Two poles formula
Contents
Theorem
The two pole formula/theorem states that for two poles some distance apart, the height of the intersection point of the lines joining the top of either pole to the foot of the opposite pole is half the harmonic mean of the heights of the two poles(and that this is true regardless of the distance between the two poles).
Proof using similar triangles
We are looking to show that for the two poles 
and 
of heights 
and 
, respectively, and 
units apart, 
.
Because 
,
![\[\frac{h}{x}=\frac{b}{p}\]](http://latex.artofproblemsolving.com/5/7/c/57cbf73dfd41475bf786fe2a2c96a44329f35034.png)
![\[x=\frac{hp}{b}.\]](http://latex.artofproblemsolving.com/e/e/b/eeb4e361add975a1d241d0b3a1a4793e8d9332be.png)
Because 
,
![\[\frac{h}{p-x}=\frac{a}{p}\]](http://latex.artofproblemsolving.com/f/5/4/f54def8e206027a72691952d2fe16d75f1d96016.png)
![\[hp=ap-ax\]](http://latex.artofproblemsolving.com/a/7/0/a70e6291e4676598e602282ef2fba1240a067b87.png)
![\[x=\frac{ap-hp}{a}.\]](http://latex.artofproblemsolving.com/c/e/0/ce0ce86d85836f8acec9a8357b48de4823478dd4.png)
Therefore,
![\[\frac{hp}{b}=\frac{ap-hp}{a}\]](http://latex.artofproblemsolving.com/5/b/0/5b0b15738588d4d518421fe277ed45040aab84f5.png)
![\[ahp=abp-bhp\]](http://latex.artofproblemsolving.com/d/4/6/d46acf4e13ae60ab1344a9399e986903c6593799.png)
![\[ahp+bhp=abp\]](http://latex.artofproblemsolving.com/5/3/0/530dc82ecfea9c940883635851a312d00d540cb4.png)
![\[hp(a+b)=abp\]](http://latex.artofproblemsolving.com/4/d/7/4d7c158932573cb0c9f0cb75e57624088822e090.png)
![\[h(a+b)=ab\]](http://latex.artofproblemsolving.com/5/8/8/588ab320b7d87137159daf55f95f17e01e9de77a.png)
![\[h=\frac{ab}{a+b},\]](http://latex.artofproblemsolving.com/2/4/0/240df553ab47438515b7ed26d814b00fde141286.png)
as desired.
![\[Q.E.D.\]](http://latex.artofproblemsolving.com/b/c/7/bc779cb200fbc15c503e7049874b39ad6f0d1f8a.png)
![\[\smiley\]](http://latex.artofproblemsolving.com/f/b/1/fb19deb279b5cc8f4c7d2ba8a426f3822480729c.png)
Coordinate/analytical geometry proof
Using the same diagram as above, if we let 
be the origin, then 
and 
are the points 
and 
respectively. Hence, 
and 
have slopes of 
and 
, respectively, and y-intercepts of 
and 
, respectively. That means their equations are 
and 
, respectively.
Setting the two equations equal to find the intersection point(recall that a point is shared by two lines if and only if it satisfies the equation of both lines), we obtain
![\[-\frac{b}{p}x+b=\frac{a}{p}x\]](http://latex.artofproblemsolving.com/7/c/e/7cec020bd93846b9d120cf217b437e1e56f90e2e.png)
![\[\frac{b}{p}x+\frac{a}{p}x=b\]](http://latex.artofproblemsolving.com/a/b/1/ab19f9f853384ca52f06c4f00286699340196621.png)
![\[(\frac{a+b}{p})x=b\]](http://latex.artofproblemsolving.com/b/5/f/b5f284ba3d6dc85e50c517418799e7926e16a28c.png)
![\[x=\frac{b}{\frac{a+b}{p}}\]](http://latex.artofproblemsolving.com/8/0/3/80333b467e60a6d679582a4ca2310be7f8afb800.png)
![\[x=\frac{bp}{a+b}.\]](http://latex.artofproblemsolving.com/9/b/c/9bc55954b9d24ca9c749258b00d6bb861356ce8a.png)
To find the ordinate, we plug in this to either equation:
![\[y=\frac{a}{p}x=\frac{a}{p} \cdot \frac{bp}{a+b}\]](http://latex.artofproblemsolving.com/4/e/9/4e9099a344ac0afa7016cf34ac2235e08b6adcf1.png)
![\[y=\frac{abp}{p(a+b)}\]](http://latex.artofproblemsolving.com/8/6/0/8604c3bfd4c78f568884c98574bef259ebfecfb9.png)
![\[y=\frac{ab}{a+b},\]](http://latex.artofproblemsolving.com/a/5/6/a56b6700ae598d8af7ec25593ecdff692dd4e0bf.png)
so 
is the ordered pair 
. Therefore, the height is 
, as desired.
