how to solve plato's cursed triangle?
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solution at http://users.goldengate.net/~jre/
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Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
1035212
I know of two:\n\nTranslation and rotation: \n\n1) translation :\n\n Take a rectangle and cut a diagonal and extended the lines  from all three side of the triangle at right angle to it  from the side that is made from cutting the rectangle until those lines equal the lenght of the side of the triangle drawn from  and it  form three boxes  around all three sides of the triangle. Now let take the big side an calulate the square  of it ,then calulate the squares of the other two sides and add themup,  it should equal the square of the big side and that proves the pythagorean theorem. \n\nRotation: \n\nSee the show : Project Mathematics by Cal tech : Entitled "The Pythagorean Theorem" it explain better  than I can, It worth watching. Very good show on Math & Geometry
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Here's my answer:\n\nhttp://mudandmuck.com/str2/triangles.JPG\n\nYou didn't say write them out in English, you said describe.  This picture describes the solution.
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Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)
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If the mirror is between Julia and the shed yeah?\n\nDraw a triangle from julia to the mirror and back to Julia's head. This gives you a right angled triangle.\n\nMirrored this would make the distance from the shed equal to the hieght of the shed, as an angle of 45 would be drawn up from the ground at the bottom of the mirror to the top of the shed. The hoziontal distance from the mirror to the shed, would equal the hieght of the shed. ie 2.5m.\n\nIs that any help?
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It helps if you know something about Pythagorean triples (integer solutions to the Pythagorean equation for right-angled triangles).\n\nIf you do, you will know that there are no primitive triples in which the difference between the legs is 3, The smallest such differences are 1, 7 and 17 (always prime, but not all primes will serve).\n\nTherefore this must be a non-primitive triple which is a multiple by three of a primitive triple with the legs differing by 1. There is a series of such triples, starting with (3, 4, 5) which can be generated from one another.\n\nSo we divide 87 by 3 and get 29 and recognise that this must be the triple (20, 21, 29), the second in the series.\n\nSo plainly the answer is length = 63 cms and width = 60 cms (i.e. 3 x 21 and 3 x 20)\n\nTHE ALGEBRA \n\nL = B + 3\nL^2 + B^2 = 87^2\n2 B^2 + 6B + 9 = 7569\nB^2 + 3B - 3780 = 0\n(B - 60) (B + 63) = 0\n\nAs rectangles in the real world have sides of positive value, ignore B = -63 and take B = 60, yielding L = 63,\n\nBut none of that is necessary if you know your triples!
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Read Plato's dialogues of Socrates.  This was his specialty....keep asking questions...make them explain it and they will sound more and more ridiculous.
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Divide all 9 balls into 3 groups of 3 balls.\n\n1) weigh any 2 groups:\n         - if they are equal, the crooked ball is in the 3rd group\n         - if not, you have found the group where the ball is\n\n2) repeat the same experiment with that group until you find the ball
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Answer: 8\nLet u = cosx, then du = -sinx\nIf you draw the triangle, you will see that sinx = sqrt(1-u^2)\nThus, du = -sqrt(1-u^2) dx and so\ndx = -du/sqrt(1-u^2)\nThe integral is now\n-sqrt(2)*sqrt(1-u)/sqrt(1-u^2) du\nFactor 1-u^2 = (1-u)(1+u)\nCancel the 1-u on the top with the 1-u on the bottom (you can do this since they are both underneath the root sign).\nNow we have\n-sqrt(2)/sqrt(1+u) du\nBring the sqrt(1+u) to the top and write it with an exponent.\n= -sqrt(2) * (1+u)^(-1/2) du\nIntegrate to get\n-sqrt(2) * 2 * sqrt(1+u)\n= -2sqrt(2)sqrt(1+u)\nNow you have to do some trick with the limits of the integral because\nu = cos(0) = 1 and u = cos(2pi) = 1\nIf you plug in those limits you will get zero.\nBecause the graph is symmetric between 0 and 2pi (at pi), you can divide it into two pieces.  Find the area of one piece and then double the area.\nu = cos(pi) = -1\n-2sqrt(2)sqrt(1+u) from 1 to -1\n-2sqrt(2)sqrt(1+-1) - -2sqrt(2)sqrt(1+1)\n= -2sqrt(2)sqrt(0 + 2sqrt(2)sqrt(2)\n= 0 + 4\n= 4\nNow double it to get the final answer\n2(4) = 8
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As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
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Plot the three boundary lines, shade the correct sides; the answers occur at the vertices of the shaded area (a triangle).  Just plug in these vertices to C = 4x + y and see what the highest and lowest outcome is.
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Diagram here:\n\nhttp://mathworld.wolfram.com/TriangularPrism.html\n\nto solve for its volume, calculate for the area of the triangle bottom and multiply it by the height of the prism.\n\nTo compute for surface area, compute for the area of the 3 rectangular faces and add the area of the 2 triangular faces.
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a right angled isosceles triangle\n\nplease give me 10 points now
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I'm assuming you are referring to the disguised triangles in the attached link... (link 1).  I'm sure there are some rigorous mathematical proofs you could supply, or automated ways to count the triangles, but I basically decided to just use the "brute force" method of counting.  The hard part is recognizing all the different triangles while not double counting...\n\nWhat I finally did was to create several copies of the diagram and tried to color the triangles.  You can color the individual triangles (24), right off the bat.  Now if you try to pair up triangles, you can pair them one way (12) or another using the flipped image (12).  Grouping them further in groups of three or four triangles, I counted another set of (7) and the flip for another (7).  Then I noticed four other ways to group four triangles (3), (3), (3), (3).  Lastly, there are two large triangles composed of the inner sixteen triangles (1) and (1).\n\nAltogether I got seventy (76) triangles.  You can double check my work using the attached image (link 2).  I think I got them all and didn't double count...
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Draw a triangle with the letter "M" INSIDE and the triangle should cut the top 2 vertices and the 3 bottom vertices.\n\nyou should get 8 triangles and the last 1 is the big triangle with "M" inside.
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Where c is the hypotenuse, and a and b are legs of the triangle:\nc² = a² + b² --- Pythagoren Therom\nc² = (12)² + (20)² --- Replace what you know...\nc² = 144 + 400 --- Find 12² and 20²...\nc² = 544 --- Add them together...\nc = √544 --- Square root of 544 equals c\nc ≈ 23.32"