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      "author": "drdawud",
      "body": "@@ -4091,88 +4091,8 @@\n %22.%0A%0A\n-We might imagine that, we could repeatedly stretch, skew, and rotate a square.  \n In t\n@@ -4112,21 +4112,17 @@\n  matrix \n-**A**\n+A\n  might r\n@@ -4184,22 +4184,20 @@\n uare\n-, meaning that\n+. Therefore,\n  any\n@@ -4211,16 +4211,45 @@\n matrix A\n+ (does it have to be square?)\n , might \n@@ -4341,17 +4341,16 @@\n  result.\n- \n  In othe\n@@ -4361,21 +4361,17 @@\n rds, If \n-**A**\n+A\n  is a 2 \n@@ -4405,23 +4405,15 @@\n ors \n-**a** and **b**\n+a and b\n , th\n@@ -4444,13 +4444,9 @@\n hat \n-**A**\n+A\n  tra\n@@ -4473,28 +4473,17 @@\n are in R\n-%3Csup%3E2%3C/sup%3E\n+2\n  into so\n@@ -4503,28 +4503,17 @@\n ram in R\n-%3Csup%3E2%3C/sup%3E\n+2\n , for in\n@@ -4523,606 +4523,715 @@\n nce:\n+%22\n %0A%0A\n-%3Ccenter%3E!%5B%5D(https://steemitimages.com/DQmUvR1c9YwTVYNovqmw7DD9oJYKtgseg2cg9ktZSXHTZDt/image.png)%3C/center%3E%0A%0ANow, most textbooks give a formula for the determinant in R%3Csup%3E2%3C/sup%3E to be:  !%5B%5D(https://steemitimages.com/DQmeG9jCdi87wAEAN9iGKWnLbdFmPhhWe3dj4vdFmcfZ1xg/image.png)%0A%0ABut why does this work?  Let us recall that we now have a parallelogram where the vector (0,1) has been changed to (a,b) and the vector (1,0) has been changed to (c,d).  That parallelogram might look like this:%0A%0A%3Ccenter%3E!%5B%5D(https://steemitimages.com/DQmQb7uj5qhSEW2JFoqw8aQBjBVYyRj4uVk22hadn3tLStq/image.png)%3C/center%3E%0A%0A\n+In practice, we could imagine that we could repeatedly stretch, skew, and rotate a square, performing several operations successively to obtain a desired parallelogram.  In this case, any square matrix **A** would represent the net result of each individual operation.  In other words, if we successively multiply several square matrices together, we would get the coordinates for the vertices of that parallelogram.  Graphically, that matrix multiplication would look something like this:%0A%0A%3Ccenter%3E!%5B%5D(https://steemitimages.com/DQmUvR1c9YwTVYNovqmw7DD9oJYKtgseg2cg9ktZSXHTZDt/image.png)%3C/center%3E%0A%0AIn this case, the vector (0,1) has been changed to (a,b) and the vector (1,0) has been changed to (c,d).  \n Now,\n@@ -5768,33 +5768,311 @@\n g)%0A%0A\n-%3Ch2%3ESome Unresolved Issue\n+This is precisely the textbook definition of the determinant of a two-by-two matrix.  The same logic can naturally be applied for a three-by-three matrix.  In this case, the determinant would yield the volume of the parallelopiped enclosed by the three column vectors of that matrix.%0A%0A%3Ch2%3EFurther Remark\n s%3C/h\n@@ -6167,16 +6167,67 @@\n egative?\n+  This is of course possible should if $bc$ %3E $ad$.\n   Well, \n@@ -6336,16 +6336,216 @@\n changed.\n+  In our normal coordinate system, the $y$ axis is found in the counterclockwise direction relative to the $x$ axis, but it is possible to change this orientation by %22mirroring%22 the coordinate system.\n   In ess\n@@ -6568,21 +6568,20 @@\n s still \n-chang\n+scal\n ed by th\n",
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      "body": "Hello steemians, it's been a very busy two weeks for me.  It was midterm week, which meant creating and grading midterms; then we were sponsoring a Math Olympiad, which meant organizing some 80-odd elementary, middle, and high school students on a bright and sunny afternoon.  All in all, it was a hectic yet somehow energizing experience.\n\n<center>![](https://steemitimages.com/DQmUiAAnrdtAvSiPnsSix5iJqYdLbvdmB8xms8TyBqQE8X4/image.png)</center>\n\nAnyways, one of my students recently asked a question which frequently comes up every time I introduce the concept of matrices and vectors for the first time:  \"*Why does the determinant of a matrix give the area of the parallelogram enclosed by its vectors?*\"\n\nThe technical answer is because det(A) represents the area or volume magnification factor for the linear the particular linear transformation given by the matrix.\n\nTo understand this a bit better, we should have a clear understanding of what is meant by a linear transformation.  In this context, a linear transformation on a vector takes place when we multiply a matrix by a vector, which is to say something like:\n\n<center>![](https://steemitimages.com/DQmSUmD652Vq9ckKeGNj4CxHPoTcT1GtzLDHfF68QqCExum/image.png)</center>\n\nNow the rules of linear transformation essentially state the following:\n\n1.  All straight lines must stay straight lines.\n2.  The distance between lines can change, but they must be evenly spaced after the transformation.\n\nIn practice this means essentially that, if you were to draw a coordinate grid, you are allowed to stretch and skew the grid, but not bend the paper, nor move the origin.  For example, this would be linear transformation:\n\n<center>![](https://steemitimages.com/DQmbPjTRL9nkCAoxWk8rToEExFhoPCVKub4vBTYACC9iin7/image.png)</center>\n\nWhile something like this would not be:\n\n![](https://steemitimages.com/DQmT5JKSgijBVMz7PX9HAzSk2KCyBD2LTnzTB5MXMCjZxw8/image.png)\n\nIn any case, let's consider the general case above.  If we multiply the matrix by the vector we get:\n\n<center>![](https://steemitimages.com/DQmb21aZ95fGjqUcLX2yL2Ke7Pjk3YFVGGmZtquEkvB9c5W/image.png)</center>\n\nWhich essentially means that for any vector, you're transforming the *x* coordinate by multiplying it by the vector [a, b]<sup>T</sup> and the *y* coordinate by multiplying it by the vector [c, d]<sup>T</sup>.  In other words, it takes the unit vectors and changes them to  [a, b]<sup>T</sup> and  [c, d]<sup>T</sup>, and because linearity is preserved, it simply means that any vector [x, y], is simply a linear combination of the two vectors transformed unit vectors.\n\n<h2>A few special cases</h2>\nIn R<sup>2</sup> we call a matrix a rotation *matrix* if it can be expressed as:\n\n<center>![](https://steemitimages.com/DQmPnmKZs5qTHinJsBJqu2XyAVzxHQ1xcm9jrAWyVYXLjfk/image.png)</center>\n\nThis rotates every point about the origin by *θ* degrees.  For instance, if I have a square with vertices at (0,0), (1,0), (1,1) and (0,1), to get the transformed square, I multiply:\n\n<center>![](https://steemitimages.com/DQmURAPHshHCjPG7efFBzQqJEs3CR2mqHcKz9aBo2NjHA9t/image.png)</center>\n\nReplacing the vector [x, y]<sup>T</sup> with the desired points.  The result looks like this:\n\n<center>![](https://steemitimages.com/DQmW8BCpQTm637pU8eQGpi5c2UtSwczyuNxVSXHqeSqctwc/image.png)</center>\n\nOne can easily verify, using coordinate geometry that the green figure is a square of side length 1.  Since this is a rotation, it does not change the area of the square, it merely rotates it 45 degrees around the origin.\n\nNext, we'll investigate a matrix with a values only along the diagonals, of the form: ![](https://steemitimages.com/DQmZjrdDCpswLQSAbpTzNzU8A5LzHHh7tN6b7cmxkSEFaFD/image.png).\n\nIn this case, it is obvious that a square will be stretched out into a rectangle with vertices at (0,0), (a, 0), (a, b) and (b, 0).  Therefore, the area of the square will now be *ab* where previously, it was 1.  \n\n<h2>Back to determinants</h2>\n\nSo, back to our original question:  \"*Why does the determinant of a matrix give the area of the parallelogram enclosed by its vectors?*\".\n\nWe might imagine that, we could repeatedly stretch, skew, and rotate a square.  In this case, any matrix **A** might represent all of those things that we're doing to the square, meaning that any given matrix A, might in fact be the net result of multiplying several matrices together to get the desired result.  In other words, If **A** is a 2 x 2 matrix with column vectors **a** and **b**, then the linearity means that **A** transforms the unit square in R<sup>2</sup> into some parallelogram in R<sup>2</sup>, for instance:\n\n<center>![](https://steemitimages.com/DQmUvR1c9YwTVYNovqmw7DD9oJYKtgseg2cg9ktZSXHTZDt/image.png)</center>\n\nNow, most textbooks give a formula for the determinant in R<sup>2</sup> to be:  ![](https://steemitimages.com/DQmeG9jCdi87wAEAN9iGKWnLbdFmPhhWe3dj4vdFmcfZ1xg/image.png)\n\nBut why does this work?  Let us recall that we now have a parallelogram where the vector (0,1) has been changed to (a,b) and the vector (1,0) has been changed to (c,d).  That parallelogram might look like this:\n\n<center>![](https://steemitimages.com/DQmQb7uj5qhSEW2JFoqw8aQBjBVYyRj4uVk22hadn3tLStq/image.png)</center>\n\nNow, we can re-arrange the parallelogram to look like this without changing the area, thus transforming it into a rectangle:\n\n<center>![](https://steemitimages.com/DQmcRArgMpTKoB15Xm4PdFo31gBTPnEVKbwstmXpYFWnXFL/image.png)</center>\n\nIt should be obvious that the height of the rectangle is *d*, and that its width is ![](https://steemitimages.com/DQmUNUigbDQbhECSF5J2Nsyv4ZaG5AavmRihmNUb8GDnfwt/image.png).  Multiplying this product, we get that:  ![](https://steemitimages.com/DQmY9x79aPRUiuxLrnbvXPJBKkmqmiM3Lpo27SvCohyJFBJ/image.png)\n\n<h2>Some Unresolved Issues</h2>\n\nThere remain a few things to be said:  Firstly, what happens if the determinant is negative?  Well, a negative determinant essentially means that the orientation of the axes relative to one another has changed.  In essence, the area is still changed by the *absoute value* of the determinant, but if one imagines that the square is drawn on a glass window, it simply means you're viewing the same drawing from the other side of the glass.\n\nSecondly, what happens when the determinant is zero?  It will turn out that both column vectors comprising the square matrix are in fact linearly dependent, because ad = bc.  For instance, if the matrix is ![](https://steemitimages.com/DQmSaEZ3z74rxixohTJtMW9C88hP3Fx9EGjhACwUqYzUUoh/image.png), then the determinant is zero, and therefore the area of the parallelogram will be zero.  This means that either all vectors will collapse onto a single line, or onto a single point.\n\n<h2>Concluding Remarks</h2>\nHere, we've seen that the determinant of a two-by-two matrix gives the area of a parallelogram enclosed by its two vectors.  In fact, we can generalize this to three dimensions, in which you would obtain the volume of the solid (called a parallelopiped), as well as higher dimensions.  We saw that this is achieved essentially because a unit square is transformed into the said parallelogram, and that geometrically, the area turns out to be precisely the determinant.",
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      "author": "drdawud",
      "body": "Hello steemians, it's been a very busy two weeks for me.  It was midterm week, which meant creating and grading midterms; then we were sponsoring a Math Olympiad, which meant organizing some 80-odd elementary, middle, and high school students on a bright and sunny afternoon.  All in all, it was a hectic yet somehow energizing experience.\n\n<center>![](https://steemitimages.com/DQmUiAAnrdtAvSiPnsSix5iJqYdLbvdmB8xms8TyBqQE8X4/image.png)</center>\n\nAnyways, one of my students recently asked a question which frequently comes up every time I introduce the concept of matrices and vectors for the first time:  \"*Why does the determinant of a matrix give the area of the parallelogram enclosed by its vectors?*\"\n\nThe technical answer is because det(A) represents the area or volume magnification factor for the linear the particular linear transformation given by the matrix.\n\nTo understand this a bit better, we should have a clear understanding of what is meant by a linear transformation.  In this context, a linear transformation on a vector takes place when we multiply a matrix by a vector, which is to say something like:\n\n<center>![](https://steemitimages.com/DQmSUmD652Vq9ckKeGNj4CxHPoTcT1GtzLDHfF68QqCExum/image.png)</center>\n\nNow the rules of linear transformation essentially state the following:\n\n1.  All straight lines must stay straight lines.\n2.  The distance between lines can change, but they must be evenly spaced after the transformation.\n\nIn practice this means essentially that, if you were to draw a coordinate grid, you are allowed to stretch and skew the grid, but not bend the paper, nor move the origin.  For example, this would be linear transformation:\n\n<center>![](https://steemitimages.com/DQmbPjTRL9nkCAoxWk8rToEExFhoPCVKub4vBTYACC9iin7/image.png)</center>\n\nWhile something like this would not be:\n\n![](https://steemitimages.com/DQmT5JKSgijBVMz7PX9HAzSk2KCyBD2LTnzTB5MXMCjZxw8/image.png)\n\nIn any case, let's consider the general case above.  If we multiply the matrix by the vector we get:\n\n<center>![](https://steemitimages.com/DQmb21aZ95fGjqUcLX2yL2Ke7Pjk3YFVGGmZtquEkvB9c5W/image.png)</center>\n\nWhich essentially means that for any vector, you're transforming the *x* coordinate by multiplying it by the vector [a, b]<sup>T</sup> and the *y* coordinate by multiplying it by the vector [c, d]<sup>T</sup>.  In other words, it takes the unit vectors and changes them to  [a, b]<sup>T</sup> and  [c, d]<sup>T</sup>, and because linearity is preserved, it simply means that any vector [x, y], is simply a linear combination of the two vectors transformed unit vectors.\n\n<h2>A few special cases</h2>\nIn R<sup>2</sup> we call a matrix a rotation *matrix* if it can be expressed as:\n\n<center>![](https://steemitimages.com/DQmPnmKZs5qTHinJsBJqu2XyAVzxHQ1xcm9jrAWyVYXLjfk/image.png)</center>\n\nThis rotates every point about the origin by *θ* degrees.  For instance, if I have a square with vertices at (0,0), (1,0), (1,1) and (0,1), to get the transformed square, I multiply:\n\n<center>![](https://steemitimages.com/DQmURAPHshHCjPG7efFBzQqJEs3CR2mqHcKz9aBo2NjHA9t/image.png)</center>\n\nReplacing the vector [x, y]<sup>T</sup> with the desired points.  The result looks like this:\n\n<center>![](https://steemitimages.com/DQmW8BCpQTm637pU8eQGpi5c2UtSwczyuNxVSXHqeSqctwc/image.png)</center>\n\nOne can easily verify, using coordinate geometry that the green figure is a square of side length 1.  Since this is a rotation, it does not change the area of the square, it merely rotates it 45 degrees around the origin.\n\nNext, we'll investigate a matrix with a values only along the diagonals, of the form: ![](https://steemitimages.com/DQmZjrdDCpswLQSAbpTzNzU8A5LzHHh7tN6b7cmxkSEFaFD/image.png).\n\nIn this case, it is obvious that a square will be stretched out into a rectangle with vertices at (0,0), (a, 0), (a, b) and (b, 0).  Therefore, the area of the square will now be *ab* where previously, it was 1.  \n\n<h2>Back to determinants</h2>\n\nSo, back to our original question:  \"*Why does the determinant of a matrix give the area of the parallelogram enclosed by its vectors?*\".\n\nWe might imagine that, we could repeatedly stretch, skew, and rotate a square.  In this case, any matrix **A** might represent all of those things that we're doing to the square, meaning that any given matrix A, might in fact be the net result of multiplying several matrices together to get the desired result.  In other words, If **A** is a 2 x 2 matrix with column vectors **a** and **b**, then the linearity means that **A** transforms the unit square in R<sup>2</sup> into some parallelogram in R<sup>2</sup>, for instance:\n\n<center>![](https://steemitimages.com/DQmUvR1c9YwTVYNovqmw7DD9oJYKtgseg2cg9ktZSXHTZDt/image.png)</center>\n\nNow, most textbooks give a formula for the determinant in R<sup>2</sup> to be:  ![](https://steemitimages.com/DQmeG9jCdi87wAEAN9iGKWnLbdFmPhhWe3dj4vdFmcfZ1xg/image.png)\n\nBut why does this work?  Let us recall that we now have a parallelogram where the vector (0,1) has been changed to (a,b) and the vector (1,0) has been changed to (c,d).  That parallelogram might look like this:\n\n<center>![](https://steemitimages.com/DQmQb7uj5qhSEW2JFoqw8aQBjBVYyRj4uVk22hadn3tLStq/image.png)</center>\n\nNow, we can re-arrange the parallelogram to look like this without changing the area, thus transforming it into a rectangle:\n\n<center>![](https://steemitimages.com/DQmcRArgMpTKoB15Xm4PdFo31gBTPnEVKbwstmXpYFWnXFL/image.png)</center>\n\nIt should be obvious that the height of the rectangle is *d*, and that its width is ![](https://steemitimages.com/DQmUNUigbDQbhECSF5J2Nsyv4ZaG5AavmRihmNUb8GDnfwt/image.png).  Multiplying this product, we get that:  ![](https://steemitimages.com/DQmY9x79aPRUiuxLrnbvXPJBKkmqmiM3Lpo27SvCohyJFBJ/image.png)\n\n<h2>Some Unresolved Issues</h2>\n\nThere remain a few things to be said:  Firstly, what happens if the determinant is negative?  Well, a negative determinant essentially means that the orientation of the axes relative to one another has changed.  In essence, the area is still changed by the *absoute value* of the determinant, but if one imagines that the square is drawn on a glass window, it simply means you're viewing the same drawing from the other side of the glass.\n\nSecondly, what happens when the determinant is zero?  It will turn out that both column vectors comprising the square matrix are in fact linearly dependent, because ad = bc.  For instance, if the matrix is ![](https://steemitimages.com/DQmSaEZ3z74rxixohTJtMW9C88hP3Fx9EGjhACwUqYzUUoh/image.png), then the determinant is zero, and therefore the area of the parallelogram will be zero.  This means that either all vectors will collapse onto a single line, or onto a single point.\n\n<h2>Concluding Remarks</h2>\nHere, we've seen that the determinant of a two-by-two matrix gives the area of a parallelogram enclosed by its two vectors.  In fact, we can generalize this to three dimensions, in which you would obtain the volume of the solid (called a parallelopiped), as well as higher dimensions.  We saw that this is achieved essentially because a unit square is transformed into the said parallelogram, and that geometrically, the area turns out to be precisely the determinant.",
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      "body": "Owl curation of both this post on its [solution](https://steemit.com/math/@drdawud/finding-the-height-of-a-sea-island-a-1800-year-old-math-problem)\nContent type:  long, (mostly) technical\nAwarded 4.5 out of 6 owls: \n![exp.png](https://steemitimages.com/DQmeLnHKbZCAfdrF2RPSCiycWZrx5ePNkaJdYgG4wPiAegf/exp.png) ![clar.png](https://steemitimages.com/DQmfH2dRuSi1jUNra9ebptNK5xxkQsHpM2boFFZT7BbFhX6/clar.png) ![form.png](https://steemitimages.com/DQmdFhHwk48hgxc2cUCW68PWj29qtvhbP9xMknePUyWg6ez/form.png)\n![spel.png](https://steemitimages.com/DQmSdZZAMgcwtaN86uYm2Q39ALBT4AcJbw3UQjmJW7ZgvrD/spel.png)![citrec21.png](https://steemitimages.com/DQmWp41QCLHuNRYKx642RAfrgPHUf6jHSLQ6BtQqtLMsf92/citrec21.png)\n\nDetails: the **originality** owl was not awarded since it requires the math to be explained in a creative/novel way.  Concerning citations and sources it would be appropriate to mention if the figures are original work or not. I assume they are but since this was not indicated only half a  **citation and source** owl was awarded.",
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      "body": "Thanks for mentioning me, and more importantly thanks for questioning, which is quite important in a time when everyone accuses everyone else of publishing \"fake news\".  \n\nAs with any knowledge, math thrives on asking good questions.",
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      "body": "I composed this poem in 2012, during the Iraq war.  At that that time, it bore the subtitle «رسالة إلى باراك أوباما» (\"A Letter to Barack Obama\"), and I dedicated it to those who had been displaced by the war.  Though six years have passed and a new President is in office, I feel that the message is still as relevant in an era of uncertain times.\n\nThe references to Karbala and Eden (classically said to have been between the Tigris and Euphrates), are intentional.\n\n<center>![](https://steemitimages.com/DQmZv3kavxfo1H4yTgxAvhG684bh7EiTJVwsdEbz25GmS3F/image.png)\nKarbala, 1932\nSource: [wikipedia.org](https://en.wikipedia.org/wiki/Karbala#/media/File:Karbala_07402u.jpg).  License:  Open Domain</center>\n\n\n<center>Come, O Adversary, while the night is young—\nFor among the flames, my heart is already ablaze.\nFire your bullets, O Impostor, while you still can,\nFor we may meet beneath the morn's dawning rays.</center>\n\n<center>Our mothers' tears have nourished the soil —  \nMartyr's blood has watered the trees of conscience;\nAnd forests of grief will grow from seeds of shrapnel,\nIn this Eden you've reduced to ashes.</center>\n\n<center>Karbala's memory hides inside my passions —  \nMy soul's fortress conceals this timeless devotion.\nSo, come, O Adversary, while there's still time\nFor I shall remain, even until the last explosion.</center>",
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      "body": "@@ -1,13 +1,14 @@\n In a \n+%5B\n previous\n@@ -8,24 +8,145 @@\n revious post\n+%5D(https://steemit.com/math/@drdawud/finding-the-height-of-a-distant-island-can-you-solve-this-1800-year-old-math-problem)\n , I wrote ab\n@@ -3119,12 +3119,437 @@\n d 150 paces.\n+%0A%0A%3Ch2%3EMore reading%3C/h2%3E%0A* %5BAncient Egyptian Mathematics%5D(http://www-groups.dcs.st-and.ac.uk/history/HistTopics/Egyptian_mathematics.html)%0A* %5BAncient Greek Mathematics%5D(http://www.math.tamu.edu/~dallen/history/greekorg/greekorg.html)%0A* %5BEuclid's Elements in Greek%5D(http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf)%0A* %5B*Haidao Suanjing* (Sea Island Calculation Manual)%5D(https://ctext.org/hai-dao-suan-jing) - Chinese Only\n",
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      "body": "In a previous post, I wrote about how ancient surveyors attempted to find the height of a sea island.  Keeping in mind, this was largely before the advent of trigonometric tables, the methods they developed were purely Euclidian in nature, relying on principles of similarity, and using algebraic geometric techniques.  In this post, I will present the solution to the Sea Island problem using Liu Hui's method.\n\n<center>![](https://steemitimages.com/DQmWvitGFq1ZCLcU3tEtAkXCEvUQkDYi2w8cvYniw4xFkFp/image.png)\nA plate from an 18th century reproduction of the Haidao Suanjing\nwikipedia; Open Domain\n</center>\n\n<h2>Liu Hui's Difference Theorem</h2>\n\n![](https://steemitimages.com/DQmcQwS9ir8wRdy1M6u2wgSDUuB88PfaNpoJ2J7JErwoHHs/image.png)\n\n\nIn modern language, the theorem might be stated:\n\n> Consider rectangle *ACEG* having diagonal *GC*.  *AC* and *CE* are divided by perpendicular lines *BF* and *DI* respectively, such that the intersection of *BF* and *DI*, *J* lies on the diagonal.  In this case, the areas of rectangles *ABJI* and *JDEF* are equal.\n\n<h3>Proof</h3>\nTriangles ACG and GCE are congruent due to *GC* being the diagonal of the rectangle.  *GIJF* and *BCDJ* are also rectangles, and since they also share the diagonal *GC*, triangles *BJC* and *CDJ* are congruent to each other, and triangles *IJG* and *FGJ* are also congruent to each other.  Therefore the sum of the area of *BJC* and *IJG* is equal to the sum of the area of *CDJ* and *FGJ*.  Subtracting these from  *ACG* and *GCE* shows that *ABJI* and *JDEF* have the same area.\n\n<h2>The Height of the Sea Island</h2>\nLiu Hui's method essentially repeats the above theorem several times in order to find the answer algebraically.\n\n![](https://steemitimages.com/DQmVZfSWNqneJE2QrnUom78PrmJEVBydcHG5iDNRSQy9843/image.png)\n\nFirst we re-draw our diagram such as to include a \"scaffolding\" around the relevant points, and observe the following\n\n* Clearly, *a*(*d*+*D*) = *b*<sub>1</sub>(*h*-*a*) (shown in red)\n* Also, *aD* = (*h*-*a*)*b*<sub>2</sub> (shown in green)\n* Therefore, *ad* + *aD* = *b*<sub>2</sub>(*h*-*a*) and \n* Rearranging and substituting *ad* = *b*<sub>1</sub>(*h*-*a*) -  (*h*-*a*)*b*<sub>2</sub>\n* Factoring the above gives: *ad* = (*h*-*a*)(*b*<sub>1</sub>-*b*<sub>2</sub>)\n* Therefore:\n\n<center>![](https://steemitimages.com/DQmRSmVnrRcX4UZTRk5CAdxKE2MTfjeWdq95irsCLfjgZ9p/image.png)</center>\n\nMoving onto the distance:\n\n* We observe that *Da* = *b*<sub>2</sub>(*h*-*a*).  Having already calculated *h* we can calculate the value numerically.\n\nPutting the measurements from the given problem, we obtain:\n\na = 30 feet\nd = 6000 feet\nb2 = 738 feet\nb1 = 762 feet\n\n<center>![](https://steemitimages.com/DQmTs7ZRdDyEG3n8kf8bp4RcSrTsZqaNGS9EqQuYQCkpUbg/image.png)</center>\n\nConverting these, we obtain ![](https://steemitimages.com/DQmcoBM51MzVM2W1zERDAraqJAU7KJyHEk9eisyDYxMqTrM/image.png), or 4 *li* and 55 paces.\n\nSimilarly, placing this number into the equation for the distance to the island gives 102 li and 150 paces.",
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