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        riddles >> easy >> Triangles, missing area
        
(Message started by: tommy cha cha on Jul 24th, 2002, 1:16am)
Title: Triangles, missing area
Post by tommy cha cha on Jul 24th, 2002, 1:16am
http://www.ocf.berkeley.edu/~wwu/images/riddles/triangle_area.gif


// 11/7/2002 4:20PM IMAGE REPRINTED HERE BY SITE ADMIN FOR YOUR CONVENIENCE

Hint: these triangles are not only different in size... they also differ also in shape.
Title: Re: Triangles, missing area
Post by D. Spig on Jul 25th, 2002, 8:33pm
Here is an even bigger hint.  They aren't triangles!  They have four sides, not three.

If you want to confirm this, check the slopes of the blue triangle verse the red triangle:

Red slope: 2/5
Blue slope: 3/8

These aren't equivalent.

This is a "hard" question only because the question lies by saying they are triangles.  Perhaps shape would be better.
Title: Re: Triangles, missing area
Post by Magneeto on Jul 26th, 2002, 1:37am
This one had me questioning the laws of math and physics(not to mention my sanity). I even drew them on graph paper and cut them out and made them fit.

A: The second triangle is larger, but only slightly. It is about 1/10 to 2/10 of a block larger on one or both sides.  If one assumes the first is 5 by 13, then the second would only have to be 5.15 by 13.15 in order to account for the extra block.  
using:
1\2 b * h = .5 * 5 * 13 = 32.5
in order to get 1 extra block 33.5
.5 * 5.15 * 13.15 = 33.86.
Yes they differ in size, but no, they do not differ in shape.  They are stated to be triangles, and so despite the pixelation I  treated them like triangles.  I ain't no math major but I don't know of any rule that says the slopes of the sub-triangles must match in order to be considered a triangle. That's weird though, you would think they would have to.  But, I can think of many examples in which the sub-triangles wouldn't match.
Still, I didn't like this question. Normally, questions like this expect you to disregard any imperfections in the picture, but this one uses those imperfections to fool you.  The key is realizing that it only takes a minor change the base and/or height to change the area by a whole block.  :P
Title: Re: Triangles, missing area
Post by l. abulafia on Jul 26th, 2002, 1:49am
Yes, Magneeto, they differ in shape.

You may have gotten the problem to work on graph paper by altering b & h, but that is not how -this- puzzle pulled the feat off.  Here, base and height are the two sides that -are- symmetrical.

It's not quite accurate to say they have four sides, either, though on a low-resolution monitor it may appear there are.

If you wish to solve -this- problem, reconsider tommy cha cha's original advice.

Or, consider these (spoiling) terms: xevnoc and evacnoc.
Title: Re: Triangles, missing area
Post by Mahender Akula on Jul 26th, 2002, 8:45am
The Blue and red square area are equal, because of dimensions.

I considered yellow and green as single entitiy (as rectangle)

So in the top picture the area comes out to be 3 x 5 = 15

In the bottom picture the area comes out to be 2 x 8 = 16

which includes the blank square.  If you take blank square out

then it is 15.

Not sure if this is the solution.

Mahender
Title: Re: Triangles, missing area
Post by hakaizou on Jul 26th, 2002, 1:11pm
D. Spig was right the first time: the shapes are NOT triangles to begin with, since the slopes of the hypotenuses of the blue and red triangles DON'T MATCH. It's an optical illusion, guys: a four-sided figure is made to look three sided.

Magneeto, the change is neither in the base nor the height, it's in the supposed "hypotenuse." See l. abulafia's hint. The figures were laid out on the grid to show that neither the "base" nor the "height" changed.
Title: Re: Triangles, missing area
Post by Vineet Kumar on Jul 26th, 2002, 1:47pm
Yes, they do have 4 sides each. They are not triangles. The question lies.

Math major or no, surely you can see that a line with 2 different slopes is not a line. If you start out in one direction, and end up in another direction, somewhere in between, there's a point where you turned. That point is the vertex between the 2 edges, and it's folly to consider it one big edge, no matter how small (but nonzero) the angle between them is. There is a vertex on what is given as the hypotenuse of each of the large "triangles". Consider the extreme case: the opposite vertex. Surely you wouldn't consider the bottom side and the right side to be the same side -- there's a vertex there joining 2 edges with different slopes. If they had the same slope, they'd be the same side (and there'd be no vertex).
Title: Re: Triangles, missing area
Post by S Spigarelli on Jul 26th, 2002, 3:32pm
Sounds like D. Spig is right.  The riddle is a big lie.  It confused me until I realized that it was a lie, not a really fair riddle.
Title: Re: Triangles, missing area
Post by william wu on Jul 29th, 2002, 7:50pm
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your common sense."

- Buddha
Title: Re: Triangles, missing area
Post by Brian Lee on Nov 7th, 2002, 2:53pm

on 07/29/02 at 19:50:52, william wu wrote:
"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your common sense."

- Buddha


Heh, you love that quote dontcha Wu? Well, it's definitely a good one... Yea I've seen this deceitful picture before on another forum... I heard its from a subway billboard somewhere in deutsche-land or something...
Title: Re: Triangles, missing area
Post by wowbagger on Nov 8th, 2002, 2:32am
Really? I haven't seen that picture IRL here in Germany ("Deutschland") yet. And yes, the quote is one you should always be aware of.


on 07/26/02 at 15:32:46, S Spigarelli wrote:
The riddle is a big lie.  It confused me until I realized that it was a lie, not a really fair riddle.

I wouldn't go that far. Of course you can argue it's a lie to state the problem using "triangle" instead of "figure" or some other term. But it wouldn't be as puzzling then, would it?

Anyway, this is really a great riddle. A friend of mine who is a teacher once posed it to his class using an overhead projector and pieces cut from a transparency. It was quite a success...
Title: Re: Triangles, missing area
Post by CrisisAverted on Dec 5th, 2002, 8:56pm
I don't feel like bothering pulling out a ruler or doing sines, cosines, and tangents, so I'll fall back on the whole is equal to the sum of its parts.  The area of each colored piece is the same in each triangle.  A two dimensional shape only has the area of its component pieces, any shape with these pieces would always have the same area, provided there was no overlap.  Compare shape one with shape two piece by piece and you can conclude that whatever shape one is, its not the same as shape two.
Title: Re: Triangles, missing area
Post by i_m_n_idiot on Jan 4th, 2003, 10:13pm
hey i saw this iin a train
ppl from my country call it mrt or mtr
Title: Re: Triangles, missing area
Post by THUDandBLUNDER on May 3rd, 2003, 7:06pm
There is a more general explanation of this apparent paradox in terms of Fibonacci numbers,

which are generated by the following recursive equation:

F(n+2) = F(n+1) + F(n), F(1) = F(2) = 1

producing the series 1,1,2,3,5,8,13,21,34,55,89............


http://www.ocf.berkeley.edu/~wwu/images/riddles/triangle_area.gif

Let the red triangle have horizontal and vertical sides of length F(n+1) and F(n-1), respectively.
Let the blue triangle have horizontal and vertical sides of length F(n+2) and F(n), respectively.

So that
in the upper figure the two triangles create a rectangle with an area of F(n)F(n+1)
and
in the lower figure the two triangles create a rectangle with an area of F(n-1)F(n+2)

Now consider the following Fibonacci identity:

F(n)F(n+1) - F(n-1)F(n+2) = +/- 1

Thus, the two rectangles are bound to differ in area by exactly one unit.

As n increases, the slopes of the hypotenuses of both triangles tend to the square of the Golden Ratio.
(phi2 = 0.3819660112...)
2/5 = 0.4 --> phi2 from above.
3/8 = 0.375 --> phi2 from below.
Title: Re: Triangles, missing area
Post by SodiumPenathol on Jan 18th, 2005, 12:20pm
If you realize that the grid isn't made up of perfect squares it seems instantly obvious that it's an optical illusion.  Just print the picture and try to fold it so that the opposite corners of the grid overlap each other perfectly.  Then try to crease the paper smoothly along the fold.  It can't be done without ripping the paper.  Actually, doing this reveals the true ratio of the sides of the 'squares' as 13:14 or 14:15 (forgot).  That's why the quote was inserted at the end.
Title: Re: Triangles, missing area
Post by THUDandBLUNDER on Jan 18th, 2005, 12:41pm

Quote:
If you realize that the grid isn't made up of perfect squares

Yes, it is.
At least, they are supposed to be perfect squares.

Title: Re: Triangles, missing area
Post by Icarus on Jan 18th, 2005, 3:16pm
Any vertical vs horizontal distortion you see is a result of the medium, not intrinsic to the puzzle. The squares are only added as a visual reference to make it easier to compare the two figures. You  can reconstruct the puzzle on a grid you know to be square, and you will see that the puzzle still remains. The missing region is still there.
Title: Re: Triangles, missing area
Post by Grimbal on Jan 19th, 2005, 3:01am
::)
Title: Re: Triangles, missing area
Post by rmsgrey on Jan 19th, 2005, 8:28am

on 01/18/05 at 12:20:57, SodiumPenathol wrote:
If you realize that the grid isn't made up of perfect squares it seems instantly obvious that it's an optical illusion.

As long as both grids are the same shape, it doesn't matter what that precise shape is - though the picture is easier to follow if the lines are still straight.

Since there's no rotation of the component pieces, they're clearly the same shapes as long as all the individual cells are equivalent.

As Grimbal's post shows, superposing the two situations makes it clear both that the two grids are identical, and where the difference between the two shapes arises.
Title: Re: Triangles, missing area
Post by OnlyAnEgg on Mar 8th, 2005, 9:41am
 Couldn't read Grimbal's post, but my approach was more like Magneeto's:  when I added the component parts of "Triangle"1, they are A=(3x5) + (8x3)/2 + (5x2)/2 = 32.0 squares.   But the entire figure gets A= (5 x 13)/2 = 32.5 squares.  Chalk up the discrepancy to inaccurate figures, or optical illusions, I don't think it matters.  My high school geometry teacher (Ed Cavey, SJ) always insisted that the proof cannot depend on the drawing, only the logic.  So the areas of the two figures agree, and there is one square difference between them.  32+1=33<>32.5
Title: Re: Triangles, missing area
Post by THUDandBLUNDER on Mar 8th, 2005, 10:09am

Quote:
Chalk up the discrepancy to inaccurate figures, or optical illusions

No, chalk it up to assuming that the 'entire figure' is a triangle, rather than a quadrilateral.


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