This question was uploaded on 17/09/21 on social media accounts.
A point inside an equilateral triangle is at a distance of 3, 4, 5 from the vertices of the triangle. Find the length of the side of the triangle.
Solution:
Method 1:
By applying cosine rule on the triangle AOB at vertices O:
\[cos(150)=\frac{3^2+4^2-s^2}{2\times3\times4}\]
\[\Rightarrow -\frac{\sqrt3}{2}=\frac{25-s^2}{24}\]
\[\Rightarrow s^2=25+12\sqrt3\]
\[\Rightarrow s=\sqrt{25+12\sqrt3}\]
Explanation:
1) Triangle O'AB is similar to triangle OCB.
2) In triangle OBO', the angle at OBO' is 60 so that angles at O and O' is also must be 60 so that it will be an equilateral triangle.
3) In triangle OAO', sides are 3, 4, 5 so that it will be a right-angled triangle.
Method 2:
Let angles at one vertex are θ and (60 - θ).
Apply cosine rule on both angles:
Let us consider angles at bottom left corner:
\[cos(\theta) = \frac{3^2+s^2-4^2}{2\times3\times s}\]
\[\Rightarrow cos(\theta) = \frac{s^2-7}{6s}\]
Also by using cos(), we can find sin():
\[\Rightarrow sin(\theta) = \frac{\sqrt{50s^2-s^4-49}}{6s}\]
Now,
\[cos(60-\theta) = \frac{3^2+s^2-5^2}{2\times3\times s}\]
\[\Rightarrow cos(60-\theta) = \frac{s^2-16}{6s}\]
We know that, cos(a - b) = cos(a)cos(b) + sin(a)sin(b):
\[\Rightarrow cos(60-\theta) = cos(60)cos(\theta) + sin(60)sin(\theta)\]
\[\Rightarrow \frac{s^2-16}{6s} = \frac{1}{2}\frac{s^2-7}{6s} + \frac{\sqrt3}{2}\frac{\sqrt{50s^2-s^4-49}}{6s}\]
\[\Rightarrow 2(s^2-16) = s^2-7 + \sqrt3\sqrt{50s^2-s^4-49}\]
\[\Rightarrow s^2-25 =\sqrt3\sqrt{50s^2-s^4-49}\]
\[\Rightarrow s^4-50s^2+625 =150s^2-3s^4-147\]
\[\Rightarrow 4s^4-200s^2+772 = 0\]
\[\Rightarrow s^4-50s^2+193 = 0\]
\[\Rightarrow (s^2-25)^2-432 = 0\]
\[\Rightarrow s^2-25 = 12\sqrt3\]
\[\Rightarrow s = \sqrt{25+12\sqrt3}\]
(Note: Negative root is neglected because it is giving the very small value of s)
Puzzle related to Geometry, Triangle, Length
