## Unity in Duality

The concept of **Yin and Yang** are used to describe how seemingly disjunct or opposing forces are interconnected and interdependent in the natural world, giving rise to each other in turn (read more).

Our 1st math question involving the great symbol, however, deals with **area**.

Which has the greater area, the inner circle containing the Yin Yang, or the outer ring?

*The 2nd and 3rd questions are harder, and will follow over the next 5-10 days!*

Mashplum| PUZZLE MASTER | Profile April 12th, 2009 - 10:25 pmThey seem to be the same. The figure is not labeled, but if the inner radius is 0.5^0.5 times the outer radius (which it probably is) then the areas are the same.

Area of ring = pi*r^2 – pi*(0.5^0.5r)^2

= pi(r^2 – 0.5r^2)

= pi(0.5r^2)

Area of yinyang = pi(0.5r^2)

suineg| PUZZLE MASTER | Profile April 12th, 2009 - 10:29 pmI think it depends: ric= radius inner circle roc= radius outer circle

the area of the inner circle is Aic= pi* square(ric)

the area of the outer circle is Aoc= pi* square(roc)

so the area of the outer ring would be Aor= Aoc – Aic–> pi* (square(roc)-square(ric))

so case1: square(roc)-square(ric)> square(ric) Aor> Aic

case2: square(roc)-square(ric)= square(ric) Aor= Aic

case3: square(roc)-square(ric)< square(ric) Aor< Aic

but graphically you see that the difference in the radius is not much, not even close to double up the ratio so you have:

roc1,414214ric then the area of the outer ring is greater than the area of the ying yang circle and viceversa, cool

Falwan| Profile April 12th, 2009 - 11:00 pmouter ring area > inner circle area.

It seems that the radius of the inner circle is twise the width of the ring.

LadyInsomnia| Profile April 13th, 2009 - 2:05 amSurprisingly, the outer ring has the greater area. Just estimating from the picture, the inner circle seems to have a radius about twice the width of the ring, so the inner radius is 2 units and the total radius is 3 units. The area of the inner circle is pi*(2^2), about 12.6. The area of the ring is pi*(3^2)-pi*(2^2), about 15.7.

APEX.JP| Profile April 13th, 2009 - 2:10 amMeasure diameter 2x circles using pixels

Calculate area of 2x circles. 66052. 32685

Subtraction for outer. 33367

ANSWER: Outer ring has greater area. By approx 2%

Shawn| PUZZLE GRANDMASTER | Profile April 13th, 2009 - 9:24 amThese look awfully close to being equal.

For the outer ring area to equal the inner circle area, you need the outer diameter squared to equal two times the inner diameter squared (OD^2 = 2 * ID^2)

When I print out the figure and measure, I get OD=75mm and ID=53mm, so:

OD^2 = 5625mm^2

2*ID^2 = 5618mm^2

I don’t know how accurate my hand measurements are – sometimes it looks like the ID is a wee little bit larger than 53mm, which would make the inner circle’s area the bigger of the two.

I am going to stick with my gut and say that the two areas are equal.

fuzzy| Profile April 13th, 2009 - 10:18 amThey are just about equal. It is very difficult to tell the exact ratio of radii of the two circles (the inner one and the big one, that contains the inner one and the ring). If we assume the inner circle has radius 1, its area is equal to pi. The big circle looks like its radius is about 1.4, and the area is 1.96pi. Thus the ring is 0.96pi, a little less than the inner cirlce. However, this is only an approximation.

aaronlau| Profile April 13th, 2009 - 11:25 amR: radius of outer circle

r: radius of inner circle

R = sqrt(2) * r , Area equal

R > sqrt(2) * r , Area of outer ring greater

R < sqrt(2) * r , Area of inner yin yang circle greater

Or is this question not algebra and we actually have to physically measure it off the screen to determine which area is greater.

RK| Founder | Profile April 13th, 2009 - 12:49 pmThere may be more than 1 correct answer, depending on how you approached the problem…

suineg| PUZZLE MASTER | Profile April 13th, 2009 - 6:23 pmSo by my first explanation I would go for the outer ring because it “seems” that the outer circle has more than 1.5 time the inner circle radius. But another “out of the box” answer could be if white means “empty” and black means “full” then the area of the inner circle is greater than the area of the outer ring because the outer ring would have “0” of area while the inner circle would have 1/2 square(ric), cool cool very cool, or maybe not jajaja cool

alexc| Profile April 13th, 2009 - 7:39 pmI think I’m over complicating a simple problem but I think i get the right answer given the nature of the problem.

No dimensions are given so there can be no definite answer.

I may print out the picture a different size to someone else so my answer may vary from someone elses, however the ratio of the radius of the inner and outer circle will remain constant.

And obviously the sizes or the ring and circle dont linearly change with a change in radius as the area of a circle is Pi x r^2

Hence, if the ratio of the radius of the ring to the radius of the yin yang stay constant there is a point of equilibrium where the area of the circle and area of the ring are equal. When the ratio of the radius’ is below this and above zero, the circle is bigger. When the ratio is above this up to infinity, the are of ring is bigger.

Since there is infinit number of possible radius ratios above the equilibirum point, the ring can be technically be infinitly larger than the yin yang symbol, however, as the ratio approaches zero the circle can be infinatly larger than the ring (depending how close you get to zero)

So technically the ring can be infinatly larger than the circle, and the circle can be infinitly larger than the ring.

Therefore they have an equal area?

michaelc| Profile April 13th, 2009 - 8:14 pmMy method was to measure with a ruler across my computer screen! Highly scientific I must say!

From my screen it looks like one was about 6.2mm and the other was about 8.75 mm or a ratio of the areas (outer circle area/inner circle area) of about 1.99 by the measurement. In this way the inner circle is slightly more area than the outer circle.

Knowing how symmetrical and balanced the Yin Yang symbol is however, I’d have to intuitively think that was a measurement error, and if I had a precise enough ruler I’d see clearly that the ratio of the areas was 2:1. Or the area of the outer ring is exactly equal to the area of the inner circle. If you let the inner diameter be 1, the outer diameter would measure the square root of 2, and the 2 areas would be in balance.

Someone| Profile April 13th, 2009 - 10:18 pmThe very way this question is worded implies a trick question. Since they can’t both be smaller, the answer is neither, they are the same size.

For both areas to be equal, the outer circles area would have to be twice as much, and therefor have a radius square root of 2 times or about 1.4

1.4 looks about right

bilbao| Profile April 14th, 2009 - 12:56 amradius of inner circle: r

thickness of outer ring: t

When t = r*(sqrt(2)-1), that is t=r*0.4142 both areas are equal.

I printed the image, measured it and got an approx. proportion of t=r*0.3913.

So I think the inner circle’s area is bigger than the outer ring’s.

hex| PUZZLE MASTER | Profile April 14th, 2009 - 5:19 amAccording to the limited available info:

Area of inner circle: Ai = 2 x Pi x Ri^2

Area of outer circle: Ao = 2 x Pi x Ro^2

Area of outer ring = Ao – Ai

Suppose Ao-Ai > Ai

then Ao – 2 x Ai > 0

then 2 x Pi x (Ro^2 – 2 x Ri^2) > 0

then 2 x Pi x (Ro + sqrt(2) x Ri) x (Ro – sqrt(2) x Ri) > 0

then Ro > sqrt(2) x Ri

Therefore if Ro > sqrt(2) x Ri, then area of outer ring will be greater than the area of the inner circle.

Hendy| Profile April 14th, 2009 - 12:15 pmThe area of the inner circle is obviously pi * (1/2 d1) ^2. The outer ring is pi * (1/2 d2) ^2 – pi * (1/2 d1) ^2. However, I cannot tell what length d1 or d2 are because every monitor and resolution setting has different dimensions. In theory, one might still come to the conclusion of which is larger based upon ratio of areas rather than on actual areas. On my monitor, d1 is approximately 2 1/4 inches and d2 is approximately 3 3/16 inches. Therefore, using handy-dandy Excel, I get the circle as 3 41/42 in^2 and the ring as 4 in^2. Thus, the ring has the larger relative area.

rdsrds2120| Profile April 14th, 2009 - 5:40 pmAssuming that the picture is drawn to scale, the radius of the larger circle is approximately 1.5x the size of the smaller circle. If we say the smaller one is 2 units, and the bigger circle 3, we find the area of the smaller circle to be 4?, and the bigger to be 9?.

9? – 4? = 5?(Area of outer ring)

5? > 4?: Therefore, the outer ring is larger in area.

RK| Founder | Profile April 14th, 2009 - 10:16 pmDepending on how you measured, the answer may vary. Hex, Mashplum, Aaronlau, Suineg, AlexC, Hendy, and Bilbao very nicely show the concept & equations for solving.

Falwan, LadyInsomnia, Bilbao and APEX.JP (and several others) used the equations and came up with their answers by measuring (with ruler or pixel count).

However, in keeping with the spirit of the question

Shawn, Fuzzy, Someone, and MichaelC correctly surmised what the puzzle was trying to (but didn’t exactly) get at: the 2 areas are equal

RK| Founder | Profile April 14th, 2009 - 10:29 pmwill put the 2nd part up in couple minutes

suineg| PUZZLE MASTER | Profile April 15th, 2009 - 8:47 pmRK, cool that teh areas are the same, but Ying Yang its also about complementary pieces, the emptyness with the fullness, the good and the bad, the 1 and the 0 so if black is 1 and white is 0 then the inner circle has the greatest area coooooooool, sorry for getting stuck with this problem but I like the possibility of many answers.