\endgroup$ Commented Nov 13, 2017 at 17:27
  • $\begingroup$ am sooo confused. Does not tell anything to fix the size of 2 white circles. $\endgroup$ Commented Sep 8, 2019 at 11:34
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    $\begingroup$ @AlwaysConfused Ah, but it does tell you something that constrains the size of the white circles, namely the length of that tangent line. And, as it happens, that (plus the fact that the centres are collinear) is enough to determine the answer, even though the size of each individual circle can vary. $\endgroup$ Commented Sep 11, 2019 at 15:12
  • $\begingroup$ So there would be only one definite big circle if we keep the length of that segment invariable? $\endgroup$ Commented Sep 11, 2019 at 15:38

3 Answers 3

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If the question has a well defined answer then

it must not depend on which of the infinitely many configurations satisfying the given description we choose.

So

we can take PR to be a diameter of the large circle, which then has area $36pi$; $2\times\frac14=\frac12$ of the circle is white, hence $\frac12$ is blue, so the blue area is $18\pi$.

That's a bit cheeky, of course. So here's a more conventional solution.

Let Q be the point of tangency in the middle of that common tangent. Let A,B be the two ends of the diameter passing through Q. Then we have $AP^2=PQ^2+d_1^2$ and $BP^2=PQ^2+d_2^2$ where $d_1,d_2$ are the diameters of the two white circles. And now note that since the angle in a semicircle is always $\pi/2$ we have $AP^2+BP^2=d^2$ where $d$ is the diameter of the outer circle. Putting this together, $d^2=2PQ^2+d1^2+d2^2$; our area is $\frac\pi4(d^2-d_1^2-d_2^2)=\frac\pi2\cdot PQ^2=\frac\pi8PR^2$ and we're done.

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  • $\begingroup$ Smart but cheaty. I added a slightly less smart but less cheaty solution. $\endgroup$ Commented Nov 13, 2017 at 17:40
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    $\begingroup$ A problem with a similar feel to it and a similar "smart" solution: A sphere has a hole bored along a diameter. The length of the hole (measured beteween its actual ends, not the now-removed poles of the sphere at the ends of that diameter) is, say, 12 units. What is the volume of what remains? Cheaty answer: Suppose the hole is of zero thickness; then 12 units is the diameter of the sphere and no material has been removed. Etc. $\endgroup$ Commented Nov 13, 2017 at 17:42
  • $\begingroup$ @GarethMcCaughan Is it not correct to say that PR is diameter of the large circle, whose centre(Q, as said by you or the middle dot in the picture) is not on PR? $\endgroup$ Commented Nov 14, 2017 at 7:08
  • $\begingroup$ In the diagram here, unless the small circles are equal PR is not a diameter. But the perpendicular to it through Q is a diameter. $\endgroup$ Commented Nov 14, 2017 at 10:58
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    $\begingroup$ @Trunk That's why the first answer is "a bit cheeky". The "more conventional solution" in the third spoiler block doesn't depend on knowing in advance that the answer doesn't depend on exactly where we draw the line PR. $\endgroup$ Commented Jun 27 at 20:07
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3circles

$$\Delta A = \pi\; (R^2 - r_1^2 - r_2^2) $$
From the diagram we can see that:
$$ 2R = 2r_1 + 2r_2 \implies R = r_1 + r_2 \;\;\;\;\;\;\; R \sin{A} = 6 \;\;\;\;\;\;\; R(1 - \cos{A}) = 2r_2 $$
$$\implies r_2 = \dfrac{R}{2}\; \left(1 - \cos{A} \right) \;\;\;\;\;\; \text {and hence} \;\;\;\;\;\; r_1 = \dfrac{R}{2}\; \left(1 + \cos{A} \right) $$
Thus:
$$ \Delta A \; = \; \pi\;\left[\; \left (r_1 + r_2 \right)^2 - r_1^2 -r_2^2 \; \right ] \; = \; 2\pi\;r_1r_2 $$
$$ = \; 2 \pi \; \dfrac{R^2}{4} \; \left (1 - \cos^2{A} \right) \; = \; \dfrac{\pi}{2} \; {R^2} \sin^2{A} $$ Since $\; R \sin{A} = 6 \; $ we can thus write
$$ \boldsymbol{ \Delta A \; = \; 18 \; \pi \;\;\;\; \text {sq units} } $$

Maybe we should rename this as the flat snowman puzzle since we can use the same amount of snow (area being analogous to mass here) to make it so it fits into a circle with the snowman's neck is level with the 12 units long chord.
Note that the snowman's proportion, i.e. the ratio $\dfrac{r_1}{r_2}$, is given by
$$ \require{cancel} \dfrac{r_1}{r_2} \; = \; \dfrac{\dfrac{\cancel{R}}{2}\; \left(1 + \cos{A} \right)} {\dfrac{\cancel{R}}{2}\; \left(1 - \cos{A} \right)} \; = \; \dfrac{\dfrac{1}{2}\; \left(1 + \cos{A} \right)} {\dfrac{1}{2}\; \left(1 - \cos{A} \right)} \; = \; \dfrac{\cos^2 \left({\dfrac{A}{2}} \right)} {\sin^2 \left({\dfrac{A}{2}} \right)} $$
$$ \boldsymbol{ \dfrac{r_1}{r_2} \; = \; \cot^2 \left({\dfrac{A}{2}} \right) } $$
This reminds me of another puzzle involving 3 enclosed circles where the diameters of the interior circles had a ratio which varied with the angles the separating chords made to a fixed chord.

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Answer is

$18\pi$

Let $d_s=$ diameter of small inner circle and $d_l=$ diameter of large inner circle,
then $d$ (diameter of outer circle) $=$

$d_s+d_l$
and, of course, $r=\dfrac{d_s+d_l}{2}.$

Using $d$ and the length of $\text{PR}$, we can find the value of $d_s$ and $d_l$ as following:

$\text{PR}$ is a chord and its equation is:
Chord length $=2\sqrt{r^2-l^2}$, where $r$ is the radius of a circle and $l$ is the perpendicular distance between a chord and the center of a circle.
After inspecting the circle, we can find that $l=d_l-r$, diameter of larger inner circle - radius of outer circle.
So,
$$\begin{align}\text{PR}=12=2\sqrt{r^2-(d_l-r)^2} \implies 6&=\sqrt{\left(\dfrac{d_s+d_l}{2}\right)^2-\left(\dfrac{2d_l}{2}-\dfrac{d_s+d_l}{2}\right)^2}\\ &=\sqrt{\left(\dfrac{d_s+d_l}{2}\right)^2-\left(\dfrac{d_l-d_s}{2}\right)^2}\\&=\sqrt{\dfrac{d_s^2+2d_sd_l+d_l^2}{4}-\dfrac{d_l^2-2d_sd_l+d_s^2}{4}}\\&=\sqrt{\dfrac{4d_sd_l}{4}}\\&=\sqrt{d_sd_l}.\end{align}$$ Then, $$36=d_sd_l.$$

Since we know the value of $d_s$ and $d_l$ now, we are ready to calculate the area of shaded area:

Area of a circle $=\pi r^2$.
Using this equation, we can find that
Area of outer circle $=\pi \left(\dfrac{d_s+d_l}{2}\right)^2=\dfrac{\pi}{4}\left(d_s+d_l\right)^2$
Area of small inner circle $=\pi \left(\dfrac{d_s}{2}\right)^2=\dfrac{\pi}{4}d_s^2$
Area of large inner circle $=\pi \left(\dfrac{d_l}{2}\right)^2=\dfrac{\pi}{4}d_l^2$
and
Area of shaded area $=$ Area of outer circle $-$ Area of small inner circle $-$ Area of large inner circle$$\begin{align}&=\dfrac{\pi}{4}\biggl[\left(d_s+d_l\right)^2-d_s^2-d_l^2\biggr]\\&=\dfrac{\pi}{4}\biggl[\left(d_s^2+2d_sd_l+d_l^2\right)-d_s^2-d_l^2\biggr]\\&=\dfrac{\pi}{4}\biggl[2 \, d_sd_l \biggr]\end{align}$$
and we already know what $d_sd_l$ is from above :)
Therefore, Area of shaded area $=\dfrac{\pi}{4}[2 \cdot 36]=18\pi$.

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