Question

The distance between the circumcentre and orthocentre of a triangle \[ABC\] is
(a) \[R\sqrt{1-8\cos A\cos B\cos C}\]
(b) \[2R\sqrt{1-4\cos A\cos B\cos C}\]
(c) \[R\sqrt{1-8\sin A\sin B\sin C}\]
(d) \[2R\sqrt{1-4\sin A\sin B\sin C}\]

Answer 1

Hint: We solve this problem by using the properties of triangles and some simple geometry. For a triangle let us assume that the orthocentre and the circumcentre as ‘H’ and ‘O’ respectively. The orthocentre and circumcentre are as shown in the figure.

Complete step-by-step solution:

We use the cosine rule of a triangle to solve this problem. The cosine rule of the following triangle is given as

\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] Similarly for the other angles.
Let us assume that the circumcentre and orthocentre as ‘O’ and ‘H’ respectively.
Let us consider the triangle \[\Delta OAG\]
Here, by using the result that is sum of all angles in a triangle is \[{{180}^{\circ }}\] we can take the angle \[\angle OAG\] as
\[\begin{align}
  & \Rightarrow \angle AOG+\angle OAG+{{90}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow \angle OAG={{90}^{\circ }}-\angle AOG.......equation(i) \\
\end{align}\]
From the definition of orthocentre we know that the angle formed by a vertex and a side passing through that vertex is equal to angle at opposite vertex of that side that is
\[\Rightarrow \angle AOG=\angle C\]
By substituting the above angle in equation (i) we get
\[\Rightarrow \angle OAG={{90}^{\circ }}-\angle C\]
Now, let us consider the triangle \[\Delta ALC\]
Here, we can take the angle \[\angle LAC\] as
\[\Rightarrow \angle LAC={{90}^{\circ }}-\angle C\]
Here, we can see that the points ‘L’ and ‘H’ are on the same line. So, we can take
\[\Rightarrow \angle HAC={{90}^{0}}-\angle C\]
Now let us consider the triangle \[\Delta OAH\] we can take the angle \[\angle OAH\] as
\[\Rightarrow \angle OAH=\angle A-\left( \angle OAG+\angle HAC \right)\]
By substituting the required angles in above equation we get
\[\begin{align}
  & \Rightarrow \angle OAH=\angle A-\left( {{180}^{\circ }}-2\angle C \right) \\
 & \Rightarrow \angle OAH=\angle A+2\angle C-{{180}^{0}}........equation(ii) \\
\end{align}\]
We know that the sum of angles of a triangle is equal to \[{{180}^{0}}\] that is
\[\Rightarrow \angle A+\angle B+\angle C={{180}^{\circ }}\]
Now, by substituting the above formula to equation (ii) we get
\[\begin{align}
  & \Rightarrow \angle OAH=\angle A+2\angle C-\left( \angle A+\angle B+\angle C \right) \\
 & \Rightarrow \angle OAH=\angle C-\angle B \\
\end{align}\]
Let us assume that the circum radius of the triangle as \['R'\].
Now, we know that, form the definition of circumradius we have
\[\Rightarrow OA=R\]
We know that the distance between orthocentre and a vertex is equal to \['2R'\] times the cosine of the angle at that vertex that is
\[\Rightarrow HA=2R\cos A\]
Now let us consider the triangle \[\Delta OAH\]
We know that the cosine rule of a triangle is given a

\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] Similarly for the other angles.
By using the cosine rule to triangle \[\Delta OAH\] we get
\[\cos \left( \angle OAH \right)=\dfrac{O{{A}^{2}}+H{{A}^{2}}-O{{H}^{2}}}{2.OA.AH}\]
Now, by substituting the required values in the above equation we get
\[\Rightarrow \cos \left( C-B \right)=\dfrac{{{R}^{2}}+{{\left( 2R\cos A \right)}^{2}}-O{{H}^{2}}}{2R\left( 2R\cos A \right)}\]
By cross multiplying we get
\[\begin{align}
  & \Rightarrow 4{{R}^{2}}\cos A\cos \left( C-B \right)={{R}^{2}}+4{{R}^{2}}{{\cos }^{2}}A-O{{H}^{2}} \\
 & \Rightarrow O{{H}^{2}}={{R}^{2}}+4{{R}^{2}}\cos A\left( \cos A-\cos \left( C-B \right) \right)........equation(iv) \\
\end{align}\]
We know that the sum of angles of a triangle is equal to \[{{180}^{0}}\] that is
\[\begin{align}
  & \Rightarrow \angle A+\angle B+\angle C={{180}^{0}} \\
 & \Rightarrow \angle B+\angle C={{180}^{0}}-\angle A \\
 & \Rightarrow \cos \left( B+C \right)=-\cos A \\
\end{align}\]
By substituting the above formula in equation (iv) we get
\[\Rightarrow O{{H}^{2}}={{R}^{2}}-4{{R}^{2}}\cos A\left( \cos \left( B+C \right)+\cos \left( B-C \right) \right)\]
We know that the formula of sum of composite angles as
\[\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B\]
By using this formula in above equation we get
\[\begin{align}
  & \Rightarrow O{{H}^{2}}={{R}^{2}}-4{{R}^{2}}\cos A\left( 2\cos B\cos C \right) \\
 & \Rightarrow O{{H}^{2}}={{R}^{2}}\left( 1-8\cos A\cos B\cos C \right) \\
 & \Rightarrow OH=R\sqrt{1-8\cos A\cos B\cos C} \\
\end{align}\]
Therefore, we can say that the distance between the orthocentre and circumcentre is given as \[R\sqrt{1-8\cos A\cos B\cos C}\]
So, option (a) is the correct answer.

Note: Students may make mistakes in remembering the formulas we used in this problem. There are a lot of standard results we used in this question. The list of formulas we used in this question are given as
(i) cosine rule

\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]
(ii) The sum of angles of a triangle is equal to \[{{180}^{0}}\] that is
\[\Rightarrow \angle A+\angle B+\angle C={{180}^{0}}\]
(iii) Sum of composite angles as
\[\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B\]
Remembering these formulas is important in this problem.