Given a point $P$ outside equilateral $\Delta ABC$ but inside $\angle ABC$, if the distance between $P$ to $BC,CA,AB$ are $h_1,h_2,h_3$ respectively, where $h_1 - h_2 + h_3 = 6$, find $[\Delta ABC]$ .
What I Tried: At first I couldn't understand if $h_1,h_2,h_3$ are just any lines touching the sides or are some specific lines like altitudes or medians (bisecting the sides of the triangle) . But since they are denoted like $h_1,h_2,h_3$ I suppose they are the altitudes. So here is a picture :-
No idea for this problem. I don't think I can use any simple geometry techniques here like angle-chasing, area of triangles, pythagorean theorem and so on, because I have been given very less info. So I am a bit stuck here.
Can anyone help me? Thank You.
$\endgroup$4 Answers
$\begingroup$By Viviani's theorem, the height of $\triangle ABC$ is $6$, so its side length is $\frac{12}{\sqrt3}=4\sqrt3$ and its area is $\frac{6×4\sqrt3}2=12\sqrt3$.
$\endgroup$ 7 $\begingroup$Connect point $P$ to $A, B, C$. Now you can see that $\triangle ABC = \triangle PAB + \triangle PBC - \triangle PAC = \frac{AB}{2}(h_1 - h_2 + h_3) = 3 AB$
Given equilateral triangle, we also know that $\triangle ABC = \frac{\sqrt3}{4}AB^2 = 3 AB$
That gives you the value of $AB = 4 \sqrt3$ and $\triangle ABC = 12 \sqrt3$
$\endgroup$ 1 $\begingroup$Consider the areas of triangles $APB, BPC, CPA$.
We have the equation
\begin{align} [\triangle ABC] &= [\triangle APB] - [\triangle CPA] + [\triangle BPC]\\ &=\frac12(ABh_3 - AC h_2 + BC h_1)\\ &=\frac{AB}2(h_1-h_2+h_3)\\ &=3AB \end{align}
We also have the relation between the side of an equilateral triangle and its area:
$$[\triangle ABC] = \frac{\sqrt 3}4AB^2$$
Now solve for $AB$ and $[\triangle ABC]$.
$\endgroup$ $\begingroup$WLOG let $y=0,y=\sqrt{3}x,y=-\sqrt{3}(x-a)$ be the eq of sides triangle with side of equilateral being $a$.
thus $$h_1+h_3-h_2=1$$ $$|k+\frac{\sqrt{3}h-k}{2}-\frac{\sqrt{3}h+k +\sqrt{3}a}{2}|=1$$
$$\sqrt{3}a/2=1$$$$a=?$$$$area(\Delta)={\sqrt{3}a^2}/4=?$$
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