Prove-that-the-radius-of-a-circle-passing-through-the-midpoints-of-the-sides-of-a-triangle-ABC-is-half-the-radius-of-a-circle-circum-scribed-about-the-triangle-

Question Number 19610 by ajfour last updated on 13/Aug/17

Prove that the radius of a circle

passing through the midpoints

of the sides of a triangle ABC is

half the radius of a circle circum -

scribed about the triangle .

Commented by ajfour last updated on 13/Aug/17

Commented by ajfour last updated on 13/Aug/17

Let B (0, 0); A (2 b, 2 c); C (2 a, 0)

R^{2} = h^{2} + k^{2} \dots . (1)

= {(2 a - h)}^{2} + k^{2} \dots (2)

= {(h - 2 b)}^{2} + {(2 c - k)}^{2} \dots (3)

from (1) and (2) :

h = a and from (1) and (3) :

{(a - 2 b)}^{2} + {(2 c - k)}^{2} = a^{2} + k^{2}

\Rightarrow {4 b}^{2} - 4 ab = (2 k - 2 c) (2 c)

or k = c - \frac{b (a - b)}{c}

hence R^{2} = a^{2} + {[\frac{b^{2} + c^{2} - ab}{c}]}^{2}

r^{2} = {(x_{0} - a)}^{2} + y_{0}^{2} \dots (i)

= {(x_{0} - b)}^{2} + {(y_{0} - c)}^{2} \dots . (ii)

= {(x_{0} - a - b)}^{2} + {(y_{0} - c)}^{2} \dots (iii)

using (ii) and (iii)

\Rightarrow x_{0} - a - b = b - x_{0}

or x_{0} = b + \frac{a}{2};

Also using (i) and (ii) :

r^{2} = {(b + \frac{a}{2} - a)}^{2} + y_{0}^{2} = {(b + \frac{a}{2} - b)}^{2} + {(y_{0} - c)}^{2}

\Rightarrow c ({2 y}_{0} - c) = \frac{a^{2}}{4} - {(b - \frac{a}{2})}^{2}

\Rightarrow c ({2 y}_{0} - c) = ab - b^{2}

\Rightarrow y_{0} = \frac{c}{2} + \frac{b (a - b)}{2 c}

\Rightarrow {(y_{0} - c)}^{2} = {[\frac{b (a - b)}{2 c} - \frac{c}{2}]}^{2}

using (ii) :

r^{2} = \frac{a^{2}}{4} + {[\frac{ab - b^{2} - c^{2}}{2 c}]}^{2}

\Rightarrow {4 r}^{2} = a^{2} + {[\frac{b^{2} + c^{2} - ab}{c}]}^{2} = R^{2}

\Rightarrow r = \frac{R}{2} .

Answered by Tinkutara last updated on 13/Aug/17

Let circumradius of original triangle is

R = \frac{a_{1} b_{1} c_{1}}{4 Δ_{1}}

New sides are \frac{a_{1}}{2}, \frac{b_{1}}{2} and \frac{c_{1}}{2} and Δ_{2}

= \frac{Δ_{1}}{4}

So R^{'} = \frac{\frac{a_{1} b_{1} c_{1}}{8}}{4 \frac{Δ_{1}}{4}} = \frac{1}{2} (\frac{a_{1} b_{1} c_{1}}{4 Δ_{1}}) = \frac{R}{2}

Commented by ajfour last updated on 13/Aug/17

Excellent!

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