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A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation of the top of the tower are $\alpha$,$\beta$,$\gamma$ from the corners of the field find the height of the tower.

Answer: $$a\sqrt{\frac{p^2+q^2+r^2-\sqrt{2p^2q^2+2p^2r^2+2q^2r^2 -p^4 -q^4 - r^4}}{2(p^4 + q^4 + r^4 -p^2q^2 -p^2r^2 - q^2r^2)}}$$ where $p= \cot \alpha$, $q = \cot \beta$ and $r = \cot \gamma$ image

My Attempt

$[.]$ denotes area $$[APB]+[BPC]+[CPA] = [ABC]$$ $$[ABC] = \frac{\sqrt{3}}{4}a^2$$ I tried finding the area of $\triangle APB$,$\triangle BPC$,$\triangle CPA$ by Heron's formula but the result was too complicated because I had to add square roots : something of the form $\sqrt{x} + \sqrt{y} + \sqrt{z} = w$. Moreover, the $x,y,z$ in this expression are also complicated.

I admit that the answer is complicated too but I am searching for an elegant approach towards the solution (something like substitution or anything else). Is there any elegant approach?

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  • $\begingroup$ If you prefix the names of trigonometric functions with a backslash in LaTeX/MathJax, they will be rendered in upright font with appropriate spacing instead of an chain of italic letters. Please compare cot\alpha → $cot\alpha$ to \cot\alpha → $\cot\alpha$. $\endgroup$
    – CiaPan
    May 4, 2022 at 13:24

1 Answer 1

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As the figure shows the distances between the base of the tower $P$ and each of the three vertices $A,B,C$ of equilateral $\triangle ABC$ is

$ PA = h \cot \alpha = h p $

$ PB = h \cot \beta = h q $

$ PC = h \cot \gamma = h r $

Now using Barycentric coordinates, and taking $A$ to be the origin, we can express point $P$ in terms of $B $ and $C$ as follows

$ P = c_1 B + c_2 C $

so that

$ AP = P = c_1 B + c_2 C , BP = (c_1 - 1) B + c_2 C , CP = c_1 B + (c_2 - 1) C $

I'll assume that the side length $a = 1$. Taking the magnitude of these vectors, and noting that $|B| = |C| = 1$ and that $B \cdot C = \dfrac{1}{2} $, then we can write the following three equations

$c_1^2 + c_2^2 + c_1 c_2 = h^2 p^2 \hspace{10pt} (*) $

$(c_1 - 1)^2 + c_2^2 + (c_1 - 1) c_2 = h^2 q^2 $

$ c_1^2 + (c_2 - 1)^2 + c_1 (c_2 - 1) = h^2 r^2 $

We want to eliminate $c_1, c_2$, so subtract each pair of equations, i.e. (1)-(2) , and (1) - (3), we get

$ 2 c_1 + c_2 = h^2 (p^2 - q^2) + 1 $

$ c_1 + 2 c_2 = h^2 (p^2 - r^2) +1$

Solving this $2 \times 2$ system, gives us

$ c_1 = \dfrac{1}{3} ( h^2 (p^2 - 2 q^2 + r^2 ) + 1 )$

$ c_2 = \dfrac{1}{3} ( h^2 (p^2 - 2 r^2 + q^2 ) + 1 )$

substituting $c_1, c_2$ into Eq. $(*)$ and expanding, gives us

$ h^4 K + h^2 L + M = 0 $

where

$K = p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2 $

$ L = - ( p^2 + q^2 + r^2 ) $

$ M = 1 $

Combining the $h^2 $ coefficients, we get the quadratic equation (in $h^2$)

From this, using the quadratic formula, the solution is

$ h^2 = \dfrac{ -L \pm \sqrt{ L^2 - 4 K } }{ 2 K } $

we have

$ - L = p^2 + q^2 + r^2 $

and

$L^2 - 4 K = p^4 + q^4 + r^4 + 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - 4 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2) $

And this reduces to

$ L^2 - 4 K = 3 ( 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4 ) $

Hence,

$ h^2 = \dfrac{ p^2 + q^2 + r^2 \pm \sqrt{3} \sqrt{2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4} }{ 2 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2)}$

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