Out[]=
Visualize the configuration. I use Canvas function to sketch the diagram
In[]:=
Canvas[]
The length of the altitude is from that
Area
1
2
·h·a
. The altitude divides the triangle into two parts and the foot divides a into two sections. Using Pythagorean theorem, let’s solve for a w.r.t b and c.
In[]:=
Uniona/.Solve
(b^2-1/a^2)
+
(c^2-1/a^2)
a,a^2
Out[]=

2
b
+
2
c
-2
-1+
2
b
2
c
,
2
b
+
2
c
+2
-1+
2
b
2
c

Also we know
Area
1
2
b·c·sin(A)
, thus
csc(A)b·c
. Our target function can be visualized with contour plot. White dashed lines are contours for
b·c
.
In[]:=
ContourPlot
2
b
+
2
c
-2
-1+
2
b
2
c
+b*c,{b,0,3},{c,0,3},​​Contours12,MeshFunctions{#1*#2&},​​Mesh{{1.01,2,3,4,5}},​​MeshStyle{White,Thick,Dashed},​​Epilog{Dashed,Blue,Line[{{0,0},{3,3}}]},​​BoundaryStyleDirective[White,Thick]
Out[]=
Now we can try to find minimum value
In[]:=
Minimize
2
b
+
2
c
-2
-1+
2
b
2
c
+b*c,b>0,c>0,b*c>1,{b,c}//FullSimplify
Out[]=

5
,b
3
1/4
5
,c
3
1/4
5

There is another way to find the minimum. If we interchange b and c , the target function is unchanged. Therefore we can use educated guess to find the minimum along
bc
.
In[]:=
PlotEvaluate
2
b
+
2
c
-2
-1+
2
b
2
c
+b*c/.bc,{c,1,2}
Out[]=
1.2
1.4
1.6
1.8
2.0
2.5
3.0
3.5
4.0
Now we have a single valued function
In[]:=
expr=Evaluate
2
b
+
2
c
-2
-1+
2
b
2
c
+b*c/.bc//FullSimplify
Out[]=
3
2
c
-2
-1+
4
c
We can also find the global minimum by
In[]:=
Minimize[{expr,c>0},c]
Out[]=

5
,c
3
1/4
5
