Inscribed Circle and Square Problem

This is a fun little geometry problem inspired by a suggestion from Yossi Rozhenko. If the red area equals the blue area, what is the value of: (large circle's radius)/(small circle's radius)?

Small correction: At 3:07 I should have said "4 minus pi."

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[MonkeySuperGamer]

3:07 bit of a voice-over mistake Presh, nice video nonetheless.

[marczeslaw]

If I weren't drunk, I could do it. Maybe. This is such a good question with not a boring aswer. Good job man!

[mi_]

First math problem solved in 2018, thx Presh

[stevethecatcouch]

For those who found this one too simple.
Circumscribe the large square. Add a larger circle and it's circumscribed square to repeat the diagram with the larger square taking the place of the smaller one. (You will now have a square, its inscribed circle, a smaller concentric circle and its inscribed square, the inscribed circle of that square, a smaller concentric circle, and finally its inscribed square. The regions between squares and their associated circles meet the conditions of the original puzzle.
Now repeat the recursion, creating an infinite sequence of circles and squares.
If the smallest circle has radius r_0, the next largest r_1, etc. What is r, the limit of r_n as n goes to infinity?

[dannyleung]

In the calculation of the area of the small square, you don't have to use the Pythagoras theorem. You just need to consider the two triangles with base 2r and height r then the area of the small square is simply 2*1/2*2r*r, which is 2r^2.

[MsAcpaul]

Happy New Year! A simplier way to do the smaller squares area without calculating the sides would be to remember that a square is a special type and has the same properties as a kite and a rhombus. These two shapes have an area formula calculated via the diagonals of the square. If the diagonals are labeled x and y then the area would be 1/2xy. Making no need to calculate the side length.

Those peeps that claim these videos are getting easier.... well from an adult tertiary maths sense they are. But for highschool problem solving building they are a great resource and teach kids (and teachers) new ways to solve problems without the tears.

[curtellingboe]

My problem set up and solution matched the video, not counting the errors made on the first attempt.

[Alasius]

A fun thing to notice is that you can inscribe 2 more circles with properties as follows:
(we denote the original squares and circle with 'inner' and 'outer' respectively)

circle with radius r0, such that: (inner square area) - pi*r0²=pi*(inner circle radius)² - (inner square area)
circle with radius r1, such that: pi*r1² = pi*(inner circle radius)² - (inner square area)

Notice how the two new circles have been defined using the same area that was used to define the inner and outer circle.

It follows that r1/r0 = (outer circle radius) / (inner circle radius)

The two new circles therefor have the same ratio as the inner and outer circle. Sadly both new circles aren't big enough to touch the inner square.

[prealgandhi]

This is the most easy question you posted.
Thanks for your videos and happy new year

[kamoroso]

Kind of interesting, you can see π has to be between 2 and 4 because of the areas of the red and blue areas, otherwise they'd have negative areas.

[stevensimpson]

Actually got another one right. Enjoy the channel. Keep them coming.

[devondevon]

1.154 or 1/sqrt 3/4 .
let the side of the large square = 2 then the area = 4 (28/7), so the radius of the circle inside is 1, so the area of the circle (large circle) is 22/7 (with radius =1), so the shaded area is the difference between square and the circle 28/7 - 22/7 =6/7. Since the shaded area of the square inscribed in the same circle is also 6/7, then difference between the two is R- S (where R is the area of the small circle and S in the area of the small square) . So R- S = 6/7. so what is the relationship between R and S? Actually S is 7/11 of R which is determine by letting the diameter of the small square = 2 then side of the small square is the square root of 2 hence the area of the small square is 2: note that radius of the circle around that small square is 1 (since the diameter is set at 2) hence the area of the circle is 22/7, so the ratio of the Area of the square to the small circle is 2/22/7 or 2 x 7/22 = 14/22 or 7/11 . So the area of the small square is 7/11 of the small circle it is inscribed in. So if R-S =6/7 (see above) then S= 7/11 R: solving for R get 11 R/11- 7/11 R= 6/7 = 4 R/11=6/7 , R =66/28 or 33/14, so the 33/14 is the Area of the small circle and 22/7 is the area of the big circle (see above), but the question didn't ask for the ratio of their area it asks for the ratio of their radius. So the big circle Area is 22/7 hence its radius is 1, . Area of the small circle is 33/14, so what is the radius of this circle? Radius square is 33/14 divided by 22/7 divided = 3/4 , so pie for this circle is 22/7 and r^2=3/4, therefore r= the square root of 3/4 or square root of 0.75 which is 0.8660254. so the radius of the small circle is 0.8660254 and again the Radius of the BIG CIRCLE is 1 and the Radius of the small circle is 0.8660254. The question of this problem is what is the ratio of the Radius of the BIG circle to the SMALL circle ANSWER= 1 / 0.8660254 = 1.154 or 1 over the square of 3/4

[deepakdas]

i solved it presh. thankyou for such a good question

[sanseng]

Much easier than your usual puzzles, new year gift may be!

[aishwarychaudhary]

Sir its answer √2/1 because if we take small circle radius as r then we can determine √2 r is length of side of small square and this is the length of large circles radius by symmetry

[sivarampavan]

First math problem I solved under 1 minute in 2018. Thanks

[nikstopics]

solved it just like Presh, I was thinking about the ''not to scale thing'', but i thought the only way that the problem could be different was if the the circles were translated somehow, but it doesn't effect the problem and happy 2018 everyone :P

[thewaytruthandlife]

S1 = the small square S2 = large now for pictural purposes tilt the innner circle and S1 45 degrees so that the red and blue sections are lined up across each other. Let the centre of the cicles be the XY planes (0, 0) then regard the upper right corner under investigation:

the area of that square is R2*R2 (that is the radius of the big circle)
the area of the circel section is then 1/4 * PI *R2*R2
the area of the red shaded upper right corner is then R2*R2 - 1/4 * PI*R2*R2

The area of the blue upper right corner is then 1/4 * PI * R1*R1 - 1/4* The inner smal square.
the area of the inner small sqauare is SQRT( R1*R1 + R1*R1) = SQRT(2*R1*R1) = R1 * SQRT(2)
so

The area of the blue upper right corner is then 1/4 * PI * R1*R1 - 1/4* R1*SQRT(2) *R1SQRT(2) =
1/4 *PI*R1*R1 - 1/4 * 2 R1*R1 = 1/4 *PI*R1*R1 - 1/2* R1*R1

working this out; I came out on


R2/R1 = SQRT ((PI/4 - 1/2) / (1 - PI/4) ) = 1.153211

lets see if thats

[chrisclews]

Happy New Year my friend, I was rubbish at maths at school. I love trying to solve your problems, it takes me day's, lol. Keep up the good work. From a old person from England.

[chinareds]

I'll use R for large radius, r for small radius. Red area is easy, the square has length 2R, so the red area is (2R)^2-pi*R^2= (R^2)*(4-pi). Blue area, square has a diagonal of 2r (45-45-90 triangle), so side length is r*sqrt(2), so the blue area is Set those equal to each other, solve for R/r and you get sqrt((pi-2)/(4-pi))