38. Minimum area of triangle touching with semicircle.

by Virgil Nicula, May 24, 2010, 1:34 AM

Kunny wrote:
Let $P$ , $Q$ be two points on the circle $x^2+y^2=1$ . Two points $P$ , $Q$ move in the first quadrant, the second quadrant respectively so that

$OP\perp OQ$ . Consider the triangle enclosed the tangent lines of $C$ at the points $P$ , $Q$ and the $x$-axis. Find the minimum area of this triangle.

Proof. Denote the intersections $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $PM=x$ ,

$QN=y$ . Observe that $\frac {PO}{RN}=$ $\frac {MO}{MN}$ $=\frac {MP}{MR}$ $\implies$ $\frac {1}{1+y}=\frac {x}{x+1}\implies$ $\boxed {\ xy=1\ }\ (1)$ . Therefore,

$[MRN]$ is minimum $\Longleftrightarrow$ $[OPM]+[OQN]$ is minimum $\Longleftrightarrow$ $x+y$ is minimum $\stackrel{(1)}{\Longleftrightarrow}$ $x=y$ .

Extension. Let $P$ , $Q$ on the circle $x^2+y^2=1$ . Points $P$ , $Q$ move in the first quadrant, the second quadrant respectively so that $m\left(\widehat{PRN}\right)=2\phi$

is constant and $\phi\le 45^{\circ}$ . Consider the triangle enclosed the tangent lines of $C$ at $P$ , $Q$ and the $x$-axis. Find the minimum area of this triangle.

Method 1. Denote $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $u=m(\widehat {POM})$ , $v=\widehat{QON}$ . Thus $u+v=2\phi$ ,

$PM=\tan u$ and $QN=\tan v$ . Therefore, $[MRN]$ - min. $\Longleftrightarrow$ $[OPM]+[OQN]$ - min. $\Longleftrightarrow$ $\tan u+\tan v$ - minimum.

Since $\left\{u\ ,\ v\right\}\subset \left[0\ ,\ 90^{\circ}\right)$ obtain $\tan u+\tan v\ge $ $2\cdot\tan\frac {u+v}{2}=$ $2\cdot\tan\phi$ with equality iff $u=v=\phi$ . In conclusion

the area of the triangle $MRN$ is minimum $\Longleftrightarrow$ $u=v=\phi$ $\Longleftrightarrow$ $OR\perp MN$ $\Longleftrightarrow$ the triangle $MRN$ is isosceles.

Remark. I used the well-known inequality $\boxed {\ \{\ u\ ,\ v\ \}\subset \left[0\ ,\ 90^{\circ}\right)\ \implies\ \tan u+\tan v\ge 2\cdot\tan\frac {u+v}{2}\ }\ (*)$ .

Indeed, denote $\tan \frac u2=p<1$ , $\tan \frac v2=q<1$ and the inequality $(*)$ becomes $\frac {2p}{1-p^2}+\frac {2q}{1-q^2}\ge\frac {p+q}{1-pq}$ $\Longleftrightarrow$

$(1-pq)^2(p+q)\ge (p+q)\left(1-p^2\right)\left(1-q^2\right)$ $\Longleftrightarrow$ $(1-pq)^2\ge (1-p^2)(1-q^2)$ $\Longleftrightarrow$ $p^2+q^2\ge 2pq$ .

Method 2. Denote the intersections $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $RM=x$ , $RN=y$ ,

$RP=RQ=k\ge 1$ . Observe that $xy\sin 2\phi=2\cdot [MRN]=x+y$ $\Longleftrightarrow$ $\boxed {\ \frac 1x+\frac 1y=\sin 2\phi\ }\ (2)$ . Since

$(x+y)\left(\frac 1x+\frac 1y\right)\ge 4$ obtain $x+y\ge \frac {4}{\sin 2\phi}$ . Prove easily that if $\triangle MRN$ is isosceles, then $x_0=y_0=\frac {1}{\sin\phi\cos\phi}$

and in this case $x_0+y_0=\frac {4}{\sin 2\phi}$ . In conclusion $\min\ [MRN]=\frac {2}{\sin 2\phi}$ when the triangle $MRN$ is isosceles.
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 27, 2015, 8:25 AM

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  • Blog created: Apr 20, 2010
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