Little explorations with HP calculators (no Prime)

04082017, 02:44 AM
(This post was last modified: 04082017 02:51 PM by Han.)
Post: #155




RE: Little explorations with the HP calculators
An analytic solution (using Gerson's ideas) is in fact possible. Not wanting to give up, I actually started with Gerson's solution, but without assuming anything about x, y, and z and tried to go as far as I could. It turns out that one can set up the formula
\[ (2x)^2 = (3y)^2 + (4z)^2  2(3y)(4z)\cos(B) \] and solve for \( \cos(B) \). Then, use this formula to deduce that \[ \sin(B) = \frac{\sqrt{(24yz)^2(9y^2+16z^24x^2)^2}}{24yz}. \] Continue this process to determine \( \sin(A) \) and \(\sin(C) \) in terms of x, y, and z. After some very tedious algebra (I ended up using Maple to do a good chunk of it to avoid mistakes in algebraic manipulations), one can deduce ratios of the areas of the subtriangles and obtain the analytic equivalence of the geometric solution. Then I then realized that one could simply just use the law of sines, which has a slightly less tedious set of algebraic manipulations because the formulas for the \( \sin(A)\), \( \sin(B)\), and \(\sin(C)\) are not really necessary. Using the law of sines: \[ \frac{2x}{\sin(B)} = \frac{4z}{\sin(A)} \Longrightarrow \sin(A) = \frac{2z}{x} \cdot \sin(B) \Longrightarrow \underbrace{x\cdot \sin(A)}_{\text{height of } \triangle ADE} = 2\cdot \underbrace{z\cdot \sin(B)}_{\text{height of } \triangle BDF} \] The area of \( \triangle ADE \) is \( \frac{1}{2} \cdot y \cdot (x\cdot \sin(A)) = \frac{1}{2} \cdot x\cdot y \cdot \sin(A) \). The area of \( \triangle BDF \) is \[ \frac{1}{2} \cdot (2 y) \cdot (z\cdot \sin(B)) = \frac{1}{2} \cdot y \cdot (2z)\cdot \sin(B) = \frac{1}{2}\cdot y \cdot (x\cdot \sin(A)) = \underbrace{\frac{1}{2}\cdot x \cdot y \cdot \sin(A)}_{\text{area of } \triangle ADE} .\] Also, \[ \frac{3y}{\sin(C)} = \frac{2x}{\sin(B)} = \frac{4z}{\sin(A)} \Longrightarrow \sin(C) = \frac{3y\sin(B)}{2x} = \frac{3y\sin(A)}{4z} \] The area of \( \triangle CEF \) is \[\frac{1}{2} \cdot x \cdot [ (3z)\cdot \sin(180^\circ  C)] = \frac{1}{2} \cdot x \cdot [ (3z)\cdot \sin(C) ] = \frac{1}{2} \cdot x \cdot 3z \cdot \frac{3y}{4z}\cdot \sin(A) = \underbrace{\frac{1}{2}\cdot x \cdot y \cdot \sin(A)}_{\text{area of } \triangle ADE} \cdot \frac{9}{4}.\] So the area of \( \triangle CEF \) is \( \frac{9}{4}\)th the area of \( \triangle ADE \). Lastly, the area of \( \triangle ABC \) gives us the following identity: \[ 24 = \frac{1}{2} \cdot (3y) \cdot [ 2x\cdot \sin(A) ] \] which implies the area of \( \triangle ADE \) is \[ \frac{1}{2}\cdot x\cdot y \cdot \sin(A) = 4 \] It should be fairly straightforward to deduce that the area of \( \triangle DEF \) is \[ 24  (4+4+9) = 7 \] This analytic solution is really the same as the geometric solution I posted earlier ( http://www.hpmuseum.org/forum/thread795...l#pid71463 ) Graph 3D  QPI  SolveSys 

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