Collinear Reflections

Geometry Level 5

The reflections of the vertices of a triangle $\Delta ABC$ about the opposite sides, are collinear.

Find the circumradius $R$ of the triangle $\Delta ABC$ .

Note: The symbols are the usual triangle notations.

\sqrt{\dfrac{a^2+b^2+c^2}{4}}

\sqrt{\dfrac{a^2+b^2+c^2}{6}}

\sqrt{\dfrac{a^2+b^2+c^2}{3}}

\sqrt{\dfrac{a^2+b^2+c^2}{5}}

2 solutions

Digvijay Singh
Feb 28, 2019

Read these two articles here and here . From this we know that the area of the reflection triangle is $\Delta'=\left(4-\left(\dfrac{OH}{R}\right)^2\right)\Delta$ where $\Delta$ is the area of the original triangle, $O$ is the circumcenter, $H$ is the orthocenter, $R$ is the circumradius.

The three reflections are collinear if $\Delta'=0$

On using some identities for the lengths $OH$ and $R$ and setting $\Delta'=0$ , we get $\boxed{R^2=\dfrac{a^2+b^2+c^2}{5}}$

Vedant Saini
Mar 7, 2019

Note that a 120-30-30 triangle satisfies the given condition and has sides $s, s, \sqrt3s$

Thus, by the sine rule, its radius equals $\large{\frac{s}{2\sin(30)}} = s$ and the option which satisfies this is the 3rd option.

Collinear Reflections

2 solutions

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