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Here is the question. It states:

A circle of radius 1 unit touches positive x -axis and positive y -axis at A and B respectively. A variable line passing through origin intersects the circle in two points D and E . If the area of the triangle DEB is maximum when the slope of the line is $m$ , then the value of $\frac{1}{m^2}$ is

When I tried plotting its graph it came out to be something like this: graph

Now When I thought, I Felt like the triangle with maximum area should be an equilateral triangle. So from the center of the circle I joined a line to B which makes 90 with the x axis. This line should also bisect the angle of the equilateral triangle so formed. Hence If I assume point D to be the point intersecting the circle on left. Then the line DB should have a slope $\tan(120) = -\sqrt{3}$. This line shall make and angle 60 with the line DE, so I calculated that and the slope comes out to be $m = 0$

However the answer to $\frac{1}{m^2}$ is $3$ and I am not able to understand how. Any hint/suggestions are appreciated. Thanks for your time

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    $\begingroup$ Apparently I failed to notice that the variable line "passes through origin". But I'm not sure how to approach it even after taking that into consideration. I can write the line as y = mx but clueless what to do next $\endgroup$ – protus Apr 16 at 15:54
  • $\begingroup$ HINT: $S_{\triangle BDE}=S_{\triangle BOE}-S_{\triangle BOD}=\frac{1}{2}OB\cdot(X_E-X_D)=\frac{1}{2}(X_E-X_D)$. If we can express $X_D$ and $X_E$ in terms of slope m, we can find and minimize $S_{\triangle BDE}$. $\endgroup$ – Star Bright Apr 16 at 19:53
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Let $D = (x_1,y_1)$ and $E = (x_2,y_2)$. Then, $$\text{ar}(\triangle DEB) = \frac{1}{2} \times \text{base } \times \text{height}\\ \text{ar}(\triangle DEB) = \frac{1}{2}(DE)(BH)$$ where $H$ is the foot of perpendicular from $B$ onto line $DE$. Suppose the equation of $DE$ is given by $y = mx$. We want to find $m$, such that the area of $\triangle DEB\ $is maximum. We know that $B = (0,1)$ and $$BH = \frac{1}{\sqrt{1+m^2}}$$ We want to find $DE$, and then maximize $\text{ar}(\triangle DEB)$. $$DE = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = |x_1-x_2|\sqrt{1+m^2}$$ since $y_1 = mx_1$ and $y_2 = mx_2$ (we know that $D$ and $E$ lie on $y = mx$). $$\text{ar}(\triangle DEB) = \frac{1}{2}|x_1 - x_2|$$ The equation of the circle is $(x-1)^2 + (y-1)^2 = 1$. Substituting $y = mx$ in this, we obtain a quadratic in $x$, with roots $x_1$ and $x_2$. With some work, you may be able to see that the equation we obtain is $$(m^2+1)x^2 - (2m+2)x + 1 = 0$$ I'm sure you know how to find $x_1 + x_2$ and $x_1x_2$ given this quadratic equation, using simply the coefficients. You'll get $$x_1+x_2 = \frac{2(m+1)}{m^2+1}$$ $$x_1x_2 = \frac{1}{m^2+1}$$ $$|x_1 - x_2| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \frac{\sqrt{8m}}{m^2+1}$$ Finally, you'll get $$\text{ar}(\triangle DEB) = \frac{\sqrt{2m}}{m^2+1} = \frac{\sqrt{2}}{m^{3/2} + m^{-1/2}}$$ To maximize $\text{ar}(\triangle DEB)$, you want to minimize the denominator, i.e. $m^{3/2} + m^{-1/2}$ subject to $0 < m < \infty$ (otherwise the line $y = mx$ will not intersect the circle). Differentiating with respect to $m$ will yield $$m = \frac{1}{\sqrt 3}$$ Another way to look at this would be to plot $$f(x) = \frac{\sqrt 2}{x^{3/2} + x^{-1/2}}$$ which attains maxima at $\displaystyle x = \frac{1}{\sqrt{3}} \approx 0.577$.

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Hope that helps!


My solution is more analytic, less geometric, but I would be interested to see more geometric approaches to this problem in other answers!

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    $\begingroup$ +1 although the answer looks long, the computations look neat and nice $\endgroup$ – Albus Dumbledore Apr 16 at 16:24
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Plugging $y=mx$ into the circle equation gives the quadratic:

$$(1+m^2)x^2-2(1+m)x+1=0$$

whose solutions are the abscissas of the intersection points $I_-$ and $I_+$ (your points $D$ and $E$):

$$x_{-}=\dfrac{1+m-\sqrt{2m}}{1+m^2} \ \ \& \ \ x_{+}=\dfrac{1+m+\sqrt{2m}}{1+m^2}$$

The corresponding ordinates are:

$$y_-=mx_- \ \ \& \ \ y_+=mx_+$$

The area of triangle $BI_-I_+$ is equal to the determinant:

$$\tfrac12 \det(\vec{I_-I_+},\vec{I_-B})=\tfrac12 \begin{vmatrix}(x_+-x_-)&(0-x_-)\\ (mx_+-mx_-)&'1-mx_-)\end{vmatrix}=\tfrac12(x_+ \ - \ x_-)=\underbrace{\dfrac{\sqrt{2m}}{1+m^2}}_{f(m)}$$

As the derivative of $f$, given by

$$f'(m)=\dfrac{1-3m^2}{2 \sqrt{m}(1+m^2)^2}$$

is positive for $m^2<\tfrac13$ and negative afterwards, the (unique) maximum occurs for $m^2=\dfrac13$, i.e.,

$$\dfrac{1}{m^2}=3$$

as desired.

Remark 1: Obtaining a slope $m=\dfrac{1}{\sqrt{3}}=\tan(\dfrac{\pi}{6})$ means that the angle made by the line issued from the origin with the $x$-axis is $\dfrac{\pi}{6}$.

Remark 2: There is a geometrical property that could be exploited: the fact that the power of the origin with respect to the circle can be expressed in two ways:

$$OI_-.OI_+=OB^2=1$$

But maybe you haven't seen the concept of power of a point with respect to a circle...

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enter image description here

First a bit of Euclidean geometry: $$S_{\triangle{DEB}} = S_{\triangle{OEB}} - S_{\triangle{ODB}} = \frac12 \times OB \times (EH - DK) = \frac12 \times 1 \times (x_E - x_D)$$ Next, we use analytic geometry to find the line that maximizes $(x_E - x_D)$.

The line $y=mx$ meets the circle $(x-1)^2+(y-1)^2=1$ at two points ($E , D$). We can plug $mx$ for $y$ in the equation of the circle to find $x_E$ and $x_D$: $$(x-1)^2+(mx-1)^2=1$$ Or $$x^2 - 2\frac{1+m}{1+m^2}x + \frac{1}{1+m^2} = 0$$ So $$x = - \frac{1+m}{1+m^2} \pm \sqrt{\frac{(1+m)^2}{(1+m^2)^2} - \frac{1}{1+m^2}}$$ The difference between the two roots (which is what we are looking for) is twice the square root term. $$x_E - x_D = 2\sqrt{\frac{(1+m)^2}{(1+m^2)^2} - \frac{1}{1+m^2}} = \frac{2\sqrt{2m}}{1+m^2}$$ What $m$ maximizes $\frac{\sqrt{m}}{1+m^2}$? We'll use the derivative to find out: $$\frac{d}{dm}(\frac{\sqrt{m}}{1+m^2}) = \frac{\frac12 \frac{1}{\sqrt m}(1+m^2) - 2m \sqrt m}{(1+m^2)^2} = 0 $$ $$\therefore \frac12 \frac{1}{\sqrt m}(1+m^2) = 2m \sqrt m$$ $$\frac12 (1+m^2) = 2m^2 $$ which simplifies to $m^2 = \frac13$ .

Did we just solve the problem?

Not yet! We still don't know whether this slope $m$ corresponds to a maximum or a minimum. An analytical way to answer this important question is to calculate the second derivative and analyze its sign, or somehow analyze the variations in the sign of the first derivative. If interested readers would like to do that as an exercise then I salut them. But here we will take a different approach:

Let's look back at the figure. At two extreme situations, the line passes through points $A$ and $B$. In these two extremes, the points $D$ and $E$ fall over each other and the difference $x_E - x_D$ is zero. In the in-between situations, such as shown in the figure, $x_E - x_D$ is greater than zero. Evidently at some point this difference has to be maximum. Et voila! Now we can say for certain that our calculated $m^2 = \frac13$ corresponds to a maximum value for $x_E - x_D$ and therefore maximum $S_{\triangle{DEB}}$ .

------ Update: This is not an alternative solution but just an observation, I am not yet sure if it leads to anywhere. Triangles $OBD$ and $OEB$ are similar (why?). It follows that $\frac{OD}{OB} = \frac{OB}{OE}$ , and because $OB=1$ we have $OD \times OE=1$ . We also have the ratio of areas of these two triangles: $$\frac{S_{\triangle OBD}}{S_{\triangle OEB}} = (\frac{OB}{OE})^2 = \frac{1}{OE^2} = \frac{1}{x_E^2 (1+m^2)}$$ I will post an update if I find a geometric solution based on these.

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