Frankenstein Jambo. Posted February 16, 2010 Share Posted February 16, 2010 http://www.realmagic.net/dp/1-4.htm Weird.. Link to comment Share on other sites More sharing options...
Homme Posted February 16, 2010 Share Posted February 16, 2010 Beats me! Link to comment Share on other sites More sharing options...
BoJack Horseman Posted February 16, 2010 Share Posted February 16, 2010 http://www.realmagic.net/dp/1-4.htm Weird.. That was the most annoying thing I've ever witnessed on KB. I'm pretty sure it's all a lie. It's not a true triangle and the hypotenuse has a curve to it, creating the extra square. Link to comment Share on other sites More sharing options...
Romanov Stole My Pension Posted February 16, 2010 Share Posted February 16, 2010 It's very simple and quite clever, but I won't ruin it for everyone. Link to comment Share on other sites More sharing options...
I P Knightley Posted February 16, 2010 Share Posted February 16, 2010 Nice Chebs! Link to comment Share on other sites More sharing options...
Tiberius Stinkfinger Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Link to comment Share on other sites More sharing options...
I P Knightley Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Gay post, imo. i ran the test "Does the smaller fit into the smaller perfectly", - attention to details absent. Link to comment Share on other sites More sharing options...
Frankenstein Jambo. Posted February 16, 2010 Author Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. I must be rather stupid then. Link to comment Share on other sites More sharing options...
chrishall Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. That's what I was gonna say. Link to comment Share on other sites More sharing options...
InNothingWeTrust Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. So basically, the space is gained by hiding the edges of the inner triangles behind the thick border line of the main triangle? Link to comment Share on other sites More sharing options...
Tiberius Stinkfinger Posted February 16, 2010 Share Posted February 16, 2010 So basically, the space is gained by hiding the edges of the inner triangles behind the thick border line of the main triangle? Thats the correct solution. Link to comment Share on other sites More sharing options...
Scotland Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Of Course. Link to comment Share on other sites More sharing options...
Walter Bishop Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Bufter post imo. Link to comment Share on other sites More sharing options...
BoJack Horseman Posted February 16, 2010 Share Posted February 16, 2010 Thats the correct solution. Then why not say that in the first place? Also, where do you copy paste that from? Link to comment Share on other sites More sharing options...
Craig_ Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Remember to quote your sources! www.daoinesidhe.net/forum/topic.php?tid=6200 Link to comment Share on other sites More sharing options...
Frankenstein Jambo. Posted February 16, 2010 Author Share Posted February 16, 2010 Remember to quote your sources! www.daoinesidhe.net/forum/topic.php?tid=6200 WHIP! Link to comment Share on other sites More sharing options...
Flux Posted February 16, 2010 Share Posted February 16, 2010 Jason Plato knows. Link to comment Share on other sites More sharing options...
topcat Posted February 16, 2010 Share Posted February 16, 2010 That was the most annoying thing I've ever witnessed on KB. I'm pretty sure it's all a lie. It's not a true triangle and the hypotenuse has a curve to it, creating the extra square. Pretty much spot on (It's a kink rather than a curve) and far more concise than LionheartF's effort. Incidentally LionheartF's idea of area "being lost in the border" is a bit of a red herring. The Kink in the "Hypotenuse" is 178.8 degrees (outwards then inwards) which is so close to 180 as to be practically inperceptable even when drawn with fine lines. the deviation from the true straight line is never more than 1/13 of a squares . The border might make it even harder to spot but the extra area is mainly hidden by being spread out in a very thin layer. Link to comment Share on other sites More sharing options...
Victorian Posted February 16, 2010 Share Posted February 16, 2010 lionheart must have eaten .lawson to gain such incredible copy&paste ability. lionheart..... a cannibal. Link to comment Share on other sites More sharing options...
Tiberius Stinkfinger Posted February 16, 2010 Share Posted February 16, 2010 lionheart must have eaten .lawson to gain such incredible copy&paste ability. lionheart..... a cannibal. Copy, paste & spellcheck actually rubber pants. As for eating .Lawson, chomping on men is much more your thing allegedly. Link to comment Share on other sites More sharing options...
southside1874 Posted February 16, 2010 Share Posted February 16, 2010 Its actually rather simple and you have to be pretty thick not to understand. The reason is, they are NOT similar triangles (Definition of similar triangles: two triangles that can be made identical but increasing the length and breadth of one triangle by the same amount (Make the triangle bigger/smaller, without warping it) and rotating it at will.) IT is obvious that these triangles are NOT similar because of the fact that, with all similar shapes, the angles are IDENTICAL. As such, a smaller similar rectangle will fit PERFECTLY into a larger similar rectangle, but there will be space remaining in a maximum of two directions. Henceforth, in order to assume that the two triangles are in fact interchangeable on the bottom right corner of the bordered triangle, i ran the test "Does the smaller fit into the smaller perfectly", and the answer returned FALSE. The smaller rectangle is 5 wide and three high, NOT INCLUSIVE OF THE BORDER. The border is openly theoretical, it should not in actuality exist. In mathematics, it is an "Empty" space, merely a border. a Two-dimensional vector of zero width, for those mathematically inclined. When trying to lay this smaller triangle in the larger one, it can be observed that the triangle, upon going five spaces width wise from the acute angle between the base and hypotenuse, it goes three and roughly three quarters way up, before reaching the border. As i mentioned before, the border does not really exist, it's only there for our benefit. As such, it can be observed that were the smaller triangle made to be of equal width of the larger, it would be quite considerable larger. As such, the shape that is created is in fact NOT A TRIANGLE AT ALL. It is, in face, a quadrilateral. In the second instance all the outer angles are >180 Degrees, or Pi radians, while in the second one the most smooth of the angles is infact <180 degrees (or Pi radians). As such, the original quadrilateral has inface for LESS AREA than the second, which is why when taking the area of the first and putting it into the second, you are left with a single unit block free. Simple really. Thats a long way around saying.........the thickness of the line that defines the perimeter of the larger triangle is equal to one square. Link to comment Share on other sites More sharing options...
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