Post Reply 
Little explorations with HP calculators (no Prime)
04-08-2017, 02:44 AM (This post was last modified: 04-08-2017 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
\[
\sin(B) = \frac{\sqrt{(24yz)^2-(9y^2+16z^2-4x^2)^2}}{24yz}.
\]
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 sub-triangles 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/thread-795...l#pid71463 )

Graph 3D | QPI | SolveSys
Find all posts by this user
Quote this message in a reply
Post Reply 


Messages In This Thread
RE: Little explorations with the HP calculators - Han - 04-08-2017 02:44 AM



User(s) browsing this thread: 3 Guest(s)