10. A challenging geometry question (STEP level)

수학 모음 (Maths collection) 전체보기

 

Question. The line $L$ has equation $y = c − mx$, with $m > 0$ and $c > 0$. It passes through the point $R (a, b)$ and cuts the axes at the points $P(p, 0)$ and $Q(0, q)$, where $a, b, p$ and $q$ are all positive. Find $p$ and $q$ in terms of $a, b$ and $m$.

 

As $L$ varies with $R$ remaining fixed, show that the minimum value of the sum of the distances of $P$ and $Q$ from the origin is $$ \begin{align} \left(a^{\frac12}+b^{\frac12}\right)^2 \end{align} $$ and find in a similar form the minimum distance between $P$ and $Q$.

 

Solution. (i) We sketch the line:

 

The graph of $y=c-mx$.

 

The points $P$ and $Q$ are the $x$- and $y$-intercepts respectively: $$ \begin{align} Q(0,q):&& q&=c \\ P(p,0):&& 0&=c-mp \qquad\Rightarrow\qquad p=\frac{c}{m} \end{align} $$ It is given that the line $L$ passes through $R(a,b)$: $$ \begin{align} b=c-ma \qquad\Rightarrow\qquad c=ma+b \end{align} $$ which gives $$ \left\{\begin{align} q&=c=ma+b \\ p&=\frac{c}{m}=\frac{ma+b}{m} \end{align}\right. $$

 

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(ii) Let $D$ be the sum of the distances of $P$ and $Q$ from the origin, i.e. $$ \begin{align} D&=OP+OQ \\ &=p+q \\ &=\frac{ma+b}{m}+ma+b \\ &=am+\frac{b}{m}+a+b \end{align} $$ We want to find the minimum of $D$ as $L$ varies with $R$ fixed. That is, we want to find the minimum of $D=p+q$ while the gradient $m$ of $L$ varies, so we differentiate $D$ with respect to $m$: $$ \begin{align} && \frac{\textrm{d}D}{\textrm{d}m} &=\frac{\textrm{d}}{\textrm{d}m}\left(am+\frac{b}{m}+a+b\right) \\ &&&=a-\frac{b}{m^2} \\ &&&=0 \\ \\ \Rightarrow&& m&=\pm\sqrt{\frac{b}{a}} \end{align} $$ To find the nature of stationary points, we find the second derivative: $$ \begin{align} && D''(m) &\equiv \frac{\textrm{d}^2D}{\textrm{d}m^2} =\frac{\textrm{d}}{\textrm{d}m}\left(a-\frac{b}{m^2}\right)=\frac{2b}{m^3} \\ \\ \Rightarrow&& D''\left(\sqrt{b/a}\right)&=\frac{2b}{\left(\frac{b}{a}\right)^{\frac32}}=\frac{2b}{\frac{b}{a}\sqrt{\frac{b}{a}}}=2a\sqrt{\frac{a}{b}}>0\;:\quad{\rm local\;min} \\ && D''\left(-\sqrt{b/a}\right)&=\frac{2b}{-\left(\frac{b}{a}\right)^{\frac32}}=-\frac{2b}{\frac{b}{a}\sqrt{\frac{b}{a}}}=-2a\sqrt{\frac{a}{b}}<0\;:\quad{\rm local\;max} \end{align} $$ So the minimum of $D$ occurs when $m=\sqrt{\frac{b}{a}}$: $$ \begin{align} D_{\rm min}&=D\left(\sqrt{b/a}\right) \\ &=a\sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}+a+b \\ &=\sqrt{ab}+\sqrt{ab}+a+b \\ &=a+2\sqrt{ab}+b \\ &=\left(\sqrt{a}+\sqrt{b}\right)^2.\quad\square \end{align} $$

 

The graph of $y=c-mx$ (red) and $y=D(m)$ (blue).

 

Aside. We can also work out the local maximum: $$ \begin{align} D_{\rm max}&=D\left(-\sqrt{b/a}\right) \\ &=-a\sqrt{\frac{b}{a}}-\frac{b}{\sqrt{\frac{b}{a}}}+a+b \\ &=-\sqrt{ab}-\sqrt{ab}+a+b \\ &=a-2\sqrt{ab}+b \\ &=\left(\sqrt{a}-\sqrt{b}\right)^2.\quad\checkmark \end{align} $$

 

 

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(iii) Let $\bar D$ be the distance of $\overline{PQ}$, i.e. $$ \begin{align} \bar D&=\overline{PQ} \\ &=\sqrt{p^2+q^2}\quad{\rm (by\;Pythagoras)} \\ &=\sqrt{\left(\frac{ma+b}{m}\right)^2+\left(ma+b\right)^2} \\ &=\vert ma+b\vert\sqrt{\frac1{m^2}+1} \\ &=\frac{ma+b}{m}\sqrt{1+m^2} \\ &=\left(a+\frac{b}{m}\right)\sqrt{1+m^2} \end{align} $$ since $a,b$ and $m$ are assumed to be positive. (For $m<0$, see comment below.) We want to find the minimum of $\bar D$ as $L$ varies with $R$ fixed. That is, we want to find the minimum of $\bar D=\sqrt{p^2+q^2}$ while the gradient $m$ of $L$ varies, so we differentiate $\bar D$ with respect to $m$: $$ \begin{align} && \frac{\textrm{d}\bar D}{\textrm{d}m} &=\frac{\textrm{d}}{\textrm{d}m}\left[\left(a+\frac{b}{m}\right)\sqrt{1+m^2}\right] \\ &&&=-\frac{b}{m^2}\sqrt{1+m^2}+\left(a+\frac{b}{m}\right)\frac{m}{\sqrt{1+m^2}} \quad{\rm (by\;product\;rule)} \\ &&&=-\frac{b}{m^2}\sqrt{1+m^2}+\frac{am+b}{\sqrt{1+m^2}} \\ &&&=\frac{m^2(am+b)-b\left(1+m^2\right)}{m^2\sqrt{1+m^2}} \\ &&&=\frac{am^3-b}{m^2\sqrt{1+m^2}} \\ &&&=0 \\ \\ \Rightarrow&& m&=\sqrt[3]{\frac{b}{a}}=\left(\frac{b}{a}\right)^{\frac13} \end{align} $$ It is given that this gives a (local) minimum - which we will nevertheless check below, $$ \begin{align} \bar D_{\rm min} &=\bar D\left(\sqrt[3]{b/a}\right) \\ &=\left(a+\frac{b}{\left(\frac{b}{a}\right)^{\frac13}}\right)\sqrt{1+\left(\frac{b}{a}\right)^{\frac23}} \\ &=\left(a+a^{\frac13}b^{\frac23}\right)\sqrt{\frac{a^{\frac23}+b^{\frac23}}{a^{\frac23}}} \\ &=a^{\frac13}\left(a^{\frac23}+b^{\frac23}\right)\sqrt{\frac{a^{\frac23}+b^{\frac23}}{a^{\frac23}}} \\ &=\left(a^{\frac23}+b^{\frac23}\right)^{\frac32} \end{align} $$

 

 

Aside. To ensure that the stationary point of $\bar D$ at $m=\left(\frac{b}{a}\right)^{\frac13}$ is a local minimum, we find the second derivative: $$ \begin{align} && \frac{\textrm{d}\bar D}{\textrm{d}m} &=\frac{am^3-b}{m^2\sqrt{1+m^2}} =\left(am-\frac{b}{m^2}\right)\frac1{\sqrt{1+m^2}} \\ \\ \Rightarrow&& \bar D''(m) \equiv \frac{\textrm{d}^2\bar D}{\textrm{d}m^2} &=\frac{\textrm{d}}{\textrm{d}m}\left[\left(am-\frac{b}{m^2}\right)\frac1{\sqrt{1+m^2}}\right] \\ &&&=\left(a+\frac{2b}{m^3}\right)\frac1{\sqrt{1+m^2}}-\left(am-\frac{b}{m^2}\right)\frac{m}{\left(1+m^2\right)^{\frac32}} \quad{\rm (by\;product\;rule)} \\ &&&=\frac{am^3+2b}{m^3\sqrt{1+m^2}}-\frac{am^3-b}{m\left(1+m^2\right)^{\frac32}} \\ &&&={\color{blue} { \frac{\left(am^3+2b\right)\left(1+m^2\right)-\left(am^3-b\right)m^2}{m^3\left(1+m^2\right)^{\frac32}} } } \\ &&&=\frac{\left(am^3+2b-am^3+b\right)m^2 + \left(am^3+2b\right)}{m^3\left(1+m^2\right)^{\frac32}} \\ &&&=\frac{am^3+3bm^2+2b}{m^3\left(1+m^2\right)^{\frac32}} \end{align} $$ Since the stationary point occurs at $m=\sqrt[3]{b/a}$ which satisfies $am^3-b=0$, the blue expression above is the convenient one to use here: $$ \begin{align} \Rightarrow&& D''\left(\sqrt[3]{b/a}\right) &= {\color{blue} {\frac{\left(a\times\frac{b}{a}+2b\right)\left[1+\left(\frac{b}{a}\right)^{\frac23}\right]}{\left(\frac{b}{a}\right)\left[1+\left(\frac{b}{a}\right)^{\frac23}\right]^{\frac32}}} } \\ &&&=\frac{3b}{\frac{b}{a}\left[1+\left(\frac{b}{a}\right)^{\frac23}\right]^{\frac12}} \\ &&&=\frac{3a}{\left( \frac{a^{\frac23}+b^{\frac23}}{a^{\frac23}} \right)^{\frac12}} \\ &&&=\frac{3a^{\frac43}}{\sqrt{a^{\frac23}+b^{\frac23}}} >0\;:\quad{\rm local\;min}\quad\checkmark \end{align} $$

 

Comment. For $m<0$, $$ \begin{align} D_{\rm min}=\bar D_{\rm min}=0 \qquad{\rm when}\qquad m=\frac{b}{a} \end{align} $$ i.e. the minimum occurs when the line $L$ goes through the origin.