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A Swiss Triangle |
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Consider a triangle with the attributes indicated in the figure below. |
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As shown, BC is twice AB, and BD = 100, and both of the angles ABD and CBD are 60 degrees. What is the length of AC? |
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One approach is to draw lines from D at 60 degrees from DB to create two equilateral triangles BDE and BDF as shown below. |
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We also draw the perpendicular from D to AB, which bisects EB. Now we see that DE = DF = EB = 100, and EG = GB = 50, and triangles AED and DFC are both similar to ABC, so AE is half DE and hence AE = 50. Also, since DF is twice AE, we know DC is twice AD, and by Pythagoras’ theorem applied to the right triangle DGE we have DG = 50√3. Then, from the right triangle DGA we have AD = 50√7, and hence AC = 150√7. |
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Obviously this simple geometrical solution depends on the special values of the parameters. For a more challenging problem, consider the more general case shown below. |
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Here we seek the length of the segment AC in terms of the arbitrary angles α and β and the specified ratio k. Equating the area of the overall triangle to the sum of the areas of the two sub-triangles, we have |
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Solving this for x gives |
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Then by the well-known (and easily proved) formula for the third edge of a triangle in terms of the other two edges and the subtended angle, we have |
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In the original specialized question we had L=100, k=2, and α = β = 60 deg, and noting that sin(60o) = sin(120o) = 1/2 and cos(120o) = −1/2 this expression gives 150√7 as expected. |
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