Dividing an Equilateral Triangle

Geometry Level 2

\frac{61}{2}

\frac{63}{2}

\frac{59}{2}

\frac{57}{2}

2 solutions

Anish Puthuraya
Mar 12, 2014

Let $\displaystyle CX = x$
$\displaystyle BX = y$

Then, applying Cosine Rule in $\displaystyle\Delta BXC$ ,

$y^2 = x^2 + 36 -2\times 6\times x\times\cos(60^0)$

$\Rightarrow y^2 = x^2 -6x+36$

Thus,

$BX^2+XC^2 = y^2+x^2$

$\Rightarrow BX^2+XC^2 = 2x^2-6x+36$

The Minimum value of this quadratic exists at $\displaystyle x = \frac{3}{2}$ (use derivatives to confirm)

$(BX^2+XC^2)_{min} = 2(\frac{3}{2})^2 - 6(\frac{3}{2}) + 36 = \boxed{\frac{63}{2}}$

You can do the same by applying Pythagoras Theorem.

Let CX=x and drop an altitude from B to AC at D.

Therefore BD=3\sqrt{3}.

Taking triangle BXD.

BX^2=BD^2+DX^2 = 27+9-6x+x^2 =36-6x+x^2 and CX^2=x^2

therefore BX^2+CX^2=2x^2-6x+36

Therefore the same procedure applies.

Malay Pandey - 7 years, 2 months ago

Abbas Mohammad
Mar 15, 2014

I did it by mistake : P