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An isosceles right angled triangle whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the coordinate axes. If it slides prove that the locus of its centroid is $(3x-y)^2 + (x-3y)^2 =\dfrac{32}{9}$ .

My attempt is as follows:-

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Let $\angle ABO=\theta$,$\angle ACB=90^\circ, \angle BAC=\angle ABC=45^\circ$

$$a=\sqrt{2}\cos\theta$$ $$b=\sqrt{2}\sin\theta$$

Line $AC$ is inclined at $45^\circ-\theta$ with respect to $x$ axis

By parametric equation, $p=\cos(45-\theta),q=\sqrt{2}\sin\theta+\sin(45-\theta)$

Line $BC$ is inclined at $180^\circ-(45^\circ+\theta)$ with respect to $x$ axis

By parametric equation, $p=\sqrt{2}\cos\theta-\cos(45+\theta),q=\sin(45+\theta)$

Let locus of centroid be $(h,k)$

$$h=\dfrac{p+a}{3}$$ $$3h=\cos(45-\theta)+\sqrt{2}\cos\theta$$ $$3h=\dfrac{1}{\sqrt{2}}\cos\theta+\dfrac{1}{\sqrt{2}}\sin\theta+\sqrt{2}\cos\theta$$ $$3h=\dfrac{3}{\sqrt{2}}\cos\theta+\dfrac{1}{\sqrt{2}}\sin\theta$$ $$3\sqrt{2}h=3\cos\theta+\sin\theta\tag{1}$$

$$k=\dfrac{q+b}{3}$$ $$k=\dfrac{2\sqrt{2}\sin\theta+\dfrac{1}{\sqrt{2}}\cos\theta-\dfrac{1}{\sqrt{2}}\sin\theta}{3}$$ $$3\sqrt{2}k=3\sin\theta+\cos\theta\tag{2}$$

$$-8\sin\theta=3\sqrt{2}h-9\sqrt{2}k$$ $$\sin\theta=\dfrac{3\sqrt{2}}{8}(3k-h)\tag{3}$$

$$\cos\theta=\dfrac{3\sqrt{2}}{8}(3h-k)\tag{4}$$

Eliminating $\theta$

$$1=\dfrac{18}{64}\left((3k-h)^2+(3h-k)^2\right)$$

$$(3x-y)^2+(y-3x)^2=\dfrac{32}{9}$$

Any different way to solve this problem ,this got very lengthy.

prat
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1 Answers1

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enter image description here

$\require{begingroup} \begingroup$ $\def\i{\mathbf i}$

Expressing the coordinates of the points $A,B,C,G$ as complex numbers in terms of a parameter $t\in[0,\sqrt2]$,

\begin{align} B&=t ,\quad A=\sqrt{2-t^2}\cdot\i ,\\ C&=\tfrac12(A+B)+\tfrac12\,(B-A)\cdot\i =\tfrac12\,(t+\sqrt{2-t^2}) +\tfrac12\,(t+\sqrt{2-t^2})\cdot\i ,\\ G&=\tfrac13(A+B+C) =\tfrac12\,t+\tfrac16\,\sqrt{2-t^2} +(\tfrac16\,t+\tfrac12\,\sqrt{2-t^2})\cdot\i , \end{align}

we have a parametric representation of the coordinates of the locus point

\begin{align} x&=\tfrac12\,t+\tfrac16\,\sqrt{2-t^2} ,\\ y&=(\tfrac16\,t+\tfrac12\,\sqrt{2-t^2}) . \end{align}

This gives the expression of $t$ in terms of $x,y$ as

\begin{align} t&=\tfrac94\,x -\tfrac34\,y . \end{align}

It follows that \begin{align} 3\,x-9\,y+\sqrt{32-9\,y^2+54\,x\,y-81\,x^2}&=0 ,\\ 45\,x^2+45\,y^2-54\,x\,y-16&=0 . \end{align}

$\endgroup$

g.kov
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