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enter image description here

Don't let the simplicity of this diagram fool you. I have been wondering about this for quite some time, but I can't think of an easy/smart way of finding it.

Any ideas?


For reference, the Area is:

$$\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}$$

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    $\begingroup$ If you consider integration as "easy". Then the area is just half of the area of rectangle minus area of circles, then subtract the little white area (calculate using integration) from that. $\endgroup$ – lEm Jul 29 '16 at 9:20
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    $\begingroup$ $$\dfrac{\text{Area}_{\square} - 2 \text{Area}_{\circ}}{2}$$ $\endgroup$ – Rodrigo de Azevedo Jul 29 '16 at 9:22
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    $\begingroup$ @RodrigodeAzevedo If you mean (area of the rectangle minus area of the two disks) halved, then that would be incorrect because the lower left corner is not shaded. $\endgroup$ – arctic tern Jul 29 '16 at 9:26
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    $\begingroup$ What is the source of the image/problem? $\endgroup$ – Mark S. Jul 29 '16 at 10:09
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    $\begingroup$ @MarkS. I found this link quora.com/What-is-the-area-of-the-shaded-part where the same problem is discussed $\endgroup$ – kai Jul 29 '16 at 12:12

12 Answers 12

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enter image description here

We observe that $\triangle PRT$ can be partitioned into five congruent sub-triangles. Therefore, the entire shaded region has area given by ... $$\begin{align} 3 u + |\text{region}\; PAT| &= 3u + |\square OAPT| - |\text{sector}\;OAT| \\[6pt] &= 3u + \frac{3}{5}\,|\triangle PRT| - |\text{sector}\;OAT| \\[6pt] &= 3\cdot\frac{1}{4} r^2 \left( 4 - \pi \right) \;+\; \frac{3}{5}\cdot r^2 \;-\; \frac{1}{2}r^2\cdot 2\theta \end{align}$$ Since $\theta = \operatorname{atan}\frac{1}{2}$, this becomes

$$r^2\left(\; \frac{18}{5} - \frac{3}{4}\pi - \operatorname{atan}\frac{1}{2} \;\right) \qquad\stackrel{r=5}{\to}\qquad 90 - \frac{75}{4}\pi - 25\;\operatorname{atan}\frac{1}{2}$$

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    $\begingroup$ Beautiful and instructive! (+1) $\endgroup$ – Markus Scheuer Jul 31 '16 at 7:25
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    $\begingroup$ @MarkusScheuer: Thanks. I actually like this one enough that I posted it to my trigonography site. :) $\endgroup$ – Blue Jul 31 '16 at 8:53
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    $\begingroup$ Nice examples and presentation on your site. One extra plus. :-) $\endgroup$ – Markus Scheuer Jul 31 '16 at 9:01
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    $\begingroup$ "We observe..." (after thinking about it for a day or so) "...that $\triangle PRT$ can be partitioned..." - very pretty, well done. $\endgroup$ – Joffan Jul 31 '16 at 16:37
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    $\begingroup$ @Joffan: It’s worth recalling the story of the very famous mathematician G.H. Hardy, who in a lecture said about some detail in a proof: “This is obvious.” After a pause, he went on: “Hmm, is it really obvious?” After another pause he left the room to consider the point, returning 20 minutes later with the verdict: “Yes, I was right, it is obvious.” —J.R. Partington (via DavidMarcus.com's Math Jokes page) Incidentally, the above was my third pass at this problem. (My second is in the edit history.) :) $\endgroup$ – Blue Jul 31 '16 at 16:50
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[Note: My second answer is much better.]

I'll focus on the unshaded region at the bottom-left.

enter image description here

By an aspect of the Inscribed Angle Theorem, we know that $\angle AOB = 2\;\angle ABP$ (justifying marking these $\theta$ and $2\theta$). By a related result, we have that $$\phi = \frac{1}{2}\left(\angle BOC - \angle AOB\right) = 45^\circ - \theta$$ Moreover, we know that $$\phi = \operatorname{atan}\frac{1}{2} \approx 26.56^\circ \qquad\to\qquad \theta = 45^\circ - \operatorname{atan}\frac{1}{2} \approx 18.43^\circ$$

From here, knowing the circle's radius, one may calculate the lower-left area as ... $$\begin{align} &|\triangle PAB| + |\triangle OAB| - |\text{sector } OAB| \\ \end{align}$$ ... from which we readily derive the area in the original question. For now, I'll leave these details to the reader.

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    $\begingroup$ What software do you use to draw figures? It's beautiful. $\endgroup$ – The Artist Jul 29 '16 at 10:36
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    $\begingroup$ @theartist Geogebra is free and makes it easy to make things like this. $\endgroup$ – Mark S. Jul 29 '16 at 10:39
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    $\begingroup$ Does the theorem need to be invoked? You have an isosceles triangle so the angles OAB and OBA must be 90 - theta. PB is a tangent and OB is a radius so perpendicular. So ABP is theta. $\endgroup$ – Carlos Jul 29 '16 at 14:10
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    $\begingroup$ @Carlos: You've basically explained why the "tangent" case of the IAT works, so it doesn't really matter whether one invokes the IAT or makes that argument explicitly. Be that as it may ... I use the related angle-subtraction result to calculate $\phi$, so invoking IAT first helps set the conceptual mood. (Note: the related result isn't entirely necessary here, either. We have $\angle ACB = \theta$ (it's an inscribed angle), and thus that $\angle OPC = \theta$ (alternate interior angles), which in turn gives $\theta + \phi = 45^\circ$. I wonder if that's a "better" way to think about this.) $\endgroup$ – Blue Jul 29 '16 at 15:37
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    $\begingroup$ Another way to find φ, perhaps a little longer but using very elementary steps: OAC = OCA = φ (by internal angles of an isosceles tri, then by alternating angles); also CAB = φ + θ (by running a horizontal line through A); so OAC = 2φ + θ, and so summing the internal angles of the triangle OAC, we get 180° = (2θ) + (2φ + θ) + (2φ + θ) = 4φ + 4θ. $\endgroup$ – Peter LeFanu Lumsdaine Jul 29 '16 at 23:11
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Put $\arctan{1\over2}=:\alpha$. Then $$\sin(2\alpha)={2\tan\alpha\over1+\tan^2\alpha}={4\over5}\ .$$ The area $A$ in question consists of three "arrow heads" plus the area shaded in the following figure. The latter is a right triangle minus a sector and a smaller triangle. We therefore obtain $$A={3\over4}(10^2- 25\pi)+{25\over2}\bigl(2-2\alpha-\sin(2\alpha)\bigr)=90-{75\over4}\pi-25\alpha\ .$$

enter image description here

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  • $\begingroup$ Clever approach. Nice and concise! (+1) $\endgroup$ – Markus Scheuer Aug 1 '16 at 6:51
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The big black roundy-corner on the bottom right has area $(10^2 - \pi\cdot 5^2)/4$ and there are 3 complete copies of it and one copy trimmed by a small roundy-triangle. We will focus on this roundy-triangle which is the same as the one in the bottom left.

So the key is to compute the area of the small white roundy-triangle at the bottom left.

To this end we must find the intersection of the diagonal and the circle that create the top of this triangle.

The equation of the circle and of the diagonal are $$ y = 1/2 x \\ (x-5)^2 + (y-5)^2 = 25 $$ Solving that with WA gives : $x=2,\ y=1$.

So now we can decompose this roundy-triangle into two parts by drawing a vertical line that goes through this intersection. This gives a true triangle (the left part) which has area $1$ and another roundy-triangle (the right part).

To compute the area of the roundy-right-triangle, we can use integration : $$ \int_2^5 -\sqrt{25 - (x-5)^2} + 5\ dx $$ See WA for the plot of this function.

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enter image description here

Considering the right triangle with legs $1-s\sin(\theta)$ and $1-s\cos(\theta)$, we have $$ 1=(1-s\cos(\theta))^2+(1-s\sin(\theta))^2\tag{1} $$ Solving for $s$, we get $$ s=\sin(\theta)+\cos(\theta)-\sqrt{2\sin(\theta)\cos(\theta)}\tag{2} $$ Setting $t=\tan(\theta)$ yields $$ \begin{align} s\sin(\theta)&=\frac{t}{1+t^2}\left(t-\sqrt{2t}+1\right)\\ s\cos(\theta)&=\frac{1}{1+t^2}\left(t-\sqrt{2t}+1\right)\\ \tan(\phi/2)&=\frac{1-s\cos(\theta)}{2-s\sin(\theta)} \end{align}\tag{3} $$ Setting $\tan(\theta)=\frac12$ gives $$ \begin{align} s\sin(\theta)&=\frac15\\ s\cos(\theta)&=\frac25\\ \tan(\phi/2)&=\frac13 \end{align}\tag{4} $$ The area of the green piece is $$ \frac\phi2-\frac12(1-s\cos(\theta))=\tan^{-1}\left(\frac13\right)-\frac3{10}\tag{5} $$ The sum of the areas of the purple and green pieces is $$ \frac12s\sin(\theta)=\frac1{10}\tag{6} $$ Therefore, the area of the purple piece is $$ \frac25-\tan^{-1}\left(\frac13\right)\tag{7} $$


$(7)$ used a circle with radius $1$. For a circle with radius $5$, we get an area of $$ 10-25\tan^{-1}\left(\frac13\right)\tag{8} $$ The area of half the $10\times20$ rectangle minus the two circles of radius $5$ is $$ 100-25\pi\tag{9} $$ Therefore, the area in the given image is the difference of $(9)$ and $(8)$: $$ \bbox[5px,border:2px solid #C0A000]{90+25\tan^{-1}\left(\frac13\right)-25\pi}\tag{10} $$


$(10)$ is a different form of the same area: $$ \begin{align} 90+25\tan^{-1}\left(\frac13\right)-25\pi &=90+25\tan^{-1}\left(\frac13\right)-25\tan^{-1}\left(1\right)-\frac{75}4\pi\\ &=90+25\tan^{-1}\left(\frac{\frac13-1}{1+\frac13\cdot1}\right)-\frac{75}4\pi\\ &=\bbox[5px,border:2px solid #C0A000]{90-25\tan^{-1}\left(\frac12\right)-\frac{75}4\pi}\tag{11} \end{align} $$

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Well, I'm assuming you have problems only with the upper right corner, since the rest is just $\frac 34 (10² - \pi5²)$. I also assume you can calculate the area of the small part of the circle, so we have

$$ A = \frac{5 \cdot 10}2 - A_1$$ $$ A_1 = \pi \cdot 5^2 - \pi \cdot (\frac52)^2 - A_{small}$$

where $A$ is the shaded area we didn't know yet, $A_1$ is the area of the triangle intersected with the circle and $A_{small}$ the area of the small portion of the circle.

Hope this helps.

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By the look of the answer, it would appear that there is no avoiding calculating the area of the roundy white triangle, or at least indirectly.

Call this area $x$ and the area of the lowest right hand corner region $a$, where $4a=100-25\pi$. Let the area of the smaller segment of the circle be $s$. The chord of this segment subtends an angle $\pi-2\arctan\frac 12$ at the centre of the circle. In which case

$$s=\frac 12\cdot25(\pi-2\arctan\frac 12-\sin(2\arctan\frac 12))$$ $$\Rightarrow s=\frac{25}{2}\pi-25\arctan\frac 12-10$$

Meanwhile, $x=25-a-s$ and the required region is $R=4a-x=5a-25+s$

Substituting in the value of $s$ gives the required answer $$R=90-18.75\pi-25\arctan \frac 12$$

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My method is just an ordinary one but it shows how the suggest answer is obtained.

It should be clear that how the lengths and angles are defined as shown.

enter image description here

$\theta = \tan^{-1}(\dfrac {1}{2})$ in radians.

By power$(A, C_2)$, $z = \dfrac {100}{\sqrt {125}}$. Then, by power $(P, C_3)$, $y = \dfrac {25}{\sqrt {125}}$.

By similar triangles, $t = … = 2$.

After transferring the leftmost black shaded part to its more suitable position (above X), the required area

$= [(10 \times 10) square] - [C_2] - [red]$

$= 100 - 25 \pi - ([purple] - [\dfrac {C_1}{4}] - [yellow])$

$= 100 - 25\pi – 25 + \dfrac {25\pi}{4} + [yellow]$

$= 75 - 18.75\pi + [yellow]$

$= 75 - 18.75\pi + ([⊿XCQ] - [grey])$

$=75 - 18.75\pi + 25 - [grey]$

$= 100 - 18.75\pi - ([⊿XYQ] + [section YPQ] $

$= 100 - 18.75\pi - ([⊿XYQ] + [sector OPQ] $

$= 100 - 18.75\pi - (10 + \dfrac {(25)2\theta}{2})$

$= 90 - 18.75\pi - 25\tan^{-1}(\dfrac {1}{2})$

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  • $\begingroup$ which program you used to draw that? $\endgroup$ – Salech Rubenstein Aug 2 '16 at 18:03
  • $\begingroup$ @SalechAlhasov Create a sketch from Geogebra. Upload it to the clipboard. $\endgroup$ – Mick Aug 3 '16 at 3:28
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Visual decomposition

$$\small \begin{align} A_\textsf{Shaded} &= A_\textsf{Square} - \color{darkorange}{A_\textsf{Circle}} - (\color{green}{A_\textsf{4-gon}}+\color{red}{A_\textsf{Triangle}})+\color{blue}{A_\textsf{CircleSegment}}\\&= 10^2 - \color{darkorange}{\pi \times 5^2} - \left(\color{green}{9}+\color{red}{1}\right)+\color{blue}{\frac{\arctan{(3/4)}}{2\pi}\pi \times 5^2} \\ &= 100 - \color{darkorange}{25\pi}-10+\color{blue}{12.5\arctan{\frac {3}{4}}}\\ &\approx 19.504…\end{align}$$

$\blacksquare $

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    $\begingroup$ +1. Interesting. I made multiple passes at this puzzle, and I even saw Christian's answer referencing a sine equal to $4/5$, but I don't think it ever really dawned on me that there was a $3$-$4$-$5$ triangle lurking. $\endgroup$ – Blue Mar 1 '17 at 4:10
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If $A=B+X$ where $B=\frac 12+\int_2^5(5-\sqrt{10x-x^2})dx$ (small white area below the diagonal) and $X$ the asked area then one has by symmetry $$2A+2(\pi\cdot5^2)=20\cdot 10\iff A+25\cdot\pi=100$$ We have$\int_2^5(5-\sqrt{10x-x^2})dx\approx 0.95624$ so $$X=100-25\cdot\pi-0.5-0.95624\approx 20.003943$$

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I posted a solution directly on 9gag:

enter image description here

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"Easy"? What's easy? My math/algebra skills are very rusty, but I'd set the problem up this way:

  1. Drop a perpendicular from the intersection of the circle and the straight line.

  2. Take the area of the triangle @30 deg.

  3. (Moving to the right) Do an integral from the perpendicular to the osculation point on the rectangle given the equation of the circle.

  4. Add the two areas: the triangle and the integral

Question is: is doing an integral "easy". Conceptually yes, mathematically ... I don't know. Depends.

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