Trisected Hypotenuse of a Triangle
Date: 12/20/98 at 14:30:43 From: Will Nuse Subject: Trisected hypotenuse of a triangle Here's the problem: In right triangle ABC, with C as the right angle, it is known that AD = DE = EB (the three segments of the hypotenuse) and that D and E are points on hypotenuse AB. It is also known that CD = 7 and that CE = 9. What is the length of AB (the hypotenuse)? I tried to work this problem using the law of cosines because at first I thought that line segments CD and CE trisected the 90 degrees into 30 degrees. However, my trig teacher showed me that there is more than one way to draw this problem, and therefore it would be almost impossible to get all three angles ACD, DCE, and ECB equal to 30 degrees. I'm stumped, along with my entire trig class. Can you help me out? Thank you very much. I appreciate your help. Will Nuse
Date: 12/22/98 at 17:05:13 From: Doctor Floor Subject: Re: Trisected hypotenuse of a triangle Hi Will, Thanks for your question! For naming, let a = BC, b = BC, and c = AB. Let's take a look at a sketch of the situation: B |\ F E | \ G D I added the points F and G that trisect BC, | \ C-H-I-A and H and I that trisect AC. It is not difficult to see that, for instance, FE = CH. But that means that CE^2 = EF^2 + FC^2, and thus 81 = ((1/3)a)^2 + ((2/3)b)^2 81 = (1/9)a^2 + (4/9)b^2 [multiply by 9] a^2 + 4b^2 = 729  In the same way DC^2 = CI^2 + ID^2, and thus 49 = ((2/3)a)^2 + ((1/3)b)^2 49 = (4/9)a^2 + (1/9)b^2 [multiply by 9] 4a^2 + b^2 = 441  Adding  and  gives that 5a^2 + 5b^2 = 1170, so then a^2 + b^2 = 234. So the hypotenuse must be sqrt(234). Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 06/03/02 at 10:26:14 From: Doctor Ian Subject: Re: Trisected hypotenuse of a triangle Grant Woodward suggests another approach. Here is a drawing of the triangle: Start with a few simple observations: cosA = b/3x cosB = a/3x a^2 + b^2 = 9x^2 Apply the Law of Cosines to triangles BCE and ACD, respectively: 9^2 = x^2 + a^2 - 2ax cosB 7^2 = x^2 + b^2 - 2bx cosA Adding these two equations gives 9^2 + 7^2 = 2x^2 + a^2 + b^2 - 2ax cosB - 2bx cosA = 2x^2 + a^2 + b^2 - 2x(a cosB + b cosA) We noted earlier that a^2 + b^2 = 9x^2, so substituting gives = 2x^2 + 9x^2 - 2x(a cosB + b cosA) = 11x^2 - 2x(a cosB + b cosA) Substituting the values we found earlier for the cosines gives = 11x^2 - 2x(a * a/(3x) + b * b/(3x)) = 11x^2 - 2x((a^2 + b^2)/(3x)) Substituting again for a^2 + b^2 gives = 11x^2 - 2x(9x^2/(3x)) = 11x^2 - 6x^2 = 5x^2 So we end up with 9^2 + 7^2 = 5x^2 9^2 + 7^2 x = sqrt( --------- ) 5 = sqrt(26) So the length of the hypotenuse must be 3 sqrt(26). - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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