The unsolvable math problem

TzarconTzarcon Join Date: 2002-02-28 Member: 259Members
<div class="IPBDescription">At least for me</div>Is there anyone here that can solve this?

Comments

  • LegionnairedLegionnaired Join Date: 2002-04-30 Member: 552Members, Constellation
    Ok, discard the triangles, and focus on the rectangles only.

    In the first one, it is a 3x5 block. 3 times 5 = 15 square units.

    In the next one, it is a 8x2 block. 8x2=16

    16-15= a 1 square unit hole.

    I own you.
  • SkriftSkrift Join Date: 2002-09-25 Member: 1329Members
    That´s easy! The answer is 8. Obviously. O_o
  • RhoadsToNowhereRhoadsToNowhere i r 8 Join Date: 2002-01-24 Member: 33Members
    You'll notice that the two triangles don't have exactly the same slope.  The big one has a slope of 3/8 (.375), but the small one has a slope of 2/5 (.4).  I've attached a picture with a straight line drawn along where you are expecting the whole line to be, and you can clearly see that the inward bulge on the top picture and the outward bulge on the bottom.  The difference between these two bulges adds up to one unit.  QED.
  • LegionnairedLegionnaired Join Date: 2002-04-30 Member: 552Members, Constellation
    Rhoads.. the difference in slope is irrelevent, since the area of each individual partition is the same. The slope of the hypotenuse of both triangles is still 13/5. The question is, how can a unit simply fall out of the problem.

    There are a total of 15 blocks in the rectangular section in the first triangle.

    However, in the bottom, those fifteen blocks, in a preset shape, are forced to fill an area of 16 units. Accounting for a 1 unit gap.
  • WindelkronWindelkron Join Date: 2002-04-11 Member: 419Members
  • LegionnairedLegionnaired Join Date: 2002-04-30 Member: 552Members, Constellation
    <!--QuoteBegin--Windelkron+Oct. 26 2002,22:24--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Quote</b> (Windelkron @ Oct. 26 2002,22:24)</td></tr><tr><td id="QUOTE"><!--QuoteEBegin-->wow. pwned.<!--QuoteEnd--></td></tr></table><span id='postcolor'><!--QuoteEEnd-->
    I thought so too.
  • TzarconTzarcon Join Date: 2002-02-28 Member: 259Members
    But, what I dont get is how if the partitions are exactly the same how one just disappears. For example:

    1+3+4+2=10
    3+1+4+2=10
    4+3+1+2=10
    2+3+1+4=10
    1+2+3+4=10

    and so on. The order of the numbers doesnt change. It's wierd how a different principle applies here

    EDIT: in the first one, you can discover the area by using 6x13/2. In the second one its the same thing, so the area should be same, I know your logic is right but I dont quite see how it could possibly be right
  • JopsJops Join Date: 2002-09-13 Member: 1312Members
    Daaaaaaa Light switch *scraches heads and readjusts overalls* Is dat what they are using now?
  • LegionnairedLegionnaired Join Date: 2002-04-30 Member: 552Members, Constellation
    <!--QuoteBegin--Tzarcon+Oct. 26 2002,22:35--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Quote</b> (Tzarcon @ Oct. 26 2002,22:35)</td></tr><tr><td id="QUOTE"><!--QuoteEBegin-->But, what I dont get is how if the partitions are exactly the same how one just disappears. For example:

    1+3+4+2=10
    3+1+4+2=10
    4+3+1+2=10
    2+3+1+4=10
    1+2+3+4=10

    and so on. The order of the numbers doesnt change. It's wierd how a different principle applies here

    EDIT: in the first one, you can discover the area by using 6x13/2. In the second one its the same thing, so the area should be same, I know your logic is right but I dont quite see how it could possibly be right<!--QuoteEnd--></td></tr></table><span id='postcolor'><!--QuoteEEnd-->
    OK, Area=L*W

    SO, lets assume that the rectangular block has to have the same area, to avoid getting the gap.

    Use the bottom as an example. 16=2x8
    Now, the height of the rectangle increased to 3.. meaning that the length would have to decrease to 5 and 1/3, in order to have the area remain constant (3*5.33333=16). But it doesn't, it decreases to 5.

    So, a 5 1/3 length of what it should be, - 5 of what is is... ends up 1/3 of a unit short of what it needs to be to have the area remain constant.

    Now, this isn't a linear problem, it's a quadradic, so multiply that by the height of the rectange... 3... and you get 1, the ammount of units missing.

    Make sense now?
  • GhostfaceKillahGhostfaceKillah Join Date: 2002-10-05 Member: 1438Members
    wow, this thread made me realize how stupid i am  <!--emo&???--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/confused.gif" border="0" valign="absmiddle" alt='???'><!--endemo-->
  • ravan46ravan46 Join Date: 2002-09-11 Member: 1309Members
    My thoughts:
    Ok, first. As Rhoads pointed out. The two smaller triangles dont have the same slope. 2/5 and 3/8. So on the top one, it smudge in a bit. And on the bottom one, it smudges out a bit.

    That means these aren't true triangles. So you cant use the area equation for them. Area= (base X hight)/2. Doing that with the given lengths is (5 X 13)/2 = <b>32.5</b>. Which is wrong, cause remember, they're not real triangles.

    To find the real area of these things we have to take the area of each individual piece...
    The green triangle = 5
    red triangle =       12
    yellow piece =       7
    green piece  =       8

    Add them up, and it comes out to 32. We know from above that 32.5 is wrong, because we used the Area of a Triangle equation on something that wasn't a triangle. So the actual area of <b>both</b> of the big "triangles" is 32.

    To simplify a bit. If you were to take both of these irregular triangles, and slap them together, they would fit quit nicely. Even with the smudges on them, they create a perfect rectangle with one missing piece.

    Remember our faulty Area of a Triable equation? If we assumed the hypotenouse was strait, the area of these shapes came out to 32.5 when the area is actually 32. Since real triangles are just half rectangles, you can see the basics of the equation. Base X Hight (the area of a rectangle) divided by 2 (cause a triangle is half a rectangle). Now we take our extra .5 that we got from the equation. Multiply that by two (to get back to a rectangle), it equals 1.
    1 extra piece.
    I.E. the 1 piece missing.

    The area of the rectangle we made from fitting the two big shapes together is 64.
    One side is 13, the other 5. 13 X 5 = 65
    <b>Minus</b> that one piece is 64.
    64 divided by two is 32.
    32 is the area of the "triangles"

    <b>SO</b>, these two things are actually two different shapes because of the smudges (lol, I don't know what else to call them), but they both have the same area, 32.

    ::dances::  <!--emo&:)--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/smile.gif" border="0" valign="absmiddle" alt=':)'><!--endemo-->
  • HBNayrHBNayr Join Date: 2002-07-13 Member: 930Members
    <!--QuoteBegin--Legionnaired+Oct. 26 2002,22:53--></span><table border="0" align="center" width="95%" cellpadding="3" cellspacing="1"><tr><td><b>Quote</b> (Legionnaired @ Oct. 26 2002,22:53)</td></tr><tr><td id="QUOTE"><!--QuoteEBegin-->OK, Area=L*W<!--QuoteEnd--></td></tr></table><span id='postcolor'><!--QuoteEEnd-->
    Skullduggery!

    <!--emo&:p--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/tounge.gif" border="0" valign="absmiddle" alt=':p'><!--endemo-->

    Rhoads was right.  Or close enough to being right.

    The area of the large triangle, as ravan46 pointed out, should be 32.5 square units.  But in counting the invidual pieces, we count 32 whole square units.  Half a unit is missing.

    The triangle isn't quite drawn with an entirely straight line.  It doesn't vary enough from a straight line to account for a whole square unit.  But it doesn't need to.  Only half on a square unit.

    If you don't believe me, open the image into MS Paint and lap one triangle on its mate.  Or just follow one of the grid lines straight down, and see the difference yourself.

    It's true that the rectangle matters, but the trick lies in the greater picture, so to speak.

    -Ryan!


    "If the colleges were better, if they really had it, you would need to get the police at the gates to keep order in the inrushing multitude. See in college how we thwart the natural love of learning by leaving the natural method of teaching what each wishes to learn, and insisting that you shall learn what you have no taste or capacity for. The college, which should be a place of delightful labor, is made odious and unhealthy, and the young men are tempted to frivolous amusements to rally their jaded spirits.  I would have the studies elective. Scholarship is to be created not by compulsion, but by awakening a pure interest in knowledge. The wise instructor accomplishes this by opening to his pupils precisely the attractions the study has for himself. The marking is a system for schools, not for the college; for boys, not for men; and it is an ungracious work to put on a professor."
    -- Ralph Waldo Emerson
  • LegionnairedLegionnaired Join Date: 2002-04-30 Member: 552Members, Constellation
    I'm afraid you are overanylizing the slope, and it's effects.

    Since the shapes presented have to have the same area, as they are comprized of the same components.

    Now, since the bottom figure appears to have a glaring hole in it, it needs another way of keeping the areas the same, which, as all of you have pointed out, is the slope problem.

    However... we are still left with the question of how that hole got there in the first place! The slope idea explains how it can be there, and serves as a proof to show the figures are comprised of the same pieces, but I'm afraid the reason for the gap, at heart, involves the laying out of the pieces into the area provided, IE, the area not taken up by the smaller triangle pieces. As I have already stated, the gap comes from 15 square units of stuff attempting to fill 16 square units of vacant space.

    Both polygons have the same area, and the reason that they can is the slope idea. But that isn't the question asked. The question was, "From where comes this hole."

    That hole comes from my theory.

    Your move.
  • ravan46ravan46 Join Date: 2002-09-11 Member: 1309Members
    I think if this question was asked on a test. (more likely a psychology test than a math test I think) Then I would assume you would get it right if you pointed out all the slope stuff, and how they can both be the same area. Legionnaired's stuff might get a little extra credit though.

    Knight B5 to C8. Check.
  • JedisarJedisar Join Date: 2002-03-03 Member: 264Awaiting Authorization
    Fancy Smancy problems, Bleh. I'll solve it simple.

    They switched two differant shaped blocks. Doing so made it so that one, which was missing a square in it, took out a square, because it didn't fit as well. Normally this square would be covered by the compactness of the triangle. Pwned.
  • DunsbyDunsby Join Date: 2002-08-01 Member: 1042Awaiting Authorization
    Its and optical illusion.
    0.o
  • GingerGinger Join Date: 2002-08-01 Member: 1043Members
    i was going to say ask SBV coz he is good at maths, he is in top set <!--emo&:(--><img src="http://www.natural-selection.org/iB_html/non-cgi/emoticons/sad.gif" border="0" valign="absmiddle" alt=':('><!--endemo-->
Sign In or Register to comment.