Algebra through geometry
by t0rajir0u, Apr 19, 2007, 6:33 AM
Problem: Given reals 
such that
Find
.
Solution: Well, this is a pretty intimidating system. You might be tempted to try a difference-of-cubes sort of thing, which does lead to a solution, but I took a different path motivated by a problem in Engel and by the observation that
:
Suppose we have triangle
with 
the Fermat point; this means 
. Let 
. Then by Law of Cosines on 
we have
We've produced a right triangle! Now, how do we go about calculating?
Well, consider the area of
. Since the triangle is right we know that this is just 
. But recall that the area of a triangle with sides 
and angle 
between them is 
. This tells us that
$\begin{eqnarray*}[ ABC ] &=& \frac{ \sqrt{6}}{2}\\ &=& [AFB]+[BFC]+[CFA] \\ &=& \frac{\sin 120^{\circ}}{2}\left( xy+yz+zx \right) \\ \end{eqnarray*}$
From which we readily deduce that
.
If you're a computational masochist you might realize that the rightness of the triangle is not necessary, and that we could've used Heron's if the right-hand constants had been something else.
Practice Problem: Find a straightedge-and-compass construction for a pentagon.
such that
Find
.Solution: Well, this is a pretty intimidating system. You might be tempted to try a difference-of-cubes sort of thing, which does lead to a solution, but I took a different path motivated by a problem in Engel and by the observation that
:Suppose we have triangle
with 
the Fermat point; this means 
. Let 
. Then by Law of Cosines on 
we have
We've produced a right triangle! Now, how do we go about calculating
?Well, consider the area of
. Since the triangle is right we know that this is just 
. But recall that the area of a triangle with sides 
and angle 
between them is 
. This tells us that$\begin{eqnarray*}[ ABC ] &=& \frac{ \sqrt{6}}{2}\\ &=& [AFB]+[BFC]+[CFA] \\ &=& \frac{\sin 120^{\circ}}{2}\left( xy+yz+zx \right) \\ \end{eqnarray*}$
From which we readily deduce that
.If you're a computational masochist you might realize that the rightness of the triangle is not necessary, and that we could've used Heron's if the right-hand constants had been something else.
Practice Problem: Find a straightedge-and-compass construction for a pentagon.
April 2009
March 2009