Viral Math Problem From China - How To Solve With A Simple Formula!

I've posted about this type of problem before. But now I present a simple formula to solve for all values of r. Thanks to Tony Hasler for figuring out this method!

Previous video (much harder way to solve this problem)

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[MindYourDecisions]

I was surprised how many people emailed me this solution too, thanks! I'm so grateful for all the wonderful insights people send me. By the way, if you email me, I probably will not reply--I receive too many emails these days. But I do try to read all emails and save good puzzle suggestions. Thanks.

[FrancisACNH]

Viral math problems be like

“This problem was given to newborn babies in China!”

[atlaskaiser]

integral is the key for all these kinds of shapes

[shane]

Presh Talwalkar sounds like a Star Wars character name

[MaxEU]

For me, the first approach that you called difficult was way more intuitive. Hence, I tried it before watching the video - successfully!

So, how does it work? Draw a line from the centre of the circle to the intersection of the circle and the diagonal line. You get a circular segment and also an isosceles triangle.

Let us call the angle at the right of the triangle beta. To solve for beta, use the outer triangle which is a right triangle:

beta = arctan(r/2r)
beta = arctan(1/2)

We also draw the height of the side at the bottom (from the intersection of the circle and the circle to the upper-right point of the rectangle) and solve for it:

sin(beta) = h/r
h = r * sin(beta)

Let us call the angle at the top of the isosceles triangle gamma. We know two angles, so it it is no problem to get the third one (let us use radians):

gamma = PI/2 - arctan(1/2)

Now, we get the following for the circular segment:

A(segm) = (PI/2 - arctan(1/2))/PI * PI*r^2
A(segm) = r^2 (PI/2 - arctan(1/2))

Moreover, we need to get c, the side at the bottom of the isosceles triangle:

cos(beta) = c/r
c = r * cos(arctan(1/2))
c = r/sqrt((1/2)^2 + 1)
c = r/sqrt(5/4)

With this, we can easily calculate the area of the triangle:

A(isoscelesTriangle) = c*h
= rh/sqrt(5/4)
= r^2 * sin(beta) / sqrt(5/4)
= r^2 * sin(arctan(1/2)) / sqrt(5/4)
= 1/2r^2 / (sqrt(5/4) * sqrt(5/4))
= 1/2 / r^2/(5/4)
= 2/5 r^2

Finally, the area of the circular part is the difference between the area of the circular segment and the area of the triangle:

A(circpart) = r^2 (PI/2 - arctan(1/2)) - 2/5r^2
= r^2 (PI/2 - 2/5 - arctan(1/2)



Okay, puh. That was the hard part! Now, let us calculate the area at the side of the circle. For this, we can simply subtract the area of the semicircle from the area of the rectangle and take the half of the resulting value:

A(side) = 1/2(2r^2 - 1/2*PI*r^2)
= r^2 (1 - 1/4*PI)

We will also need the area of the large triangle:

A(triangle) = 1/2 * 2 * r^2
= r^2

Great! The red area is:

A = A(triangle) - A(side) - A(circpart)
= r^2 - r^2 (1 - 1/4*PI) - r^2 (PI/2 - 2/5 - arctan(1/2))
= r^2 (1 - 1 + 1/4*PI - PI/2 + 2/5 + arctan(1/2))
= r^2 (-1/4*PI + 2/5 + arctan(1/2))

According to WolframAlpha, this expression is identical to the expression determined by you. Hence, my solution is also correct!



PS: What am I doing at 3 o'clock?!

[et]

For any center angle given, within a disk, you can get the area of the circular sector, and the triangle contained, and the difference of the areas of the circular sector and the triangle (circular segment) is what you need here. The angle is 2*pi-2*arctan(1/2), and then the difference is calculated, and the rest is simple.

[birupakhyaroychowdhury]

We can also find out the area using definite integral....

[reznovvazileski]

This is an option, yet in calculus they let you take integral(y1(x)dx) + integral(y2(x)dx)) from 0 to (x at y1(x)=y2(x)) to x = r. Nice job teach :')

[justpaulo]

Almost the way I solved it. But I did it in a slightly easier way still.

- At 2:39, instead of drawing the diagonal line all the way down I stopped at the semicircle.
- Then I calculate the area of the red right angle triangle with base (2r/5) and height (r/5)
- For the remainder of the red area I consider the rectangle with height (r) and width (3r/5) to which I subtract the right angle triangle with base (3r/5) and height (4r/5) and also subtract the area of the circular sector.
- To find the circular sector angle I used the inscribed angle to central angle relationship (central = 2*inscribed). Theta will be π/2 - 2*atan(r/2r)

[jgkonig]

Once you solved it for r=4 (in your first video) it is easy to get the general solution. The area will always be proportional to r^2. For example: Area=kr^2+b. Now, it is easy to see that for r=0, Area=0. This means that b=0. Then taking in consideration that for r=4 the Area=1.252, then k=1.252/4^2=0.078. Then the general solution is Area=0.078*r^2

[hoangkimviet]

The Chinese math exercises challenge Chinese students with the very mathematical professional level :-0

[stephanforseilles]

"It's really great when geometry works so simply." I need a stiff drink.

[kshin]

I did it another way and got the answer: r^2 [2/5 - Pi/4 + tan-1(1/2)]
which gives the same value for the r^2 coefficient.

[johnchessant]

This is really easy. The area, of course, is a constant times r^2. Since you know the answer for r = 4, you can find the constant.

[commonmancrypto]

Imagine the problem as an 8x8 box. Draw another line from the bottom left corner to the middle of the top. Notice the little triangle you've made in the bottom left. Assume it has a flat bottom and calculate it's area....it's also import to note the length of it's shortest side. Now, because you know the size of the circles, you can determine what percentage of that big circle is under that little flat triangle line...with that information you can determine how much the arch would deviate from the flat line. Now you can calculate the area of the little triangle with the rounded bottom. So, 1/4 square minus 1/4 circle minus 1/4 little rounded triangle....and then divide by 2 for your answer.

[raiu]

Why should the purple shape (Circle part) be hard ?....

Purple Shape = Area of Halfcircle - Area of The Other Part of the Circle (dunno the specific name, looks like a Pie).

Purple = (r^2*pi)/2 - (pi*r^2*arctan(1/2))/90

[solarfluxman]

Your answer can be further simplified:
Area = r^2 (0.4 - arctan (1/3 ))

[DamienConcordel]

1 point and 1 alternative method:
Point: Remember that 3²+4²=5², so once you know that the intersection is at 2r/5, you can easily figure the coordinate by going from the center of the semicircle and noticing that as the point is 3r/5 away in one direction, it has to be 4r/5 away in the other.
Alternative method: I used the line going VERTICALLY through the intersection and therefore had a blue right triangle and a green rectangle, with two pieces to subtract: another right triangle with a simple area (see point above) and the circular sector, which I figured with arcsine instead of arctan.

[alienpioneer]

It is possible to do it pure geometrically in this particular case, if we cut the rectangle in half and play with areas .
The answer i got is 1.252 with the formula R^2/2*[0.5-alpha+0.5*sin(alpha)], where alpha = atan(2)-atan(1/2).

[bluerizlagirl]

Surely it does not matter what r is anyway, since the red shapes will always be similar however big or small you draw the whole thing, and so the area will always be equal to the same thing times r**2? If you double r, you quadruple the area; if you treble r, you increase the area ninefold, and so forth.