Jtdylktuy

This is a popular problem spreading around. Solve for the shaded reddish/orange area. (more precisely: the area in hex color #FF5600)

$ABCD$ is a square with a side of $10$, $APD$ and $CPD$ are semicircles, and $ADQB$ is a quarter circle. The problem is to find the shaded area $DPQ$.

I was able to solve it with coordinate geometry and calculus, and I verified the exact answer against a numerical calculation on Desmos.

Ultimately the result is 4 terms and not very complicated. So I was wondering: Is there was a way to solve this using trigonometry? Perhaps there is a way to decompose the shapes I am not seeing.

A couple of years ago there was a similar "Find the shaded area" problem for Chinese students. I was able to solve that without calculus, even though it was quite an involved calculation.

Disclosure: I run the YouTube channel MindYourDecisions. I plan to post a video on this topic. I'm okay posting only the calculus solution, but it would be nice to post one using only trigonometry as many have not taken calculus. I will give proper credit to anyone that helps, thanks!

Update: Thanks for everyone's help! I prepared a video for this and presented 3 methods of solving it (the short way like Achille Hui's answer, a slightly longer way like David K and Seyed's answer, and a third way using calculus). I thanked those people in the video on screen, see around 1:30 in this link: https://youtu.be/cPNdvdYn05c.

The area can be simplified to $75tan^{-1}left(frac12right) - 25 approx 9.773570675060455 $.

It come down to finding the area of lens $DP$ and $DQ$ and take difference.

What you need is the area of the lens formed by intersecting two circles, one centered at $(a,0)$ with radius $a$, another centered on $(0,b)$ with radius $b$. It is given by the expression.

$$begin{align}Delta(a,b)
stackrel{def}{=} & overbrace{a^2tan^{-1}left(frac{b}{a}right)}^{I} + overbrace{b^2tan^{-1}left(frac{a}{b}right)}^{II} - ab\
= & (a^2-b^2) tan^{-1}left(frac{b}{a}right) + frac{pi}{2} b^2 - ab
end{align}
$$

In above expression,

• 
  $I$ is area of the circular sector span by the lens at $(a,0)$ (as a convex hull).


• 
  $II$ is area of the circular sector span by the lens at $(0,b)$ (as a convex hull).


• 
  $ab$ is area of union of these two sectors, a right-angled kite with sides $a$ and $b$.

Apply this to problem at hand, we get

$$begin{align}verb/Area/(DPQ)
&= verb/Area/({rm lens}(DQ)) - verb/Area/({rm lens}(DP))\[5pt]
&= Delta(10,5) - Delta(5,5)\
&= left((10^2-5^2)tan^{-1}left(frac12right) + 5^2cdotfrac{pi}{2} - 5cdot 10right)
- left( 5^2cdotfrac{pi}{2} - 5^2right)\
&= 75tan^{-1}left(frac12right) - 25
end{align}
$$

The area is equal to difference between the area of two lenses.

It is easy to find the area of lenses like the one I did in this question before: How to find the shaded area

Let $E$ be the midpoint of the edge $CD.$
Then $triangle ADE$ and $triangle AQE$ are congruent right triangles,
and we find that $angle DAQ = 2arctanleft(frac12right).$
Moreover, $angle CEQ = angle DAQ$ and therefore
$angle DEQ = pi - 2arctanleft(frac12right).$
And of course each of the arcs from $D$ to $P$ has angle $fracpi2.$

Knowing the radius and angle of an arc you can find the area of the circular segment bounded by the arc and the chord between the arc's endpoints without calculus.
The area of the red region is the sum of the areas of the segments bounded by the arcs between $D$ and $Q,$
minus the sum of the areas of the segments bounded by the arcs between $D$ and $P.$
Note that one of the arcs from $D$ to $Q$ has radius $10$, but the other three arcs all have radius $5.$

For fun, I did the old chemist trick of printing out the diagram, cutting it apart, then weighing the pieces on a milligram scale. No calculus!

The total diagram weighed 720 mg, and the sliver weighed 77 mg. Then, $frac{77,mathrm{mg}}{720,mathrm{mg}}cdot 10^2,mathrm{cm}^2approx 10.7,mathrm{cm}^2$ is the estimated area. This is about $9.5%$ greater than the analytical solution. Not that good, but still not bad for something quick.

One source of error was the extra weight of the toner on the sliver, which printed out rather dark gray. If I knew where my compasses were, I could make a more accurate construction.

The complete solution can be watched here : https://youtu.be/4Yrk-UNfAis

I think this is a good step to find the result without any coordinates, while it is actually not the full solution.

You have six non intersecting subareas, say:

• S1 is DPD


• S2 is DQPD


• S3 is DCQD


• S4 is CBQC


• S5 is BAPQD


• S6 is ADPA

Also say that L is the length of the square.

You can at least state these equations :

• S1+S2+S3+S4+S5+S6 = $L^2$
  


• S1+S6 = $frac{1}{2}pileft(frac{L}{2}right)^2$
  


• S1+S2+S5+S6 = $frac{pi L^2}{4}$
  


• S1+S2+S3 = $frac{1}{2}pileft(frac{L}{2}right)^2$
  


• S3+S4 = $frac{(2L)^2-pi L^2}{4}$
  


• S2+S5 = $frac{pi L^2-pileft(frac{L}{2}right)^2}{4}$
  

Alas these are not independent, but I'm pretty sure that you can find six independent ones like this.