You visited us 1 times! Enjoying our articles? Unlock Full Access!

Circum Radius

In ABC, circu...

Question

In $△ A B C$ , circumradius, $R = 3$ . Let $O$ be the circumcentre and $H$ be the orthocentre, then the value of $\frac{1}{64} (A H^{2} + B C^{2}) (B H^{2} + A C^{2}) (C H^{2} + A B^{2})$ is $3^{k}$ , where $k$ is

Open in App

Solution

In
$△ A E H, cos (90 - C) = \frac{A E}{A H} \Rightarrow sin C = \frac{A E}{A H} \Rightarrow A H = \frac{A E}{sin C}$
In
$△ A B E cos A = \frac{A E}{c} \Rightarrow A E = c cos A ∴ A H = \frac{c cos A}{sin C} \Rightarrow A H = \frac{2 R sin C cos A}{sin C} = 2 R cos A$
Similarily $B H = 2 R cos B$ and $C H = 2 R cos C$
Now,
$\frac{1}{64} (A H^{2} + B C^{2}) (B H^{2} + A C^{2}) (C H^{2} + A B^{2}) = 3^{k} \frac{1}{64} ((2 R cos A)^{2} + (2 R sin A)^{2}) ((2 R cos B)^{2} + (2 R sin B)^{2}) ((2 R cos C)^{2} + (2 R sin C)^{2}) = 3^{k}$
$\Rightarrow \frac{1}{64} (64 R^{6}) = 3^{k} R^{6} = 3^{k}$
since $R = 3 \Rightarrow k = 6$

Suggest Corrections

Similar questions

△ A B C

2

\frac{1}{64} (A H^{2} + B C^{2}) (B H^{2} + A C^{2}) (C H^{2} + A B^{2})

△ A B C

R = 3

O

H

\frac{1}{64} (A H^{2} + B C^{2}) (B H^{2} + A C^{2}) (C H^{2} + A B^{2})

3^{k}

k

A B C

O

H

N

O H

- - \to N A + - - \to N B + - - \to N C

A B C

O

H

N

O H

\to N A + \to N B + \to N C

A B C

O

H

N

O H

- - \to N A + - - \to N B + - - \to N C

Join BYJU'S Learning Program