Finding the maximum value of area of triangle PXY as X varies.

Question

The circle C has radius 1 and touches the line L at P. The point X lies on C and Y is the foot of the perpendicular from X to L. Find the maximum value of the area of triangle PXY as X varies.

Let the circle be $x^2+(y-1)^2=1$ and the point P be $(0,0)$

Let X be $(\cos\theta,1+\sin\theta)\implies$ area of triangle PXY=$\frac12\cos\theta(1+\sin\theta)$

The answer is $\frac{3\sqrt3}8$, that seems to come at $\theta=30^\circ$ but why?

Also, is there a general way to solve this question without letting a particular circle.

Answer 1

Let $\angle XPY=\theta$. Because of symmetry, WLOG $\theta<\dfrac\pi2$. By the law of sines, $PX=R\cdot\sin\theta=2\sin\theta$. So $PY=2\cos\theta\sin\theta$ and $XY=2\sin^2\theta$. By AM-GM inequality, \begin{align*} &[PXY]=\frac{PY\cdot XY}2=2\sin^3\theta\cdot\cos\theta\\[1em] ={}&6\sqrt3\sqrt{\frac13\sin^2\theta\cdot\frac13\sin^2\theta\cdot\frac13\sin^2\theta\cdot\cos^2\theta}\\[1em] \le{}&6\sqrt3\sqrt{\left(\tfrac{\textstyle\frac13\sin^2\theta+\frac13\sin^2\theta+\frac13\sin^2\theta+\cos^2\theta}{\textstyle4}\right)^4}=\frac{3\sqrt3}8. \end{align*}

Answer 2

$A=\frac12 \cos \theta(1+\sin \theta)$

$A'=-\frac12 \sin \theta+\frac 24 \cos 2\theta$

$\cos 2\theta -\sin \theta=1-2\sin^2 \theta-\sin \theta=0$

which gives $\sin \theta = -1, \frac 12$

Maximum of A is for $\sin \theta=\frac 12$

Or

$\theta =30^o$

Putting in A we get $A=\frac{3\sqrt 3}8$

Answer 3

Answering your question on the general case:

Let $O$ be the centre of the circle and let $\angle POX=\theta$.
As mentioned in the question, $\angle XYP=90^\circ$.
Since $L$ is a tangent to the circle at $P$ and $OP$ is a radius, $L$ is perpendicular to $OP$ and so $\angle OPY=90^\circ$.
$\therefore OPYX$ is a trapezium and $\angle OXY=180^\circ-\theta$. Since $OX$ and $OP$ are radii of $C$, $OX=OP$, meaning that $OXP$ is an isosceles triangle, so $\angle OXP=90^\circ-\frac{\theta}{2}$.
$\therefore\angle PXY=180^\circ-\theta-\left(90^\circ-\frac{\theta}{2}\right)=90^\circ-\frac{\theta}{2}.$ Let $M$ be the midpoint of $PX$.
Since $OPX$ is isosceles, $OM$ bisects $XP$ and so triangles $OMX$ and $PYX$ are similar.
By the cosine rule, $PX=\sqrt{2-2\cos\theta}$, so the scale factor is $\sqrt{2-2\cos\theta}$.
$\therefore[PXY]=(2-2\cos\theta)[OMX]$, where $[ABC]$ denotes the area of a triangle $ABC$.
$[OMX]=\frac12[OPX]=\frac14\sin\theta$.
$\therefore[PXY]=\frac14\sin\theta(2-2\cos\theta)=\frac12\sin\theta(1-\cos\theta)$.
$\frac{d}{d\theta}\frac12\sin\theta(1-\cos\theta)=\frac12(\sin^2\theta+\cos\theta-\cos^2\theta)$
$\frac12(\sin^2\theta+\cos\theta-\cos^2\theta)=0\Rightarrow -2\cos^2\theta+\cos\theta+1=0\Rightarrow\cos\theta=1$ or $-\frac12$
$\cos\theta=1\Rightarrow \theta=0^\circ$ minimises the area of the triangle and $\cos\theta=-\frac12\Rightarrow \theta=120^\circ$ maximises the area of the triangle.
$\therefore[PXY]_{max}=\frac12\sin120^\circ(1-\cos120^\circ)=\frac12\cdot\frac{\sqrt3}2\cdot\frac32=\frac{3\sqrt3}8$