Lets have a straight talk!

Geometry Level 4

A straight line L L with negative slope passes through the point ( 8 , 2 ) (8,2) and cuts the positive coordinate axes at points P P and Q Q . Find the the absolute minimum value of O P + O Q OP+OQ , as L L varies where O O is the origin.

This is a part of the set JEE Fest .


The answer is 18.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chew-Seong Cheong
Jun 28, 2017

Similar solution with @Dheeman Kuaner 's

Suppose that line L L cuts the x x - and y y -axis at P P and Q Q respectively. Let O P = a OP=a , O Q = b OQ=b and Q P O = θ \angle QPO = \theta . Then we have:

{ tan θ = 2 a 8 a = 2 tan θ + 8 tan θ = b a b = a tan θ = 2 + 8 tan θ \begin{cases} \tan \theta = \dfrac 2{a-8} & \implies a = \dfrac 2{\tan \theta} + 8 \\ \tan \theta = \dfrac ba & \implies b = a \tan \theta = 2 + 8 \tan \theta \end{cases}

Then we have: O P + O Q = a + b = 2 tan θ + 10 + 8 tan θ OP+OQ = a+b = \dfrac 2{\tan \theta} + 10 + 8 \tan \theta . Using AM-GM inequality 2 tan θ + 8 tan θ 2 16 = 8 \dfrac 2{\tan \theta} + 8 \tan \theta \ge 2 \sqrt {16} = 8 . Therefore O P + O Q 18 OP+OQ \ge \boxed{18}

Tom Engelsman
Jul 11, 2017

Let m R + m \in \mathbb{R^{+}} be the slope of line L in question, which is represented as y 2 = m ( x 8 ) y = m x + ( 8 m + 2 ) . y - 2 = -m(x - 8) \Rightarrow y = -mx + (8m+2). The corresponding y and x-intercepts, P P and Q Q , are given as:

P ( 0 , 8 m + 2 ) P(0, 8m+2) ; Q ( 8 + 2 m , 0 ) Q(8 + \frac{2}{m}, 0)

and O P + O Q = L ( m ) = ( 8 m + 2 ) + ( 8 + 2 m ) = 10 + 8 m + 2 m . OP + OQ = L(m) = (8m+2) + (8 + \frac{2}{m}) = 10 + 8m + \frac{2}{m}. as a function of the slope m . m. Differentiating L ( m ) L(m) produces:

L ( m ) = 8 2 m 2 = 0 8 m 2 = 2 m = 1 2 L'(m) = 8 - \frac{2}{m^{2}} = 0 \Rightarrow 8m^2 = 2 \Rightarrow m = \frac{1}{2} (since we require m > 0 ) . m > 0).

and L ( m ) = 4 m 3 L ( 1 2 ) > 0 L''(m) = \frac{4}{m^{3}} \Rightarrow L''(\frac{1}{2}) > 0 (hence, a minimum).

Thus, the minimum value of O P + O Q OP + OQ is L ( 1 2 ) = 10 + 8 ( 1 2 ) + 2 1 2 = 18 . L(\frac{1}{2}) = 10 + 8(\frac{1}{2}) + \frac{2}{\frac{1}{2}} = \boxed{18}.

Dheeman Kuaner
Jun 28, 2017

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...