## The Final Cut

Here’s the 3rd ‘Unity in Duality‘ challenge!

By drawing one NON-CURVED line, can you separate the Yin/Yang into 4 sections of equal size, but different shape? (ignore the 2 eyes/spots)

Here’s the 3rd ‘Unity in Duality‘ challenge!

By drawing one NON-CURVED line, can you separate the Yin/Yang into 4 sections of equal size, but different shape? (ignore the 2 eyes/spots)

Shawn| PUZZLE GRANDMASTER | Profile April 17th, 2009 - 8:06 amSlanted line from top left to bottom right – shoot the yinyang between the eyes!

scottk| Profile April 17th, 2009 - 10:52 ami would try drawing the S shape, just opposite of the current one to form 2 circles and 2 shapes on the right and left that seem to be of similar size

RK| Founder | Profile April 17th, 2009 - 12:07 pmwording of problem changed, please see above

bizarette18| PUZZLE MASTER | Profile April 17th, 2009 - 2:30 pmFrom north-west to south-east?

fuzzy| Profile April 17th, 2009 - 3:08 pmInscribe the yin-yang into a square. Now, take the diagonal line that goes from top-left to bottom-right corners. This line cuts both shapes into equal-size parts, thus creating 4 shapes of equal size.

RK| Founder | Profile April 17th, 2009 - 3:51 pmwanted to add: Bonus Points to those who can prove their answer

rdsrds2120| Profile April 17th, 2009 - 4:00 pmI say that yes, it can. Here is why I think this.

Let’s say you draw a diameter, anywhere on the circle, and you were to turn it. Since the Yin Yang symbol’s two ‘fish’ are exactly the same, if you can make one side’s area completely in half, it would then hold true for the other half of the symbol. Using this knowledge, and the face that there are an infinite number angles within a circle, it then makes it possible to divide the area of one side in half with a radius, and thus you can complete the line by drawing the other half.

hex| PUZZLE MASTER | Profile April 17th, 2009 - 5:08 pmA STRAIGHT line at 45 degrees counterclockwise from the vertical will divide the Yin/Yang into 4 equal areas.

To arrive to this conclusion (and claim my bonus points ), let’s assume that the angle is alpha:

Ro = 2 x Ri

Area1 = alpha / 360 x Pi x Ro^2 = alpha / 360 x 4 x Pi x Ri^2

Area2 = Pi x Ri^2 / 2

Area3 = (90 – alpha) / 360 x Pi x Ro^2 = (90 – alpha) / 360 x 4 x Pi x Ri^2

Area4 = Pi x Ro^2 / 4 – Pi x Ri^2 / 2 = Pi x Ri^2 / 2

Area1 + Area2 = (4 x alpha / 360 + 1/2) x Pi x Ri^2

Area3 + Area4 = (4 x (90 – alpha) / 360 + 1/2) x Pi x Ri^2

Set Area1 + Area2 = Area3 + Area4

then alpha = 45 degrees

brianu| Profile April 17th, 2009 - 6:38 pma line from the center of the ying-yang which goes up and left (quad II)135 degrees and down and right (quad IV) 315 degrees.

Let the area of the ying-yang equal pi (a radius of 1).

Each piece, black and white then equals pi/2 because the area is equally divided.

So each new section should equal pi/4.

Using a radius line from the center of the ying-yang extending upwards 90 degrees, it creates a semicircle of area pi/8:

(pi*r^2)/2 = 0.25pi/2 = pi/8

Then we need an additional pi/8 to have the area of pi/4 previously noted.

The ying-yang as a whole can be divided into eight sections using radius lines which are set in increments of 45 degrees. there are four places where this can be taken as whole parts of the teardrop shapes, in quadrants II and IV.

So the sections of these quadrants which are attached to the semicircles described above create a dividing line as described in the opening.

APEX.JP| Profile April 17th, 2009 - 8:34 pmANSWER: Straight line bisecting the centrepoint rotated 45deg. anti-clockwise from vertical.

METHOD:

Break the symbol down to (4x) equal 90deg. quadrants.

Delineated by horizontal & vertical centrelines (crosshairs).

The Lower Right (and Upper Left) quadrants are ‘full’ quadrants. each containing a single geometric shape with area equal to 1/4 that of the area of the whole symbol.

The Lower Left (and Upper Right) quadrants are each comprised of (2x) geometric shapes.

being a semi-circle with radius 1/2 that of the radius of the whole symbol.

being the ‘tail’ or ‘wave’ shape.

The combined area of + (for each quadrant) = 1/4 the area of the whole symbol.

PROOF:

To find the area for the whole symbol use: A=Pi.r^2

To find the area for use: 0.5A=(Pi.r^2)/4

Therefore area of = 1/8 area of the whole symbol.

Therefore area of =

SUMMARY:

As the Upper Right & Lower Left quadrants each contain equal portions of area (by colour) the straight line bisector should be rotated to a position equi-distant between them.

bizarette18| PUZZLE MASTER | Profile April 17th, 2009 - 10:51 pmBy non-curved as opposed to straight, do you mean the line can zig-zag so all 4 shapes are different rather than 2 pairs?

RK| Founder | Profile April 17th, 2009 - 11:45 pmHi there Bizarette18- to clarify, the line should not zig-zag. It is straight.

aaronlau| Profile April 17th, 2009 - 11:47 pmA line 3/4 pi clockwise from the vertical passing thru the center, taking reference from the picture.

Equation proof:

Let A be the angle from vertical to required line

Let r be the radius of the “inner” circle

PI x r2 / 2 + 0.5 x (2r)2 x A = 0.5 x (2r)2 x (PI – A) – PI x r2 / 2

Solve for A = PI / 4

Ari| Profile April 18th, 2009 - 12:36 pmOkay, now the wording makes sense

A picture tells more than a thousand words and all that, so…

http://arikemppainen.com/misc/.....lution.jpg

Good one, cheers!

Jimmy Anders| PUZZLE MASTER | Profile April 18th, 2009 - 1:12 pmThe answer is a line (with negative slope) going through

the center at a 11.25 degree angle.

Draw a horizontal line through the center of the yin-yang.

Consider the top half. The area of the white side is greater

than the area of the black side by (1/16)*pi*r^2, where r is

the radius of the total yiu-yang. This is the size of the

circle created by going from the center of the black dot and

making a radius the distance to the curved boundary line,

which is circular. So, if the horizontal line is rotated

about the center so that (1/32)*pi*r^2 is taken from the white

side and added to the black side, then both will have the same

area by this argument:

White = Black + Extra =>

White – (1/2)Extra = Black + (1/2)Extra

Now, (1/32)*pi*r^2 is 1/32 of the area of the full yin-yang.

Therefore, the angle the horizontal must be rotated is 1/32 of

the way round the circle, which is [360 degrees / 32] = 11.25

degrees. Since we want to take from the white and add to the

black, this must by 11.25 degrees clock-wise. Because of the

rotational symmetry of the yin-yang, this will also make the

bottom quarters equal.

suineg| PUZZLE MASTER | Profile April 19th, 2009 - 9:30 amcool, I think the line has to be a diagonal that pass for the centre of the big circle, and also has to be in higher point of the circle in relation with the diameter in the part where is missing the little circles. So it goes from the picture left up—> right down like this \ but with less inclination jajaja. Now the demostration ok this is hard:

you have to find and angle B that fulfill this condition:

Area of the circlar sector= 1/2* Area of the little circle

where: r–> radius of the big circle

Area of the circular sector: (pi* square(r)* square(B)) /360

Area of the little circle: pi*square(1/2*r)–> 1/4* pi square(r)

so when you put both equations into the condition you get:

square(B)=360/4–> square(B)=90–> B= 9,486832 degrees its the angle of the circular sectors cool, I think this is the demostration, hope that, coolkit

suineg| PUZZLE MASTER | Profile April 19th, 2009 - 9:35 amups sorry my bad and error in my last demostration has to be like half the area of the little circle jajaj ok so:

B= squareroot(360/8)–> squareroot(45) —> B= 6,7082 degrees ok, sorry but I am distracted always, cool

alexc| Profile April 19th, 2009 - 6:14 pmbecause the shape is symmetrical it is only necessary to focus on one colour, I’ll chose the white section.

Starting from the center of the yin yang, draw a line going to the top left of the circle, the angle this make with the vertical axis = x. Now it is just a matter of stating the remaining sectional areas.

The radius of the yin yang is r, hence the radius of the semi circles are r/2.

The area of white above the split line is:

X/360 * Pi * r^2 + Pi * r^2 /8

Area below the split line

(180-X)/360 * Pi * r^2 – Pi * r^2 /8

These two secions equal each other and the Pi*r^2 cancels out. Solve for x

We then get x = 45 degrees

Mashplum| PUZZLE MASTER | Profile April 19th, 2009 - 7:07 pmI think the line should be the function f(x) = -x, if center of the figure is at (0,0) on the coordinate plane. No proof yet.

KMESON| Profile April 19th, 2009 - 11:37 pmDraw the diameter that is at 45 degrees to the tangent of the curve at the center of the circle (y=-x in the picture above).

Just equate the two areas: r^2/8 + th r^2/2 == (pi-th)r^2/2 – r^2/8

Then th = pi/4 => 45 degrees

Shawn| PUZZLE GRANDMASTER | Profile April 20th, 2009 - 7:54 amWouldn’t let me get away with “Sure, just draw a line,” huh?

Shawn| PUZZLE GRANDMASTER | Profile April 20th, 2009 - 11:02 amFirst off, all 4 sections will not be of “different shape.” There will be 2 pairs of matching shapes, one black and one white.

Draw a line at a 45 degree angle that passes through the centerpoint of the large circle, starting in the upper left quadrant and ending in the lower right.

This is hard to describe without a drawing, but I’ll give it a go:

Imagine that the semicircle the surrounds the dot in each shape is continued to form a complete circle. You would then have a white circle in the top half of the drawing and a black circle in the bottom. Because these circles meet in the middle, their diameters are 1/2 of the large circle, and therefore their areas are exactly 1/4 of the area of the large circle. This means that the remaining area of each color is also equal to 1/4 of the area of the large circle. I’ll focus on the white “yin.”

Because the overall shape is symmetrical, the line that we draw must pass through the centerpoint of the large circle.

It is easy to see that by drawing a line that is tangent to the small circle, we will end up with unequal areas. We will have the small circle plus the bits that are chopped off by the tangent line and remain on the “small circle side” of the tangent.

The correct line will lop off a bit of the small circle, which will be compensated for by the addition of a bit of area between the small and large circle in the extreme upper left quadrant.

D = diameter of large circle

R = radius of large circle (= diameter of little circle)

R/2 = radius of little circle

SO, by drawing a line through the center of the large circle at 45 degrees from top left to bottom right, we can calculate that the bit of the circle that is lopped off is [R^2*(pi-2)]/16. We do this by dividing the small white circle into 4 pieces, and calculating the area of one of those pieces, specifically the lower left piece through which the 45 degree line passes ([pi*(R/2)^2]/4). We then subtract out the triangle formed by the 45 degree line (0.5*(R/2)^2).

Now, all we have to do is prove that the bit of the white shape that is included with the small circle in the upper left quadrant is equal to [R^2*(pi-2)]/16, the amount that has been lopped off. If true, this would make the total area of the shape equal to the total area of the small circle, which we have already demonstrated is exactly 1/4 of the total area of the large circle, which is exactly what we are looking for.

Because the angle is 45, the area of the slice of pie formed by the line and the vertical is 1/8 of the area of the large circle, or pi*R^2/8. The section of the this slice of pie that is inhabited by the small circle is then equal to the area of a small semicircle minus the lopped off bit, or [pi*(R/2)^2]/2 – [R^2*(pi-2)]/16 = [R^2*(pi+2)]/16

Therefore, the additional white shape in the upper left has an area equal to the total large slice of pie (pi*R^2/8) less the portion of the slice of pie that is inhabited by the small circle ([R^2*(pi+2)]/16.

pi*R^2/8 – [R^2*(pi+2)]/16 = [R^2*(pi-2)]/16

This proves that the area of the additional white shape in the upper left quadrant is indeed equal to the area of the lopped off bit of the small circle, and thus cancel each other out. The total area of the white shape thus created by the 45 degree line is pi*(R/2)^2, and so the remaining cornucopia-shaped white section must also have this area.

Given the symmetry of the drawing, it is easy to show that the black shape reacts in the same manner, and that 4 sections of equal area, two black and two white, will be produced.

diegote| Profile April 20th, 2009 - 12:36 pmI think the answer is no.

You can have equal area but with 2 pairs (same shape) of figures using a straight line that passes trough the center and 45° counterclockwise. Using a line out of the center does not allow me to have 4 figures with equal area.

diegote| Profile April 20th, 2009 - 12:54 pmTo reenforce my previous enty:

I think a correct solution would derive in this 2 situations:

1) The line passes trough the center: It permits us to overlap the 2 parts it divides, and they should match.

There’s only one cut for equal areas, this corresponds to 4 figures with equal area but only 2 different shapes. Any other angle of this line gives different areas. Discarded.

2) The line does not pass through the center: The little portion must be 1/4 of the total area and the remaining must show the other 3 pieces.

The only way to have a little portion with only one piece is a vertical line at some exact measure, but it leaves a big part with a complete yang (or ying) and one or two pieces (depending on the measure I didn’t calculate) that have obviously different area. Discarded.

Awainting for comments.

bilbao| Profile April 21st, 2009 - 9:29 amI have sent 3 images to RK to support my explanations:

The easiest way to get different shapes/equal area would be via a straight line at 45º as shown in image1.jpg.

But this way you don´t get 4 different shapes. Instead you get 2 + 2.

So I tried to get 4 all different shapes/equal area. The simplest way would be as shown in image2.jpg.

But this solution may not be considered ONE line, but a composition of two lines, although it may be drawn without raising the hand.

Finally I decided to find ONE line (that could be expressed as a continuous and differentiable function) that divides the yinyang in 4 all different sections. My starting point was image1.jpg. After some reasoning and after I tried a few functions I concluded that it had to be simmetrical to axis y-y, it had to cross (0,0), it had to show a minimum/maximum at (0,0) and that areas left under and above axis x-x should be equal (A). The best function I have found that follows these rules is shown in image3.jpg. It is y=ax*sin(bx). Still I need to find a and b and valuate the areas. This is as far as I got with this puzzle. I am trying to get a and b, but it will take me some time.

bilbao| Profile April 21st, 2009 - 9:33 amI send to RK a new image with my results:

These are 3 real solutions for ‘image3 case’. (calculated with maple)

First one is x*sin(x)

Second one is 1/2*x*sin(x)

Third one is 1/4*x*sin(x)

There are more solutions by varying the fraction (1/8, 1/16,…. The smaller the fraction the closer the function comes to a straight line.

Once I established the function, for each case I had to find the radius for the yinyang circle that would satisfy that areas below and above x axis are equal. To do that I calculated areas via integration.

RK| Founder | Profile April 21st, 2009 - 11:02 amOk- quite a few bonus points to dole out this am : )

Gunslinging Shawn has a really nice explanation, and Hex, Brianu, AlexC, APEX.JP, Aaronlau, Jimmy Anders, Suineg, Ari (good picture to help) and Bilbao got the proof.

Admittedly, the question could have been interpreted as all 4 pieces had to be different in shape from each other. Diegote picked up on this, as well as Bilbao. This can be accomplished with a curved line but not a straight one. For more info on this approach, you’ll want to read Bilbao’s explanation and refer to the images he sent (below). .

Great job otherwise!

RK| Founder | Profile April 21st, 2009 - 2:38 pmHere are Bilbao’s images:

http://smart-kit.com/wp-conten.....iginal.jpg

http://smart-kit.com/wp-conten.....inyang.jpg

http://smart-kit.com/wp-conten.....inyang.jpg

http://smart-kit.com/wp-conten.....yang-4.jpg