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A point $P$ is given on the circumference of a circle of radius $r$. The chord $QR$ is parallel to the tangent at $P$. Find the maximum area of triangle $PQR$.

I tried solving it, I assumed $QP$ makes an angle $\theta$ with the tangent at $P$. But it does not help to formulate the area equation. I did not understand what is the importance of chord being parallel to tangent in this question? How should I do this question. Please help me.

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  • $\begingroup$ Why downvote members,what is wrong in the question? $\endgroup$ – Brahmagupta Sep 30 '15 at 10:54
  • $\begingroup$ @BLAZE,sir only this much data is given in the question.Its answer given is $\frac{3\sqrt3 r^2}{4}$ $\endgroup$ – Brahmagupta Sep 30 '15 at 10:56
  • $\begingroup$ Since this is asking for the maximum area,so i put supremum infimum tag. $\endgroup$ – Brahmagupta Sep 30 '15 at 10:58
  • $\begingroup$ But do you know what Supremum and Infimum actually mean? $\endgroup$ – BLAZE Sep 30 '15 at 11:01
  • $\begingroup$ I suppose supremum infimum problems mean maxima minima problems.Am i wrong? $\endgroup$ – Brahmagupta Sep 30 '15 at 11:02
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If $O$ is the center of the given circle of radius $r$ and $\widehat{QPR}=\theta\leq\frac{\pi}{2}$, $\widehat{QOR}=2\theta$ and $\widehat{QOP}=\widehat{ROP}=\pi-\theta$, hence: $$[QPR]=[QOR]+[QOP]+[ROP]=\frac{r^2}{2}\left(\sin(2\theta)+2\sin(\theta)\right) $$ and the $RHS$ achieves its maximum at $\theta=\frac{\pi}{3}$ by differentiation. On the other hand, $\theta\geq\frac{\pi}{2}$ implies $[QPR]\leq r^2$, hence the maximum area is achieved by the equilateral triangle.

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  • $\begingroup$ Sir $[QPR]=[QOP]+[ROP]-[QOR]$.,but you have written $[QPR]=[QOP]+[ROP]+[QOR]$.Am i right?@Jack D'Aurizio $\endgroup$ – Brahmagupta Sep 30 '15 at 11:17
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    $\begingroup$ @Brahmagupta: if $\widehat{QPR}$ is an acute angle, $O$ lies inside $PQR$. $\endgroup$ – Jack D'Aurizio Sep 30 '15 at 11:21
  • $\begingroup$ Sir have you taken two seperate cases,first when $\theta\leq \frac{\pi}{2}$ and second when $\theta\geq \frac{\pi}{2}$.but when we take the second case,same equation $[QPR]=\frac{r^2}{2}(\sin2\theta+2\sin\theta)$ comes,so why do you say second case gives max area QPR as $r^2$ and i did not understand why maximum area is achieved by equilateral triangle.First case i have understood. @Jack D'Aurizio $\endgroup$ – Brahmagupta Sep 30 '15 at 11:37
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    $\begingroup$ @Brahmagupta: because the area of the equilateral triangle is strictly greater than $r^2$. $\endgroup$ – Jack D'Aurizio Sep 30 '15 at 11:38
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Notice, since chord $QR$ is parallel to the tangent at $P$ hence the point $P$ will lie on the perpendicular bisector passing through the center $O$ of the circle. Thus $$PQ=PR\iff \angle PQR=\angle PRQ$$

Let $\angle PQR=\angle PRQ=\alpha \ (\forall \ \ 0<\alpha<\pi)$

then sum of angles subtended b the all the sides $PQ$, $QR$ & $PR$ at the center $O$ is $2\pi$ hence, $$\angle PQR+\angle PRQ+\angle QOR=2\pi\implies \angle QOR=2\pi-\alpha-\alpha=2\pi-2\alpha$$ Hence, the area of $\triangle PQR$ is given as $$=(\text{area of}\ \triangle POQ)+(\text{area of}\ \triangle QOR)+(\text{area of}\ \triangle POR)$$ $$A=\frac{1}{2}r^2\sin\alpha+\frac{1}{2}r^2\sin(2\pi-2\alpha)+\frac{1}{2}r^2\sin\alpha$$ $$A=\frac{1}{2}r^2(2\sin \alpha-\sin2\alpha)\tag 1$$ $$\frac{dA}{d\alpha}=\frac{1}{2}r^2(2\cos \alpha-2\cos2\alpha)=r^2(\cos \alpha-\cos2\alpha)$$ $$\implies \frac{d^2A}{d\alpha^2}=r^2(2\sin 2\alpha-\sin\alpha)\tag 2$$ For maximum or minimum, we have $\frac{dA}{d\alpha}=0$ $$\cos \alpha-\cos 2\alpha=0\implies 2\alpha=2k\pi\pm \alpha$$ $$\alpha=2k\pi\implies \alpha=0, \pm2\pi, \pm4\pi \ldots $$ or $$\alpha=\frac{2k\pi}{3}\implies \alpha=0,\pm\frac{2\pi}{3},\pm\frac{4\pi}{3}\ldots $$ But, $0<\alpha<\pi$ hence, we accept $\alpha=\frac{2\pi}{3}$, we get $$\left(\frac{d^2A}{d\alpha^2}\right)_{\alpha=2\pi/3}=r^2(2\sin 240^\circ-\sin 120^\circ)=-\frac{3\sqrt 3r^2}{2}<0$$ Hence, the area is maximum at $\alpha=\frac{2\pi}{3}$

Hence, the maximum area of $\triangle PQR$ is given by substituting $\alpha=\frac{2\pi}{3}$ in (1) $$A_{\text{max}}=\frac{1}{2}r^2\left(2\sin \frac{2\pi}{3}-\sin \frac{4\pi}{3}\right)=\frac{3\sqrt 3r^2}{4}$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{A_{\text{max}}=\frac{3\sqrt 3\ r^2}{4}}}$$

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“Chord $QR$ is parallel to the tangent at $P$” implies “the altitude of $⊿PQR$ through $P$ must go through $C$, the center of the circle”.

This further means $⊿PQR$ is symmetrical about the line $PC$. Then, $A = [⊿PQR]$ is a function of $h$ only as shown below:-

$A = … = (R + h) \sqrt {R^2 – h^2} = (R + h)^{\frac {3}{2}} \cdot (R – h)^{\frac {1}{2}}$

Following the standard procedures in finding $A_{max}$, we have $h = \dfrac {R}{2}$

From which, it is not difficult to see that $⊿PQR$ is equilateral with a side of length $= \dfrac {\sqrt 3R}{2}$ and area $= \dfrac {3 \sqrt 3R^2}{4}$

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