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enter image description here

What fraction of the larger semicircle is filled? The two smaller semicircles are of equal size.

This is a puzzle originally set by Catriona Agg, who is a puzzle setting genius.

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  • $\begingroup$ Hmm, I'm seeing a 90-60-30 triangle but nothing more than that :/ $\endgroup$ – Bubbler Dec 22 '20 at 13:18
  • $\begingroup$ Maybe this is a better fit in MSE rather than PSE? $\endgroup$ – Anonymous Dec 22 '20 at 14:28
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    $\begingroup$ @Anonymous I know an answer so I don't think it works for MSE. $\endgroup$ – Anush Dec 22 '20 at 14:48
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The fraction is

$\frac{2}{3}$

Without loss of generality let the radius of the smaller circle equal 1.enter image description here




Edit:

Bubbler has observed in a comment that from here it is faster to continue like this enter image description here

We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$.

We also have

$$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^2+r^2-2 \sqrt{3} r-2 r+2 \sqrt{3}+4$$ $$r=\frac{6+2\sqrt{3}}{2+2\sqrt{3}}=\sqrt{3}$$




Since $\angle HIG=90^\circ$ and $\arcsin(\frac{1}{2})=30^\circ$, we haveenter image description here

By the pythagorean theorem, ${HD}^2=HI^2+ID^2=1^2+(1+\sqrt{3})^2=5+2 \sqrt{3}$enter image description here

Now label the centre of the large circle J. We have $JA=JD=r$, and since JH bisects the chord AB, $\angle JHA=90^\circ $.enter image description here By the pythagorean theorem $$JH^2+1^2=r^2$$

And by the cosine rule $$JH^2=JD^2+HD^2-2\cdot HD\cdot JD\cdot\cos(IDH)$$ $$JH^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Combining these 2 equations, $$JH^2=r^2-1^2=(r^2)+(5+2\sqrt{3})-2\cdot r\cdot \sqrt{5+2\sqrt{3}} \cos(IDH)$$ Solving for r produces $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \cos(IDH)}$$ We can use the pythagorean theorem to show that $\cos(IDH)=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}$ which now gives $$r=\frac{6+2\sqrt{3}}{2\sqrt{5+2\sqrt{3}}\cdot \frac{\sqrt{4+2\sqrt{3}}}{\sqrt{5+2\sqrt{3}}}}=\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}$$ Squaring both sides,

$$r^2=\left(\frac{6+2\sqrt{3}}{2\sqrt{4+2\sqrt{3}}}\right)^2=\frac{48+24\sqrt{3}}{4(4+2\sqrt{3})}=3$$ $$r^2=3$$ $$r=\sqrt{3}$$

And the ratio of the areas is

$$\frac{\ \frac{2\cdot \pi \cdot (1^2)}{2}\ }{\ \frac{\pi \cdot (\sqrt{3})^2}{2}\ }=\frac{2}{3}$$

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    $\begingroup$ Wow, I failed to spot that JHA is a right triangle (which works because H bisects BA and A and B are on the outer circle, whose center is J). But then actually you don't need cosine rules; using Pythagoras twice on JIH and JHA would have been enough (which gives $(1+\sqrt{3}-r)^2+1^2+1^2=r^2$, which reduces to a nice linear equation). $\endgroup$ – Bubbler Dec 23 '20 at 0:14
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    $\begingroup$ I think bubbler's explanation of why JHA is a right triangle should be added to the explanation. It isn't immediately obvious at a glance. $\endgroup$ – Chris Dec 23 '20 at 10:38
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    $\begingroup$ +1 for coming up with the most complicated expression for $\sqrt 3$ I've seen. $\endgroup$ – Paul Sinclair Dec 24 '20 at 16:24
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UPDATE: Added more formal proof, see bottom

Here is a less rigorous but maybe more intuitive answer:

enter image description here
First of all it is easy to see that the small half circles are compatible with a hexagonal packing. Indeed, we can start by taking the mirror image of the entire set wrt its base line to obtain a triangle of touching circles. So we have the familiar grid with units 1 and $\sqrt 3$. If we insert a circle with radius the larger grid unit and we center it at a small circle center then it is easy to parse the picture.
The puzzle is constructed around an "off center" copy of that circle (fat red circle). What one needs to grasp is that by aligning a side of the circle with the corresponding side of a small circle (top in the visualization) we will also have the center on a corner of a grid square (teal square). 45° symmetry as regards intersections with the small circle in the teal cell follows immediately and from there it is a straight-forward calculation to confirm that the large circle cuts the small one at opposite points.

More formal proof:

![enter image description here All in units of the smaller radius. We start by mirroring. This yields the regular triangle DD'X with height $CX=\sqrt 3$ or, equivalently, the 30-60-90 triangle DCX. We can shift this to the right by one unit aligning former circle centers D and X with right ends of circumferences Ds and Xs. We observe that CsDsDC is a unit square. Now we define YZ as the diameter of the unit circle around D that is perpendicular to DCs. Pythagoras on triangle DYCs gives $YCs=\sqrt 3$ and similar for $ZCs$, in other words Y,Z,Xs lie on a circle around Cs.

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    $\begingroup$ what is your answer? How did you make this grid if you didn't know the ratio of small:big circles $\endgroup$ – Ankit Dec 23 '20 at 0:08
  • $\begingroup$ @Ankit Well, the radii are 1 and $\sqrt 3$, calculating the ratio of areas is left as an exercise to the interested reader. The ratio is obtained by making the educated guess and then confirming that it works and is unique. $\endgroup$ – Paul Panzer Dec 23 '20 at 4:36
  • $\begingroup$ @PaulPanzer Having by now spent entirely too long on this whole puzzle, I see your more-formal proof includes translating triangle DCX one unit to the right. But I don't see how this is proved to be a meaningful thing that then means Xs is actually on the circumference of the large circle? (It obviously is, but I don't understand this as a proof) $\endgroup$ – Ed. Jan 29 at 16:25
  • $\begingroup$ @Ed. Well, the problem is posed in a kind of visual way, and part of the givens is that the right side small semi circle touches the large semi circle at the right end of their shared diameter. By our shifty (pun intended) construction Xs clearly is on the right end of the right small semi circle which by assumption coincides with the large semi circle. $\endgroup$ – Paul Panzer Jan 29 at 17:16
  • $\begingroup$ @PaulPanzer My answer below now proves that it's one unit :-) $\endgroup$ – Ed. Feb 2 at 19:58
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Here's something resembling a solution using mostly just image editing magic and Pythagoras:

We can continue the pattern to the left by adding a copy of the image to itself, mirrored over the vertical line crossing the centre of the obliquely halved circle.

Here's the result of that operation:

enter image description here

Adding up the parts gives the diameter of the bigger circle in small circle radius units:

$ 1 + \sqrt{3} + \sqrt3 - 1 = 2\sqrt3$

Which gives us the ratio of radiuses

$\frac{1}{\sqrt3}$

Which we can square to get the ratio of areas

$\frac{1}{3}$

And as long as we remember that there are two of the smaller semicircles in the picture, we get the final result:

the covered area is two thirds of the larger semicircle.

Yes, there might conceivably be a gap or some overlap between the seemingly tangent circles in the picture. <handwaving type="energetic">If there were, the answer would be something else, which wouldn't fit the known distances between optimally packed circles. </handwaving>

enter image description here

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  • $\begingroup$ Using the coordinates from Oray's answer, doesn't this assume that the point C is halfway between I and A? If it isn't, then the large duplicate circle won't be tangent to the original small whole circle. $\endgroup$ – Mike Dec 23 '20 at 0:56
  • $\begingroup$ @Mike Point C is definitely halfway between A and mirrored A (since it's on the pivot line), but mirrored A being equal to point I is kind of left out, which is what the handwaving part is all about. $\endgroup$ – Bass Dec 23 '20 at 3:46
  • $\begingroup$ OK that's fair enough. $\endgroup$ – Mike Dec 23 '20 at 3:59
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This answer is a bit of a sanity-check for the other answers posted.

Note: Depending on how you view the image, its dimensions might be different than my measurements here, but the individual dimensions here don't matter; rather, it's the ratio between them that matter.

To find the ratio of the area of the small semicircles to the large semicircle: $$\text{ratio} = \frac{\text{Area of the small semicircles}}{\text{Area of the large semicircle}}$$ ...we really need to know only two things: (1), the diameter of the small semicircles, and (2), the diameter of the large semicircle.

image

After measuring it with a ruler (yes, that's right), I found the diameter of the small semicircles to be $3$ inches, whilst the diameter of the large one is $5.25$ inches: img with dimensions

We need to compute their respective areas, and to do that, we need to know their respective radii. That's half of their diameters, so the radius of the small semicircles is $\frac{3 \text{in}}{2} = 1.5 \text{in}$ , while the radius of the large is $\frac{5.25\text{in}}{2} = 2.625\text{in}$

The area of a semicircle is just half the area of a full circle with the same radius, ($A = \frac{r^2\pi}{2}$), so to get the radius of the semicircles, we just compute the area of full circles with the same radius, and halve them.

For the small semicircles: $$\text{Area of the small semicircles} = \frac{1.5\text{in}^2\pi}{2} × 2$$ ($× 2$ since there are two small semicircles)

For the large one: $$\text{Area of the large semicircle} = \frac{2.625\text{in}^2\pi}{2}$$

Thus, $$\frac{\text{Area of the small semicircles}}{\text{Area of the large semicircle}}$$ $$= \dfrac{\dfrac{1.5\text{in}^2\pi}{2} × 2}{\dfrac{2.625\text{in}^2\pi}{2}}$$ $$= \frac{32}{49}. $$

And since

$$\frac{32}{49} \approx 0.653 \approx 0.\overline{6} = \frac{2}{3},$$

...we see that the portion of the large semicircle that is shaded is likely

$$\dfrac{2}{3}$$

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    $\begingroup$ Sanity checks always welcome! $\endgroup$ – Anush Dec 24 '20 at 14:21
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Slightly late to the party. This finds $r$ of the larger circle (where the smaller circle is a unit radius) using $FHD$ as a circular segment on that larger circle:

circles image

Some generalities, not all of which are used:

$$\begin{align} AB & = BC = BJ = DE = EF = EK = EG = GB = 1 \\ AC & = EB = 2 \end{align}$$

Use Pythagoras to establish relative location of $E$, $B$ and $K$:

$$\begin{align} ∠EKB & = π/2 \\ EB^2 & = EK^2 + BK^2 \\ BK & = √(EB^2 - EK^2) \\ & = √(4 - 1) = √3 \end{align}$$

Cartesian coordinates for all the points identified except $D$, $F$, $L$ as they are more complicated:

$$\begin{align} A & = (2r, 0) \\ B & = (2r-1, 0) \\ C & = (2r-2, 0) \\ E & = (2r-1-√3, 1) \\ G & = (B + E)/2 \\ & = 1/2(2r-1 + 2r-1-√3, 1) \\ & = 1/2(4r-2-√3, 1) \\ & = (2r-1-√3/2, 1/2) \\ H & = (r, 0) \\ J & = (2r-1, 1) \\ K & = (2r-1-√3, 0) \end{align}$$

Establish $EH$ using Pythagoras, in order to get the apothem of the segment:

$$\begin{align} HK & = H - K \\ & = r - (2r-1-√3) \\ & = r - 2r + 1 + √3 \\ & = 1 + √3 - r \\ HK^2 & = (1 + √3 - r)^2 \\ & = (1 + √3)^2 - 2r(1 + √3) + r^2 \\ & = 1 + 2√3 + 3 - 2r(1 + √3) + r^2 \\ & = 4 + 2√3 - 2r(1 + √3) + r^2 \\ EH^2 & = HK^2 + EK^2 \\ & = 4 + 2√3 - 2r(1 + √3) + r^2 + 1 \\ & = 5 + 2√3 - 2r(1 + √3) + r^2 \\ & = r^2 - 2r(1+√3) + 5+2√3 \\ EH = & √(r^2 - 2r(1+√3) + 5+2√3) \end{align}$$

$r$ in terms of $h$ (the sagitta, $EL$) and $c$ (the chord length, $DF$) via the intersecting chord theorem:

$$\begin{align} r & = c^2/8h + h/2 = HL \\ c & = DF = 2 \\ h & = EL = HL - EH \\ & = r - √(r^2 - 2r(1+√3) + 5+2√3) \\ r & = 4/8h + h/2 = 1/2h + h/2 \\ 2hr & = h^2 + 1 \\ h^2 - 2hr + 1 & = 0 \end{align}$$

Expand out $h^2$ and $2hr$ because then it is very clean to complete the equation above:

$$\begin{align} h^2 & = r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) + r^2 - 2r(1+√3) + 5+2√3 \\ & = 2r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) - 2r(1+√3) + 5+2√3 \\ 2hr & = 2r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) \end{align}$$

Then we have:

$$\begin{align} -2r(1+√3) + 5+2√3 + 1 & = 0 \\ -2r(1+√3) + 6+2√3 & = 0 \\ -r(1+√3) + 3+√3 & = 0 \\ r(1+√3) - (3+√3) & = 0 \\ r(1+√3) & = (3+√3) \\ r & = (√3+3)/(1+√3) \\ & = √3(√3+3)/√3(1+√3) \\ & = √3(√3+3)/(√3+3) \\ & = √3 \end{align}$$

Feeding $r$ back into $EH$:

$$\begin{align} r & = √3 \\ EH & = √(r^2 − 2r(1+√3) + 5 + 2√3) \\ & = √(3 − 2√3(1+√3) + 5 + 2√3) \\ & = √(8 − (2√3+6) + 2√3) \\ & = √(8 − 2√3 - 6 + 2√3) \\ & = √(8 - 6) \\ & = √2 \end{align}$$

Since $EK$ is 1, and $∠EKH$ is a right-angle, $KH$ is also 1, and $∠EHK = π/4$.

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